phys1112 - electricity and magnetism lecture notes · 2.2. the electric field 9 (2) if q1, q2 are...

PHYS1112 - Electricity and MagnetismLecture Notes

Dr. Jason Chun Shing PunDepartment of Physics

The University of Hong Kong

January 2005

Contents

1 Vector Algebra 11.1 Definitions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11.2 Vector Algebra . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11.3 Components of Vectors . . . . . . . . . . . . . . . . . . . . . . . . 21.4 Multiplication of Vectors . . . . . . . . . . . . . . . . . . . . . . . 41.5 Vector Field (Physics Point of View) . . . . . . . . . . . . . . . . 61.6 Other Topics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6

2 Electric Force & Electric Field 82.1 Electric Force . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 82.2 The Electric Field . . . . . . . . . . . . . . . . . . . . . . . . . . . 92.3 Continuous Charge Distribution . . . . . . . . . . . . . . . . . . . 122.4 Electric Field Lines . . . . . . . . . . . . . . . . . . . . . . . . . . 182.5 Point Charge in E-field . . . . . . . . . . . . . . . . . . . . . . . . 212.6 Dipole in E-field . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22

3 Electric Flux and Gauss’ Law 253.1 Electric Flux . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 253.2 Gauss’ Law . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 283.3 E-field Calculation with Gauss’ Law . . . . . . . . . . . . . . . . . 283.4 Gauss’ Law and Conductors . . . . . . . . . . . . . . . . . . . . . 31

4 Electric Potential 364.1 Potential Energy and Conservative Forces . . . . . . . . . . . . . 364.2 Electric Potential . . . . . . . . . . . . . . . . . . . . . . . . . . . 404.3 Relation Between Electric Field E and Electric Potential V . . . . 454.4 Equipotential Surfaces . . . . . . . . . . . . . . . . . . . . . . . . 48

5 Capacitance and DC Circuits 515.1 Capacitors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 515.2 Calculating Capacitance . . . . . . . . . . . . . . . . . . . . . . . 515.3 Capacitors in Combination . . . . . . . . . . . . . . . . . . . . . . 545.4 Energy Storage in Capacitor . . . . . . . . . . . . . . . . . . . . . 55

i

5.5 Dielectric Constant . . . . . . . . . . . . . . . . . . . . . . . . . . 575.6 Capacitor with Dielectric . . . . . . . . . . . . . . . . . . . . . . . 585.7 Gauss’ Law in Dielectric . . . . . . . . . . . . . . . . . . . . . . . 605.8 Ohm’s Law and Resistance . . . . . . . . . . . . . . . . . . . . . . 615.9 DC Circuits . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 645.10 RC Circuits . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 69

6 Magnetic Force 736.1 Magnetic Field . . . . . . . . . . . . . . . . . . . . . . . . . . . . 736.2 Motion of A Point Charge in Magnetic Field . . . . . . . . . . . . 756.3 Hall Effect . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 766.4 Magnetic Force on Currents . . . . . . . . . . . . . . . . . . . . . 78

7 Magnetic Field 817.1 Magnetic Field . . . . . . . . . . . . . . . . . . . . . . . . . . . . 817.2 Parallel Currents . . . . . . . . . . . . . . . . . . . . . . . . . . . 867.3 Ampere’s Law . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 887.4 Magnetic Dipole . . . . . . . . . . . . . . . . . . . . . . . . . . . . 927.5 Magnetic Dipole in A Constant B-field . . . . . . . . . . . . . . . 937.6 Magnetic Properties of Materials . . . . . . . . . . . . . . . . . . 94

8 Faraday’s Law of Induction 988.1 Faraday’s Law . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 988.2 Lenz’ Law . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 998.3 Motional EMF . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1008.4 Induced Electric Field . . . . . . . . . . . . . . . . . . . . . . . . 104

9 Inductance 1079.1 Inductance . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1079.2 LR Circuits . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1109.3 Energy Stored in Inductors . . . . . . . . . . . . . . . . . . . . . . 1129.4 LC Circuit (Electromagnetic Oscillator) . . . . . . . . . . . . . . . 1139.5 RLC Circuit (Damped Oscillator) . . . . . . . . . . . . . . . . . . 115

10 AC Circuits 11610.1 Alternating Current (AC) Voltage . . . . . . . . . . . . . . . . . . 11610.2 Phase Relation Between i, V for R,L and C . . . . . . . . . . . . . 11710.3 Single Loop RLC AC Circuit . . . . . . . . . . . . . . . . . . . . . 11910.4 Resonance . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12110.5 Power in AC Circuits . . . . . . . . . . . . . . . . . . . . . . . . . 12110.6 The Transformer . . . . . . . . . . . . . . . . . . . . . . . . . . . 123

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11 Displacement Current and Maxwell’s Equations 12511.1 Displacement Current . . . . . . . . . . . . . . . . . . . . . . . . . 12511.2 Induced Magnetic Field . . . . . . . . . . . . . . . . . . . . . . . . 12711.3 Maxwell’s Equations . . . . . . . . . . . . . . . . . . . . . . . . . 128

iii

Chapter 1

Vector Algebra

1.1 Definitions

A vector consists of two components: magnitude and direction .(e.g. force, velocity, pressure)

A scalar consists of magnitude only.(e.g. mass, charge, density)

1.2 Vector Algebra

Figure 1.1: Vector algebra

~a +~b = ~b + ~a

~a + (~c + ~d) = (~a + ~c) + ~d

1.3. COMPONENTS OF VECTORS 2

1.3 Components of Vectors

Usually vectors are expressed according to coordinate system. Each vector canbe expressed in terms of components.

The most common coordinate system: Cartesian

~a = ~ax + ~ay + ~az

Magnitude of ~a = |~a| = a,

a =√

a2x + a2

y + a2z

Figure 1.2: φ measured anti-clockwisefrom position x-axis

~a = ~ax + ~ay

a =√

a2x + a2

y

ax = a cosφ; ay = a sinφ

tanφ =ay

ax

Unit vectors have magnitude of 1

a =~a

|~a| = unit vector along ~a direction

i j k are unit vectors alongl l lx y z directions

~a = ax i + ay j + az k

Other coordinate systems:

1.3. COMPONENTS OF VECTORS 3

1. Polar Coordinate:

Figure 1.3: Polar Coordinates

~a = ar r + aθ θ

2. Cylindrical Coordinates:

Figure 1.4: Cylindrical Coordinates

~a = ar r + aθ θ + az z

r originated from nearest point onz-axis (Point O’)

3. Spherical Coordinates:

Figure 1.5: Spherical Coordinates

~a = ar r + aθ θ + aφ φ

r originated from Origin O

1.4. MULTIPLICATION OF VECTORS 4

1.4 Multiplication of Vectors

1. Scalar multiplication:

If ~b=m~a ~b,~a are vectors; m is a scalarthen b=m a (Relation between magnitude)

bx=m ax

by=m ay

Components also follow relation

i.e.~a = ax i + ay j + az k

m~a = max i + may j + maz k

2. Dot Product (Scalar Product):

Figure 1.6: Dot Product

~a ·~b = |~a| · |~b| cosφ

Result is always a scalar. It can be pos-itive or negative depending on φ.

~a ·~b = ~b · ~a

Notice: ~a ·~b = ab cosφ = ab cosφ′

i.e. Doesn’t matter how you measureangle φ between vectors.

i · i = |i| |i| cos0 = 1 · 1 · 1 = 1

i · j = |i| |j| cos90 = 1 · 1 · 0 = 0

i · i = j · j = k · k = 1

i · j = j · k = k · i = 0

If ~a = ax i + ay j + az k~b = bx i + by j + bz k

then ~a ·~b = axbx + ayby + azbz

~a · ~a = |~a| · |~a| cos0 = a · a = a2

1.4. MULTIPLICATION OF VECTORS 5

3. Cross Product (Vector Product):

If ~c = ~a×~b,then c = |~c| = a b sinφ

~a×~b 6= ~b× ~a !!!

~a×~b = −~b× ~a

Figure 1.7: Note: How angle φ is mea-sured

• Direction of cross product determined from right hand rule.

• Also, ~a×~b is ⊥ to ~a and ~b, i.e.

~a · (~a×~b) = 0

~b · (~a×~b) = 0

• IMPORTANT:

~a× ~a = a · a sin0 = 0

|i× i| = |i| |i| sin0 = 1 · 1 · 0 = 0

|i× j| = |i| |j| sin90 = 1 · 1 · 1 = 1

i× i = j × j = k × k = 0

i× j = k; j × k = i; k × i = j

~a×~b =

∣∣∣∣∣i j kax ay az

bx by bz

∣∣∣∣∣ = (ay bz − az by) i

+(az bx − ax bz) j

+(ax by − ay bx) k

1.5. VECTOR FIELD (PHYSICS POINT OF VIEW) 6

4. Vector identities:

~a× (~b + ~c) = ~a×~b + ~a× ~c

~a · (~b× ~c) = ~b · (~c× ~a) = ~c · (~a×~b)

~a× (~b× ~c) = (~a · ~c)~b− (~a ·~b)~c

1.5 Vector Field (Physics Point of View)

A vector field ~F(x, y, z) is a mathematical function which has a vector outputfor a position input.

(Scalar field ~U(x, y, z))

1.6 Other Topics

Tangential Vector

Figure 1.8: d~l is a vector that is always tangential to the curve C with infinitesimallength dl

Surface Vector

Figure 1.9: d~a is a vector that is always perpendicular to the surface S withinfinitesimal area da

1.6. OTHER TOPICS 7

Some uncertainty! (d~a versus − d~a)

Two conventions:

• Area formed from a closed curve

Figure 1.10: Direction of d~a determined from right-hand rule

• Closed surface enclosing a volume

Figure 1.11: Direction of d~a going from inside to outside

Chapter 2

Electric Force & Electric Field

2.1 Electric Force

The electric force between two chargesq1 and q2 can be described byCoulomb’s Law.

~F12 = Force on q1 exerted by q2

~F12 = 14πε0

· q1q2r212· r12

where r12 =~r12

|~r12| is the unit vector which locates particle 1 relative to particle 2.

i.e. ~r12 = ~r1 − ~r2

• q1, q2 are electrical charges in units of Coulomb(C)

• Charge is quantizedRecall 1 electron carries 1.602× 10−19C

• ε0 = Permittivity of free space = 8.85× 10−12C2/Nm2

COULOMB’S LAW:

(1) q1, q2 can be either positive or negative.

2.2. THE ELECTRIC FIELD 9

(2) If q1, q2 are of same sign, then the force experienced by q1 is in directionaway from q2, that is, repulsive.

(3) Force on q2 exerted by q1:

~F21 =1

4πε0

· q2q1

r221

· r21

BUT:

r12 = r21 = distance between q1, q2

r21 =~r21

r21

=~r2 − ~r1

r21

=−~r12

r12

= −r12

∴ ~F21 = −~F12 Newton’s 3rd Law

SYSTEM WITH MANY CHARGES:

The total force experienced by chargeq1 is the vector sum of the forces on q1

exerted by other charges.

~F1 = Force experienced by q1

= ~F1,2 + ~F1,3 + ~F1,4 + · · ·+ ~F1,N

PRINCIPLE OF SUPERPOSITION:

~F1 =N∑

j=2

~F1,j

2.2 The Electric Field

While we need two charges to quantify the electric force, we define the electricfield for any single charge distribution to describe its effect on other charges.


Total force ~F = ~F1 + ~F2 + · · ·+ ~FN

The electric field is defined as

limq0→0

~F

q0

= ~E

(a) E-field due to a single charge qi:

From the definitions of Coulomb’s Law, theforce experienced at location of q0 (point P)

~F0,i =1

4πε0

· q0qi

r20,i

· r0,i

where r0,i is the unit vector along the direction from charge qi to q0,

r0,i = Unit vector from charge qi to point P

= ri (radical unit vector from qi)

Recall ~E = limq0→0

~F

q0∴ E-field due to qi at point P:

~Ei =1

4πε0

· qi

r2i

· ri

where ~ri = Vector pointing from qi to point P,thus ri = Unit vector pointing from qi to point PNote:

(1) E-field is a vector.

(2) Direction of E-field depends on both position of P and sign of qi.

(b) E-field due to system of charges:

Principle of Superposition:In a system with N charges, the total E-field due to all charges is thevector sum of E-field due to individual charges.


i.e. ~E =∑

i

~Ei =1

4πε0

∑

i

qi

r2i

ri

(c) Electric Dipole

System of equal and opposite chargesseparated by a distance d.

Figure 2.1: An electric dipole. (Direction of~d from negative to positive charge)

Electric Dipole Moment

~p = q~d = qdd

p = qd

Example: ~E due to dipole along x-axis

Consider point P at distance x along the perpendicular axis of the dipole ~p :

~E = ~E+ + ~E−↑ ↑

(E-field (E-fielddue to +q) due to −q)

Notice: Horizontal E-field components of ~E+ and ~E− cancel out.

∴ Net E-field points along the axis oppo-site to the dipole moment vector.

2.3. CONTINUOUS CHARGE DISTRIBUTION 12

Magnitude of E-field = 2E+ cos θ

∴ E = 2(

E+ or E− magnitude!︷︸︸︷1

4πε0

· q

r2

)cos θ

But r =

√(d

2

)2+ x2

cos θ =d/2

r

∴ E =1

4πε0

· p

[x2 + (d2)2]

32

(p = qd)

Special case: When x À d

[x2 + (d

2)2]

32 = x3[1 + (

d

2x)2]

32

• Binomial Approximation:

(1 + y)n ≈ 1 + ny if y ¿ 1

E-field of dipole +1

4πε0

· p

x3∼ 1

x3

• Compare with1

r2E-field for single charge

• Result also valid for point P along any axis with respect to dipole

2.3 Continuous Charge Distribution

E-field at point P due to dq:

d ~E =1

4πε0

· dq

r2· r


∴ E-field due to charge distribution:

~E =

ˆ

V olume

d~E =

ˆ

V olume

1

4πε0

· dq

r2· r

(1) In many cases, we can take advantage of the symmetry of the system tosimplify the integral.

(2) To write down the small charge element dq:

1-D dq = λ ds λ = linear charge density ds = small length element2-D dq = σ dA σ = surface charge density dA = small area element3-D dq = ρ dV ρ = volume charge density dV = small volume element

Example 1: Uniform line of charge

charge perunit length= λ

(1) Symmetry considered: The E-field from +z and −z directions cancel alongz-direction, ∴ Only horizontal E-field components need to be considered.

(2) For each element of length dz, charge dq = λdz∴ Horizontal E-field at point P due to element dz =

|d ~E| cos θ =1

4πε0

· λdz

r2︸︷︷︸

dEdz

cos θ

∴ E-field due to entire line charge at point P

E =

L/2ˆ

−L/2

1

4πε0

· λdz

r2cos θ

= 2

L/2ˆ

0

λ

4πε0

· dz

r2cos θ


To calculate this integral:

• First, notice that x is fixed, but z, r, θ all varies.

• Change of variable (from z to θ)

(1)z = x tan θ ∴ dz = x sec2 θ dθx = r cos θ ∴ r2 = x2 sec2 θ

(2) Whenz = 0 , θ = 0

z = L/2 θ = θ0 where tan θ0 =L/2

x

E = 2 · λ

4πε0

θ0ˆ

0

x sec2 θ dθ

x2 sec2 θ· cos θ

= 2 · λ

4πε0

θ0ˆ

0

1

x· cos θ dθ

= 2 · λ

4πε0

· 1

x· (sin θ)

∣∣∣θ0

0

= 2 · λ

4πε0

· 1

x· sin θ0

= 2 · λ

4πε0

· 1

x· L/2√

x2 + (L2)2

E =1

4πε0

· λL

x√

x2 + (L2)2

along x-direction

Important limiting cases:

1. x À L : E +1

4πε0

· λL

x2

But λL = Total charge on rod∴ System behave like a point charge

2. L À x : E +1

4πε0

· λL

x · L2

Ex =λ

2πε0x

ELECTRIC FIELD DUE TO INFINITELY LONG LINE OF CHARGE


Example 2: Ring of Charge

E-field at a height z above a ring ofcharge of radius R

(1) Symmetry considered: For every charge element dq considered, there exists

dq′ where the horizontal ~E field components cancel.⇒ Overall E-field lies along z-direction.

(2) For each element of length dz, charge

dq = λ · ds↑ ↑

Linear Circularcharge density length element

dq = λ · R dφ, where φ is the anglemeasured on the ring plane

∴ Net E-field along z-axis due to dq:

dE =1

4πε0

· dq

r2· cos θ


Total E-field =

ˆdE

=

ˆ 2π

0

1

4πε0

· λR dφ

r2· cos θ (cos θ =

z

r)

Note: Here in this case, θ, R and r are fixed as φ varies! BUT we want toconvert r, θ to R, z.

E =1

4πε0

· λRz

r3

ˆ 2π

0

dφ

E =1

4πε0

· λ(2πR)z

(z2 + R2)3/2along z-axis

BUT: λ(2πR) = total charge on the ring

Example 3: E-field from a disk of surface charge density σ

We find the E-field of a disk byintegrating concentric rings ofcharges.


Total charge of ring

dq = σ · (2πr dr︸︷︷︸Area of the ring

)

Recall from Example 2:

E-field from ring: dE =1

4πε0

· dq z

(z2 + r2)3/2

∴ E =1

4πε0

ˆ R

0

2πσr dr · z(z2 + r2)3/2

=1

4πε0

ˆ R

0

2πσzr dr

(z2 + r2)3/2

• Change of variable:

u = z2 + r2 ⇒ (z2 + r2)3/2 = u3/2

⇒ du = 2r dr ⇒ r dr = 12du

• Change of integration limit:

r = 0 , u = z2

r = R , u = z2 + R2

∴ E =1

4πε0

· 2πσz

ˆ z2+R2

z2

1

2u−3/2du

BUT:

ˆu−3/2du =

u−1/2

−1/2= −2u−1/2

∴ E =1

2ε0

σz (−u−1/2)∣∣∣z2+R2

z2

=1

2ε0

σz

( −1√z2 + R2

+1

z

)

E =σ

2ε0

[1− z√

z2 + R2

]

2.4. ELECTRIC FIELD LINES 18

VERY IMPORTANT LIMITING CASE:

If R À z, that is if we have an infinite sheet of charge with charge den-sity σ:

E =σ

2ε0

[1− z√

z2 + R2

]

' σ

2ε0

[1− z

R

]

E ≈ σ

2ε0

E-field is normal to the charged surfaceFigure 2.2: E-field due to an infi-nite sheet of charge, charge den-sity = σ

Q: What’s the E-field belows the charged sheet?

2.4 Electric Field Lines

To visualize the electric field, we can use a graphical tool called the electricfield lines.

Conventions:

1. The start on position charges and end on negative charges.

2. Direction of E-field at any point is given by tangent of E-field line.

3. Magnitude of E-field at any point is proportional to number of E-field linesper unit area perpendicular to the lines.

2.5. POINT CHARGE IN E-FIELD 21

2.5 Point Charge in E-field

When we place a charge q in an E-field ~E, the force experienced by the charge is

~F = q ~E = m~a

Applications: Ink-jet printer, TV cathoderay tube.

Example:

Ink particle has mass m, charge q (q < 0 here)

Assume that mass of inkdrop is small, what’s the deflection y of the charge?

Solution:

First, the charge carried by the inkdrop is negtive, i.e. q < 0.

Note: q ~E points in opposite direction of ~E.

Horizontal motion: Net force = 0

∴ L = vt (2.1)

2.6. DIPOLE IN E-FIELD 22

Vertical motion: |q ~E| À |m~g|, q is negative,

∴ Net force = −qE = ma (Newton’s 2nd Law)

∴ a = −qE

m(2.2)

Vertical distance travelled:

y =1

2at2

2.6 Dipole in E-field

Consider the force exerted on the dipole in an external E-field:

Assumption: E-field from dipole doesn’t affect the external E-field.

• Dipole moment:

~p = q~d

• Force due to the E-field on +veand −ve charge are equal andopposite in direction. Total ex-ternal force on dipole = 0.

BUT: There is an external torque onthe center of the dipole.

Reminder:

Force ~F exerts at point P.The force exerts a torque~τ = ~r × ~F on point P withrespect to point O.

Direction of the torque vector ~τ is determined from the right-hand rule.


Reference: Halliday Vol.1 Chap 9.1 (Pg.175) torqueChap 11.7 (Pg.243) work done

Net torque ~τ

• direction: clockwisetorque

• magnitude:

τ = τ+ve + τ−ve

= F · d

2sin θ + F · d

2sin θ

= qE · d sin θ

= pE sin θ

~τ = ~p× ~E

Energy Consideration:

When the dipole ~p rotates dθ, the E-field does work.

Work done by external E-field on the dipole:

dW = −τ dθ

Negative sign here because torque by E-field acts to decrease θ.

BUT: Because E-field is a conservative force field 1 2 , we can define apotential energy (U) for the system, so that

dU = −dW

∴ For the dipole in external E-field:

dU = −dW = pE sin θ dθ

∴ U(θ) =

ˆdU =

ˆpE sin θ dθ

= −pE cos θ + U0

1more to come in Chap.4 of notes2ref. Halliday Vol.1 Pg.257, Chap 12.1


set U(θ = 90) = 0,∴ 0 = −pE cos 90 + U0

∴ U0 = 0

∴ Potential energy:

U = −pE cos θ = −~p · ~E

Chapter 3

Electric Flux and Gauss’ Law

3.1 Electric Flux

Latin: flux = ”to flow”

Graphically:Electric flux ΦE represents the number of E-field linescrossing a surface.

Mathematically:

Reminder: Vector of the area ~A is perpendicular to the area A.

For non-uniform E-field & surface, direction of the area vector ~A is notuniform.

d ~A = Area vector forsmall area elementdA

3.1. ELECTRIC FLUX 26

∴ Electric flux dΦE = ~E · d ~A

Electric flux of ~E through surface S: ΦE =

ˆ

S

~E · d ~A

ˆ

S

= Surface integral over surface S

= Integration of integral over all area elements on surface S

Example:

~E =1

4πε0

· −2q

r2r =

−q

2πε0R2r

For a hemisphere, d ~A = dA r

ΦE =

ˆ

S

−q

2πε0R2r · (dA r) (∵ r · r = 1)

= − q

2πε0R2

ˆ

S

dA

︸︷︷︸2πR2

=−q

ε0

For a closed surface:

Recall: Direction of area vector d ~Agoes from inside to outside of closedsurface S.

3.1. ELECTRIC FLUX 27

Electric flux over closed surface S: ΦE =

˛

S

~E · d ~A

˛

S

= Surface integral over closed surface S

Example:

Electric flux of charge q over closedspherical surface of radius R.

~E =1

4πε0

· qr2

r =q

4πε0R2r at the surface

Again, d ~A = dA · r

∴ ΦE =

˛

S

~E︷︸︸︷q

4πε0R2r ·

d ~A︷︸︸︷dA r

=q

4πε0R2

˛

S

dA

︸︷︷︸Total surface area of S = 4πR2

ΦE =q

ε0

IMPORTANT POINT:If we remove the spherical symmetry of closed surface S, the total number of

E-field lines crossing the surface remains the same.∴ The electric flux ΦE

3.2. GAUSS’ LAW 28

ΦE =

˛

S

~E · d ~A =

˛

S′

~E · d ~A =q

ε0

3.2 Gauss’ Law

ΦE =

˛

S

~E · d ~A =q

ε0

for any closed surface S

And q is the net electric charge enclosed in closed surface S.

• Gauss’ Law is valid for all charge distributions and all closed surfaces.(Gaussian surfaces)

• Coulomb’s Law can be derived from Gauss’ Law.

• For system with high order of symmetry, E-field can be easily determined ifwe construct Gaussian surfaces with the same symmetry and applies Gauss’Law

3.3 E-field Calculation with Gauss’ Law

(A) Infinite line of charge

Linear charge density: λCylindrical symmetry.E-field directs radially outward from therod.Construct a Gaussian surface S in theshape of a cylinder, making up of acurved surface S1, and the top andbottom circles S2, S3.

Gauss’ Law:

˛

S

~E · d ~A =Total charge

ε0

=λL

ε0

3.3. E-FIELD CALCULATION WITH GAUSS’ LAW 29

˛

S

~E · d ~A =

ˆ

S1

~E · d ~A

︸︷︷︸~E‖d ~A

+

ˆ

S2

~E · d ~A +

ˆ

S3

~E · d ~A

︸︷︷︸= 0 ∵ ~E⊥d ~A

∴ E

ˆ

S1

dA

︸︷︷︸Total area of surface S1

=λL

ε0

E(2πrL) =λL

ε0

∴ E =λ

2πε0r(Compare with Chapter 2 note)

~E =λ

2πε0rr

(B) Infinite sheet of charge

Uniform surface charge density:σPlanar symmetry.E-field directs perpendicular tothe sheet of charge.Construct Gaussian surface S inthe shape of a cylinder (pillbox) of cross-sectional area A.

Gauss’ Law:

˛

S

~E · d ~A =Aσ

ε0

ˆ

S1

~E · d ~A = 0 ∵ ~E ⊥ d ~A over whole surface S1

ˆ

S2

~E · d ~A +

ˆ

S3

~E · d ~A = 2EA ( ~E ‖ d ~A2, ~E ‖ d ~A3)

3.3. E-FIELD CALCULATION WITH GAUSS’ LAW 30

Note: For S2, both ~E and d ~A2 point up

For S3, both ~E and d ~A3 point down

∴ 2EA =Aσ

ε0

⇒ E =σ

2ε0

(Compare with Chapter 2 note)

(C) Uniformly charged sphereTotal charge = QSpherical symmetry.

(a) For r > R:

Consider a spherical Gaussian surface S ofradius r:

~E ‖ d ~A ‖ r

Gauss’ Law:

˛

S

~E · d ~A =Q

ε0

˛

S

E · dA =Q

ε0

E

˛

S

dA

︸︷︷︸surface area of S = 4πr2

=Q

ε0

∴ ~E =Q

4πε0r2r ; for r > R

(b) For r < R:

Consider a spherical Gaussian surface S ′ ofradius r < R, then total charge included q isproportional to the volume included by S ′

∴q

Q=

Volume enclosed by S ′

Total volume of sphere

3.4. GAUSS’ LAW AND CONDUCTORS 31

q

Q=

4/3 πr3

4/3 πR3⇒ q =

r3

R3Q

Gauss’ Law:

˛

S′~E · d ~A =

q

ε0

E

˛

S′dA

︸︷︷︸surface area of S ′ = 4πr2

=r3

R3

1

ε0

·Q

∴ ~E =1

4πε0

· Q

R3r r ; for r ≤ R

3.4 Gauss’ Law and Conductors

For isolated conductors, charges are freeto move until all charges lie outside thesurface of the conductor. Also, the E-field at the surface of a conductor is per-pendicular to its surface. (Why?)

Consider Gaussian surface S of shape of cylinder:

˛

S

~E · d ~A =σA

ε0


BUT

ˆ

S1

~E · d ~A = 0 ( ∵ ~E ⊥ d ~A )ˆ

S3

~E · d ~A = 0 ( ∵ ~E = 0 inside conductor )

ˆ

S2

~E · d ~A = E

ˆ

S2

dA

︸︷︷︸Area of S2

( ∵ ~E ‖ d ~A )

= EA

∴ Gauss’ Law ⇒ EA =σA

ε0

∴ On conductor’s surface E =σ

ε0

BUT, there’s no charge inside conductors.

∴ Inside conductors E = 0 Always!

Notice: Surface charge density on a conductor’s surface is not uniform.

Example: Conductor with a charge insideNote: This is not an isolated system (because of the charge inside).

Example:


I. Charge sprayed on a conductor sphere:

First, we know that charges all moveto the surface of conductors.

(i) For r < R:Consider Gaussian surface S2

˛

S2

~E · d ~A = 0 ( ∵ no charge inside )

⇒ E = 0 everywhere.

(ii) For r ≥ R:Consider Gaussian surface S1:

˛

S1

~E · d ~A =Q

ε0

E

˛

S1

d ~A

︸︷︷︸4πr2

=Q

ε0

(

For a conductor︷︸︸︷~E ‖ d ~A ‖ r︸︷︷︸

Spherically symmetric

)

E =Q

4πε0r2

II. Conductor sphere with hole inside:


Consider Gaussian surface S1: Totalcharge included = 0

∴ E-field = 0 inside

The E-field is identical to the case of asolid conductor!!

III. A long hollow cylindrical conductor:

Example:Inside hollow cylinder ( +2q )

Inner radius aOuter radius b

Outside hollow cylinder ( −3q )

Inner radius cOuter radius d

Question: Find the charge on each surface of the conductor.

For the inside hollow cylinder, charges distribute only on the sur-face.∴ Inner radius a surface, charge = 0and Outer radius b surface, charge = +2q

For the outside hollow cylinder, charges do not distribute only onoutside.∵ It’s not an isolated system. (There are charges inside!)

Consider Gaussian surface S ′ inside the conductor:E-field always = 0

∴ Need charge −2q on radius c surface to balance the charge of innercylinder.So charge on radius d surface = −q. (Why?)

IV. Large sheets of charge:Total charge Q on sheet of area A,


∴ Surface charge density σ =Q

A

By principle of superposition

Region A: E = 0 E = 0

Region B: E =Q

ε0AE =

Q

ε0ARegion C: E = 0 E = 0

Chapter 4

Electric Potential

4.1 Potential Energy and Conservative Forces

(Read Halliday Vol.1 Chap.12)Electric force is a conservative force

Work done by the electric force ~F as acharge moves an infinitesimal distance d~salong Path A = dW

Note: d~s is in the tangent direction of the curve of Path A.

dW = ~F · d~s

∴ Total work done W by force ~F in moving the particle from Point 1 to Point 2

W =

ˆ 2

1

~F · d~sPath A

ˆ 2

1

= Path Integral

Path A= Integration over Path A from Point 1 to Point 2.

4.1. POTENTIAL ENERGY AND CONSERVATIVE FORCES 37

DEFINITION: A force is conservative if the work done on a particle bythe force is independent of the path taken.

∴ For conservative forces,

ˆ 2

1

~F · d~s =

ˆ 2

1

~F · d~sPath A Path B

Let’s consider a path starting at point1 to 2 through Path A and from 2 to 1through Path C

Work done =

ˆ 2

1

~F · d~s +

ˆ 1

2

~F · d~sPath A Path C

=

ˆ 2

1

~F · d~s −ˆ 2

1

~F · d~sPath A Path B

DEFINITION: The work done by a conservative force on a particle when itmoves around a closed path returning to its initial position is zero.

MATHEMATICALLY, ~∇× ~F = 0 everywhere for conservative force ~F

Conclusion: Since the work done by a conservative force ~F is path-independent,we can define a quantity, potential energy, that depends only on theposition of the particle.

Convention: We define potential energy U such that

dU = −W = −ˆ

~F · d~s

∴ For particle moving from 1 to 2ˆ 2

1

dU = U2 − U1 = −ˆ 2

1

~F · d~s

where U1, U2 are potential energy at position 1, 2.


Example:

Suppose charge q2

moves from point 1to 2.

From definition: U2 − U1 = −ˆ 2

1

~F · d~r

= −ˆ r2

r1

F dr ( ∵ ~F ‖ d~r )

= −ˆ r2

r1

1

4πε0

q1q2

r2dr

( ∵ˆ

dr

r2= −1

r+ C ) =

1

4πε0

q1q2

r

∣∣∣∣∣r2

r1

−∆W = ∆U =1

4πε0

q1q2

(1

r2

− 1

r1

)

Note:

(1) This result is generally true for 2-Dimension or 3-D motion.

(2) If q2 moves away from q1,then r2 > r1, we have

• If q1, q2 are of same sign,then ∆U < 0, ∆W > 0(∆W = Work done by electric repulsive force)

• If q1, q2 are of different sign,then ∆U > 0, ∆W < 0(∆W = Work done by electric attractive force)

(3) If q2 moves towards q1,then r2 < r1, we have

• If q1, q2 are of same sign,then ∆U 0, ∆W 0

• If q1, q2 are of different sign,then ∆U 0, ∆W 0


(4) Note: It is the difference in potential energy that is important.

REFERENCE POINT: U(r = ∞) = 0

∴ U∞ − U1 =1

4πε0

q1q2

(1

r2

− 1

r1

)

↓∞

U(r) =1

4πε0

· q1q2

r

If q1, q2 same sign, then U(r) > 0 for all rIf q1, q2 opposite sign, then U(r) < 0 for all r

(5) Conservation of Mechanical Energy:For a system of charges with no external force,

E = K + U = Constant

(Kinetic Energy) (Potential Energy)

or ∆E = ∆K + ∆U = 0

Potential Energy of A System of Charges

Example: P.E. of 3 charges q1, q2, q3

Start: q1, q2, q3 all at r = ∞, U = 0

Step1: Move q1 from ∞ to its position ⇒ U = 0

Step2:

Move q2 from ∞ to new position ⇒

U =1

4πε0

q1q2

r12

Step3:

Move q3 from ∞ to new position ⇒ Total P.E.

U =1

4πε0

[q1q2

r12

+q1q3

r13

+q2q3

r23

]

Step4: What if there are 4 charges?

4.2. ELECTRIC POTENTIAL 40

4.2 Electric Potential

Consider a charge q at center, we consider its effect on test charge q0

DEFINITION: We define electric potential V so that

∆V =∆U

q0

=−∆W

q0

( ∴ V is the P.E. per unit charge)

• Similarly, we take V (r = ∞) = 0.

• Electric Potential is a scalar.

• Unit: V olt(V ) = Joules/Coulomb

• For a single point charge:

V (r) =1

4πε0

· q

r

• Energy Unit: ∆U = q∆V

electron− V olt(eV ) = 1.6× 10−19

︸︷︷︸charge of electron

J

Potential For A System of Charges

For a total of N point charges, the po-tential V at any point P can be derivedfrom the principle of superposition.

Recall that potential due to q1 at

point P: V1 =1

4πε0

· q1

r1

∴ Total potential at point P due to N charges:

V = V1 + V2 + · · ·+ VN (principle of superposition)

=1

4πε0

[q1

r1

+q2

r2

+ · · ·+ qN

rN

]


V =1

4πε0

N∑

i=1

qi

ri

Note: For ~E, ~F , we have a sum of vectorsFor V, U , we have a sum of scalars

Example: Potential of an electric dipole

Consider the potential ofpoint P at distance x > d

2

from dipole.

V =1

4πε0

[+q

x− d2

+−q

x + d2

]

Special Limiting Case: x À d

1

x∓ d2

=1

x· 1

1∓ d2x

' 1

x

[1± d

2x

]

∴ V =1

4πε0

· q

x

[1 +

d

2x− (1− d

2x)

]

V =p

4πε0x2(Recall p = qd)

For a point charge E ∝ 1

r2V ∝ 1

r

For a dipole E ∝ 1

r3V ∝ 1

r2

For a quadrupole E ∝ 1

r4V ∝ 1

r3

Electric Potential of Continuous Charge Distribution

For any charge distribution, we write the electri-cal potential dV due to infinitesimal charge dq:

dV =1

4πε0

· dq

r


∴ V =

ˆ

chargedistribution

1

4πε0

· dq

r

Similar to the previous examples on E-field, for the case of uniform chargedistribution:

1-D ⇒ long rod ⇒ dq = λ dx2-D ⇒ charge sheet ⇒ dq = σ dA3-D ⇒ uniformly charged body ⇒ dq = ρ dV

Example (1): Uniformly-charged ring

Length of the infinitesimal ring element= ds = Rdθ

∴ charge dq = λ ds

= λR dθ

dV =1

4πε0

· dq

r=

1

4πε0

· λR dθ√R2 + z2

The integration is around the entire ring.

∴ V =

ˆ

ring

dV

=

ˆ 2π

0

1

4πε0

· λR dθ√R2 + z2

=λR

4πε0

√R2 + z2

ˆ 2π

0

dθ

︸︷︷︸2π

Total charge on thering = λ · (2πR) V =

Q

4πε0

√R2 + z2

LIMITING CASE: z À R ⇒ V =Q

4πε0

√z2

=Q

4πε0|z|


Example (2): Uniformly-charged disk

Using the principle of superpo-sition, we will find the potentialof a disk of uniform charge den-sity by integrating the potential ofconcentric rings.

∴ dV =1

4πε0

ˆ

disk

dq

r

Ring of radius x: dq = σ dA = σ (2πxdx)

∴ V =

ˆ R

0

1

4πε0

· σ2πx dx√x2 + z2

=σ

4ε0

ˆ R

0

d(x2 + z2)

(x2 + z2)1/2

V =σ

2ε0

(√

z2 + R2 −√

z2)

=σ

2ε0

(√

z2 + R2 − |z|)Recall:

|x| =

+x; x ≥ 0−x; x < 0

Limiting Case:

(1) If |z| À R

√z2 + R2 =

√z2

(1 +

R2

z2

)

= |z| ·(1 +

R2

z2

) 12 ( (1 + x)n ≈ 1 + nx if x ¿ 1 )

' |z| ·(1 +

R2

2z2

)(|z|z2

=1

|z| )

∴ At large z, V ' σ

2ε0

· R2

2|z| =Q

4πε0|z| (like a point charge)

where Q = total charge on disk = σ · πR2


(2) If |z| ¿ R

√z2 + R2 = R ·

(1 +

z2

R2

) 12

' R(1 +

z2

2R2

)

∴ V ' σ

2ε0

[R− |z|+ z2

2R

]

At z = 0, V =σR

2ε0

; Let’s call this V0

∴ V (z) =σR

2ε0

[1− |z|

R+

z2

2R2

]

V (z) = V0

[1− |z|

R+

z2

2R2

]

The key here is that it is the difference between potentials of two pointsthat is important.⇒ A convenience reference point to compare in this example is thepotential of the charged disk.∴ The important quantity here is

V (z)− V0 = −|z|R

V0 +½

½½½z2

2R2V0

neglected as z ¿ R

V (z)− V0 = −V0

R|z|

4.3. RELATION BETWEEN ELECTRIC FIELD E AND ELECTRICPOTENTIAL V 45

4.3 Relation Between Electric Field E and Elec-

tric Potential V

(A) To get V from ~E:Recall our definition of the potential V:

∆V =∆U

q0

= −W12

q0

where ∆U is the change in P.E.; W12 is the work done in bringing chargeq0 from point 1 to 2.

∴ ∆V = V2 − V1 =− ´ 2

1~F · d~s

q0

However, the definition of E-field: ~F = q0~E

∴ ∆V = V2 − V1 = −ˆ 2

1

~E · d~s

Note: The integral on the right hand side of the above can be calculatedalong any path from point 1 to 2. (Path-Independent)

Convention: V∞ = 0 ⇒ VP = −ˆ P

∞~E · d~s

(B) To get ~E from V :

Again, use the definition of V :

∆U = q0∆V = −W︸︷︷︸Work done

However,

W = q0~E︸︷︷︸

Electric force

· ∆~s

= q0 Es ∆s

where Es is the E-field component alongthe path ∆~s.

∴ q0∆V = −q0Es∆s


∴ Es = −∆V

∆s

For infinitesimal ∆s,

∴ Es = −dV

ds

Note: (1) Therefore the E-field component along any direction is the neg-tive derivative of the potential along the same direction.

(2) If d~s ⊥ ~E, then ∆V = 0

(3) ∆V is biggest/smallest if d~s ‖ ~E

Generally, for a potential V (x, y, z), the relation between ~E(x, y, z) and Vis

Ex = −∂V

∂xEy = −∂V

∂yEz = −∂V

∂z

∂

∂x,

∂

∂y,

∂

∂zare partial derivatives

For∂

∂xV (x, y, z), everything y, z are treated like a constant and we only

take derivative with respect to x.

Example: If V (x, y, z) = x2y − z

∂V

∂x=

∂V

∂y=

∂V

∂z=

For other co-ordinate systems

(1) Cylindrical:

V (r, θ, z)

Er = −∂V

∂r

Eθ = −1

r· ∂V

∂θ

Ez = −∂V

∂z


(2) Spherical:

V (r, θ, φ)

Er = −∂V

∂r

Eθ = −1

r· ∂V

∂θ

Eφ = − 1

r sin θ· ∂V

∂φ

Note: Calculating V involves summation of scalars, which is easier thanadding vectors for calculating E-field.∴ To find the E-field of a general charge system, we first calculateV , and then derive ~E from the partial derivative.

Example: Uniformly charged diskFrom potential calculations:

V =σ

2ε0

(√

R2 + z2 − |z| ) for a point alongthe z-axis

For z > 0, |z| = z

∴ Ez = −∂V

∂z=

σ

2ε0

[1− z√

R2 + z2

] (Compare withChap.2 notes)

Example: Uniform electric field(e.g. Uniformly charged +ve and −ve plates)

Consider a path going from the −veplate to the +ve platePotential at point P, VP can be deducedfrom definition.

i.e. VP − V− = −ˆ s

0

~E · d~s (V− = Potential of−ve plate)

= −ˆ s

0

(−E ds) ∵ ~E, d~s pointingopposite directions

= E

ˆ s

0

ds = Es

Convenient reference: V− = 0

∴ VP = E · s

4.4. EQUIPOTENTIAL SURFACES 48

4.4 Equipotential Surfaces

Equipotential surface is a surface on which the potential is constant.⇒ (∆V = 0)

V (r) =1

4πε0

· +q

r= const

⇒ r = const

⇒ Equipotential surfaces arecircles/spherical surfaces

Note: (1) A charge can move freely on an equipotential surface without anywork done.

(2) The electric field lines must be perpendicular to the equipotentialsurfaces. (Why?)On an equipotential surface, V = constant⇒ ∆V = 0 ⇒ ~E ·d~l = 0, where d~l is tangent to equipotential surface∴ ~E must be perpendicular to equipotential surfaces.

Example: Uniformly charged surface (infinite)

Recall V = V0 − σ

2ε0

|z|↑

Potential at z = 0Equipotential surface means

V = const ⇒ V0 − σ

2ε0

|z| = C

⇒ |z| = constant


Example: Isolated spherical charged conductors

Recall:

(1) E-field inside = 0

(2) charge distributed on theoutside of conductors.

(i) Inside conductor:

E = 0 ⇒ ∆V = 0 everywhere in conductor

⇒ V = constant everywhere in conductor

⇒ The entire conductor is at the same potential

(ii) Outside conductor:

V =Q

4πε0r

∵ Spherically symmetric (Just like a point charge.)BUT not true for conductors of arbitrary shape.

Example: Connected conducting spheres

Two conductors con-nected can be seen as asingle conductor


∴ Potential everywhere is identical.

Potential of radius R1 sphere V1 =q1

4πε0R1

Potential of radius R2 sphere V2 =q2

4πε0R2

V1 = V2

⇒ q1

R1

=q2

R2

⇒ q1

q2

=R1

R2

Surface charge density

σ1 =q1

4πR21︸︷︷︸

Surface area of radius R1 sphere

∴σ1

σ2

=q1

q2

· R22

R21

=R2

R1

∴ If R1 < R2, then σ1 > σ2

And the surface electric field E1 > E2

For arbitrary shape conductor:

At every point on the conductor,we fit a circle. The radius of thiscircle is the radius of curvature.

Note: Charge distribution on a conductor does not have to be uniform.

Chapter 5

Capacitance and DC Circuits

5.1 Capacitors

A capacitor is a system of two conductors that carries equal and oppositecharges. A capacitor stores charge and energy in the form of electro-static field.

We define capacitance as

C =Q

VUnit: Farad(F)

where

Q = Charge on one plate

V = Potential difference between the plates

Note: The C of a capacitor is a constant that depends only on its shape andmaterial.i.e. If we increase V for a capacitor, we can increase Q stored.

5.2 Calculating Capacitance

5.2.1 Parallel-Plate Capacitor

5.2. CALCULATING CAPACITANCE 52

(1) Recall from Chapter 3 note,

| ~E| = σ

ε0

=Q

ε0A

(2) Recall from Chapter 4 note,

∆V = V+ − V− = −ˆ +

−~E · d~s

Again, notice that this integral is independent of the path taken.∴ We can take the path that is parallel to the ~E-field.

∴ ∆V =

ˆ −

+

~E · d~s

=

ˆ −

+

E · ds

=Q

ε0A

ˆ −

+

ds

︸︷︷︸Length of path taken

=Q

ε0A· d

(3) ∴ C =Q

∆V=

ε0A

d

5.2.2 Cylindrical Capacitor

Consider two concentric cylindrical wireof innner and outer radii r1 and r2 re-spectively. The length of the capacitoris L where r1 < r2 ¿ L.

5.2. CALCULATING CAPACITANCE 53

(1) Using Gauss’ Law, we determine that the E-field between the conductorsis (cf. Chap3 note)

~E =1

2πε0

· λ

rr =

1

2πε0

· Q

Lrr

where λ is charge per unit length

(2)

∆V =

ˆ −

+

~E · d~s

Again, we choose the path of integration so that d~s ‖ r ‖ ~E

∴ ∆V =

ˆ r2

r1

E dr =Q

2πε0L

ˆ r2

r1

dr

r︸︷︷︸ln(

r2r1

)

∴ C =Q

∆V= 2πε0

L

ln(r2/r1)

5.2.3 Spherical Capacitor

For the space between the two conductors,

E =1

4πε0

· Q

r2; r1 < r < r2

∆V =

ˆ −

+

~E · d~s

Choose d~s ‖ r =

ˆ r2

r1

1

4πε0

· Q

r2dr

=Q

4πε0

[1

r1

− 1

r2

]

C = 4πε0

[r1r2

r2 − r1

]

5.3. CAPACITORS IN COMBINATION 54

5.3 Capacitors in Combination

(a) Capacitors in Parallel

In this case, it’s the potential differenceV = Va − Vb that is the same across thecapacitor.

BUT: Charge on each capacitor different

Total charge Q = Q1 + Q2

= C1V + C2V

Q = (C1 + C2)︸︷︷︸Equivalent capacitance

V

∴ For capacitors in parallel: C = C1 + C2

(b) Capacitors in Series

The charge across capacitors arethe same.

BUT: Potential difference (P.D.) across capacitors different

∆V1 = Va − Vc =Q

C1

P.D. across C1

∆V2 = Vc − Vb =Q

C2

P.D. across C2

∴ Potential difference

∆V = Va − Vb

= ∆V1 + ∆V2

∆V = Q (1

C1

+1

C2

) =Q

C

where C is the Equivalent Capacitance

∴1

C=

1

C1

+1

C2

5.4. ENERGY STORAGE IN CAPACITOR 55

5.4 Energy Storage in Capacitor

In charging a capacitor, positive chargeis being moved from the negative plateto the positive plate.⇒ NEEDS WORK DONE!

Suppose we move charge dq from −ve to +ve plate, change in potential energy

dU = ∆V · dq =q

Cdq

Suppose we keep putting in a total charge Q to the capacitor, the total potentialenergy

U =

ˆdU =

ˆ Q

0

q

Cdq

∴ U =Q2

2C=

1

2C∆V 2

(∵ Q=C∆V )

The energy stored in the capacitor is stored in the electric field between theplates.

Note : In a parallel-plate capacitor, the E-field is constant between the plates.

∴ We can consider the E-field energy

density u =Total energy stored

Total volume with E-field

∴ u =U

Ad︸︷︷︸Rectangular volume

Recall

C =ε0A

d

E =∆V

d⇒ ∆V = Ed

∴ u =1

2(

C︷︸︸︷ε0A

d) · (

(∆V )2︷︸︸︷Ed )2 ·

1V olume︷︸︸︷

1

Ad

5.4. ENERGY STORAGE IN CAPACITOR 56

u =1

2ε0E

2 Energy per unit volumeof the electrostatic field

↑can be generally applied

Example : Changing capacitance

(1) Isolated Capacitor:Charge on the capacitor plates remains constant.

BUT: Cnew =ε0A

2d=

1

2Cold

∴ Unew =Q2

2Cnew

=Q2

2Cold/2= 2Uold

∴ In pulling the plates apart, work done W > 0

Summary :Q → Q C → C/2

(V = QC

) ⇒ V → 2V E → E (E = Vd)

12ε0E

2 = u → u U → 2U (U = u · volume)

(2) Capacitor connected to a battery:Potential difference between capacitor plates remains constant.

Unew =1

2Cnew∆V 2 =

1

2· 1

2Cold∆V 2 =

1

2Uold

∴ In pulling the plates apart, work done by battery < 0

Summary :

Q → Q/2 C → C/2V → V E → E/2u → u/4 U → U/2

5.5. DIELECTRIC CONSTANT 57

5.5 Dielectric Constant

We first recall the case for a conductor being placed in an external E-field E0.

In a conductor, charges are free to moveinside so that the internal E-field E ′ setup by these charges

E ′ = −E0

so that E-field inside conductor = 0.

Generally, for dielectric, the atoms andmolecules behave like a dipole in an E-field.

Or, we can envision this so that in the absence of E-field, the direction of dipolein the dielectric are randomly distributed.

5.6. CAPACITOR WITH DIELECTRIC 58

The aligned dipoles will generate an induced E-field E ′, where |E ′| < |E0|.We can observe the aligned dipoles in the form of induced surface charge.

Dielectric Constant : When a dielectric is placed in an external E-field E0,the E-field inside a dielectric is induced.E-field in dielectric

E =1

Ke

E0

Ke = dielectric constant ≥ 1

Example :

Vacuum Ke = 1Porcelain Ke = 6.5Water Ke ∼ 80Perfect conductor Ke = ∞Air Ke = 1.00059

5.6 Capacitor with Dielectric

Case I :

Again, the charge remains constant after dielectric is inserted.

BUT: Enew =1

Ke

Eold

∴ ∆V = Ed ⇒ ∆Vnew =1

Ke

∆Vold

∴ C =Q

∆V⇒ Cnew = Ke Cold

For a parallel-plate capacitor with dielectric:

C =Keε0A

d

5.6. CAPACITOR WITH DIELECTRIC 59

We can also write C =εA

din general with

ε = Ke ε0 (called permittivity of dielectric)

(Recall ε0 = Permittivity of free space)

Energy stored U =Q2

2C;

∴ Unew =1

Ke

Uold < Uold

∴ Work done in inserting dielectric < 0

Case II : Capacitor connected to a battery

Voltage across capacitor plates remains constant after insertion of dielec-tric.In both scenarios, the E-field inside capacitor remains constant(∵ E = V/d)

BUT: How can E-field remain constant?ANSWER: By having extra charge on capacitor plates.

Recall: For conductors,

E =σ

ε0

(Chapter 3 note)

⇒ E =Q

ε0A(σ = charge per unit area = Q/A)

After insertion of dielectric:

E ′ =E

Ke

=Q′

Keε0A

But E-field remains constant!

∴ E ′ = E ⇒ Q′

Keε0A=

Q

ε0A

⇒ Q′ = KeQ > Q

5.7. GAUSS’ LAW IN DIELECTRIC 60

∴ Capacitor C = Q/V ⇒ C ′ → KeCEnergy stored U = 1

2CV 2 ⇒ U ′ → KeU

(i.e. Unew > Uold)

∴ Work done to insert dielectric > 0

5.7 Gauss’ Law in Dielectric

The Gauss’ Law we’ve learned is applicable in vacuum only. Let’s use the capac-itor as an example to examine Gauss’ Law in dielectric.

Free chargeon plates

±Q ±Q

Induced chargeon dielectric

0 ∓Q′

Gauss’ Law Gauss’ Law:˛

S

~E · d ~A =Q

ε0

˛

S

~E ′ · d ~A =Q−Q′

ε0

⇒ E0 =Q

ε0A(1) ∴ E ′ =

Q−Q′

ε0A(2)

However, we define E ′ =E0

Ke

(3)

From (1), (2), (3) ∴Q

Keε0A=

Q

ε0A− Q′

ε0A

∴ Induced charge density σ′ =Q′

A= σ

(1− 1

Ke

)< σ

where σ is free charge density.

Recall Gauss’ Law in Dielectric:

ε0

˛

S

~E ′ · d ~A = Q − Q′

↑ ↑ ↑E-field in dielectric free charge induced charge

5.8. OHM’S LAW AND RESISTANCE 61

⇒ ε0

˛

S

~E ′ · d ~A = Q−Q[1− 1

Ke

]

⇒ ε0

˛

S

~E ′ · d ~A =Q

Ke

˛

S

Ke~E ′ · d ~A =

Q

ε0

Gauss’ Lawin dielectric

Note :

(1) This goes back to the Gauss’ Law in vacuum with E =E0

Ke

for dielectric

(2) Only free charges need to be considered, even for dielectric where thereare induced charges.

(3) Another way to write: ˛

S

~E · d ~A =Q

ε

where ~E is E-field in dielectric, ε = Keε0 is Permittivity

Energy stored with dielectric:

Total energy stored: U =1

2CV 2

With dielectric, recall C =Keε0A

d

V = Ed

∴ Energy stored per unit volume:

ue =U

Ad=

1

2Keε0E

2

and udielectric = Keuvacuum

∴ More energy is stored per unit volume in dielectric than in vacuum.

5.8 Ohm’s Law and Resistance

ELECTRIC CURRENT is defined as the flow of electric charge through across-sectional area.


i =dQ

dtUnit: Ampere (A)= C/second

Convention :

(1) Direction of current is the direction of flow of positive charge.

(2) Current is NOT a vector, but the current density is a vector.

~j = charge flow per unit time per unit area

i =

ˆ~j · d ~A

Drift Velocity :

Consider a current i flowing througha cross-sectional area A:

∴ In time ∆t, total charges passing through segment:

∆Q = q A(Vd∆t)︸︷︷︸Volume of chargepassing through

n

where q is charge of the current carrier, n is density of charge carrierper unit volume

∴ Current: i =∆Q

∆t= nqAvd

Current Density: ~j = nq~vd

Note : For metal, the charge carriers are the free electrons inside.∴ ~j = −ne~vd for metals∴ Inside metals, ~j and ~vd are in opposite direction.

We define a general property, conductivity (σ), of a material as:

~j = σ ~E


Note : In general, σ is NOT a constant number, but rather a function of positionand applied E-field.

A more commonly used property, resistivity (ρ), is defined as ρ =1

σ

∴ ~E = ρ~j

Unit of ρ : Ohm-meter (Ωm)where Ohm (Ω) = Volt/Ampere

OHM’S LAW:Ohmic materials have resistivity that are independent of the applied electric field.i.e. metals (in not too high E-field)

Example :

Consider a resistor (ohmic material) oflength L and cross-sectional area A.

∴ Electric field inside conductor:

∆V =

ˆ~E · d~s = E · L ⇒ E =

∆V

L

Current density: j =i

A

∴ ρ =E

j

ρ =∆V

L· 1

i/A

∆V

i= R = ρ

L

A

where R is the resistance of the conductor.

Note: ∆V = iR is NOT a statement of Ohm’s Law. It’s just a definition forresistance.

5.9. DC CIRCUITS 64

ENERGY IN CURRENT:

Assuming a charge ∆Q enterswith potential V1 and leaves withpotential V2 :

∴ Potential energy lost in the wire:

∆U = ∆QV2 −∆QV1

∆U = ∆Q(V2 − V1)

∴ Rate of energy lost per unit time

∆U

∆t=

∆Q

∆t(V2 − V1)

Joule’s heating P = i ·∆V =Power dissipatedin conductor

For a resistor R, P = i2R =∆V 2

R

5.9 DC Circuits

A battery is a device that supplies electrical energy to maintain a current in acircuit.

In moving from point 1 to 2, elec-tric potential energy increase by∆U = ∆Q(V2 − V1) = Work done by E

Define E = Work done/charge = V2 − V1

5.9. DC CIRCUITS 65

Example :

Va = Vc

Vb = Vd

assuming(1) perfect conducting wires.

By Definition: Vc − Vd = iR

Va − Vb = E

∴ E = iR ⇒ i =ER

Also, we have assumed(2) zero resistance inside battery.

Resistance in combination :

Potential differece (P.D.)

Va − Vb = (Va − Vc) + (Vc − Vb)

= iR1 + iR2

∴ Equivalent Resistance

R = R1 + R2 for resistors in series1

R=

1

R1

+1

R2

for resistors in parallel

5.9. DC CIRCUITS 66

Example :

For real battery, there is aninternal resistance thatwe cannot ignore.

∴ E = i(R + r)

i =E

R + r

Joule’s heating in resistor R :

P = i · (P.D. across resistor R)

= i2R

P =E2R

(R + r)2

Question: What is the value of R to obtain maximum Joule’s heating?

Answer: We want to find R to maximize P.

dP

dR=

E2

(R + r)2− E2 2R

(R + r)3

SettingdP

dR= 0 ⇒ E2

(R + r)3[(R + r)− 2R] = 0

⇒ r −R = 0

⇒ R = r

5.9. DC CIRCUITS 67

ANALYSIS OF COMPLEX CIRCUITS:

KIRCHOFF’S LAWS:

(1) First Law (Junction Rule):Total current entering a junction equal to the total current leaving thejunction.

(2) Second Law (Loop Rule):The sum of potential differences around a complete circuit loop is zero.

Convention :

(i)

Va > Vb ⇒ Potential difference = −iR

i.e. Potential drops across resistors

(ii)

Vb > Va ⇒ Potential difference = +E

i.e. Potential rises across the negative plate of the battery.

Example :

5.9. DC CIRCUITS 68

By junction rule:i1 = i2 + i3 (5.1)

By loop rule:

Loop A ⇒ 2E0 − i1R− i2R + E0 − i1R = 0 (5.2)

Loop B ⇒ −i3R− E0 − i3R− E0 + i2R = 0 (5.3)

Loop C ⇒ 2E0 − i1R− i3R− E0 − i3R− i1R = 0 (5.4)

BUT: (5.4) = (5.2) + (5.3)General rule: Need only 3 equations for 3 current

i1 = i2 + i3 (5.1)

3E0 − 2i1R− i2R = 0 (5.2)

−2E0 + i2R− 2i3R = 0 (5.3)

Substitute (5.1) into (5.2) :

3E0 − 2(i2 + i3)R− i2R = 0

⇒ 3E0 − 3i2R− 2i3R = 0 (5.4)

Subtract (5.3) from (5.4), i.e. (5.4)−(5.3)

3E0 − (−2E0)− 3i2R− i2R = 0

⇒ i2 =5

4· E0

R

Substitute i2 into (5.3) :

−2E0 +(5

4· E0

R

)R− 2i3R = 0

5.10. RC CIRCUITS 69

⇒ i3 = −3

8· E0

R

Substitute i2, i3 into (5.1) :

i1 =(5

4− 3

8

) E0

R=

7

8· E0

R

Note: A negative current means that it is flowing in opposite direction from theone assumed.

5.10 RC Circuits

(A) Charging a capacitor with battery:

Using the loop rule:

+E0 − iR︸︷︷︸P.D.across R

− Q

C︸︷︷︸P.D.across C

= 0

Note: Direction of i is chosen so that the current represents the rate atwhich the charge on the capacitor is increasing.

∴ E = R

i︷︸︸︷dQ

dt+

Q

C1st orderdifferential eqn.

⇒ dQ

EC −Q=

dt

RC

Integrate both sides and use the initial condition:t = 0, Q on capacitor = 0

ˆ Q

0

dQ

EC −Q=

ˆ t

0

dt

RC


− ln(EC −Q)∣∣∣Q

0=

t

RC

∣∣∣t

0

⇒ − ln(EC −Q) + ln(EC) =t

RC

⇒ ln( 1

1− QEC

)=

t

RC

⇒ 1

1− QEC

= et/RC

⇒ Q

EC= 1− e−t/RC

⇒ Q(t) = EC(1− e−t/RC)

Note: (1) At t = 0 , Q(t = 0) = EC(1− 1) = 0

(2) As t →∞ , Q(t →∞) = EC(1− 0) = EC= Final charge on capacitor (Q0)

(3) Current:

i =dQ

dt

= EC( 1

RC

)e−t/RC

i(t) =ER

e−t/RC

i(t = 0) =ER

= Initial current = i0

i(t →∞) = 0

(4) At time = 0, the capacitor acts like short circuit when there iszero charge on the capacitor.

(5) As time → ∞, the capacitor is fully charged and current = 0, itacts like a open circuit.


(6) τc = RC is called the time constant. It’s the time it takes forthe charge to reach (1− 1

e) Q0 ' 0.63Q0

(B) Discharging a charged capacitor:

Note: Direction of i is chosen so that the current represents the rate atwhich the charge on the capacitor is decreasing.

∴ i = −dQ

dt

Loop Rule:Vc − iR = 0

⇒ Q

C+

dQ

dtR = 0

⇒ dQ

dt= − 1

RCQ

Integrate both sides and use the initial condition:t = 0, Q on capacitor = Q0

ˆ Q

Q0

dQ

Q= − 1

RC

ˆ t

0

dt

⇒ ln Q− ln Q0 = − t

RC

⇒ ln( Q

Q0

)= − t

RC

⇒ Q

Q0

= e−t/RC

⇒ Q(t) = Q0 e−t/RC

(i = −dQ

dt) ⇒ i(t) =

Q0

RCe−t/RC

(At t = 0) ⇒ i(t = 0) =1

R· Q0

C︸︷︷︸Initial P.D. across capacitor

i0 =V0

R


At t = RC = τ Q(t = RC) =1

eQ0 ' 0.37Q0

Chapter 6

Magnetic Force

6.1 Magnetic Field

For stationary charges, they experienced an electric force in an electric field.For moving charges, they experienced a magnetic force in a magnetic field.

Mathematically, ~FE = q ~E (electric force)~FB = q~v × ~B (magnetic force)

Direction of the magnetic force determined from right hand rule.

Magnetic field ~B : Unit = Tesla (T)1T = 1C moving at 1m/s experiencing 1N

Common Unit: 1 Gauss (G) = 10−4T ≈ magnetic field on earth’s surface

Example: What’s the force on a 0.1C charge moving at velocity ~v = (10j −20k)ms−1 in a magnetic field ~B = (−3i + 4k)× 10−4T

~F = q~v × ~B

6.1. MAGNETIC FIELD 74

= +0.1 (10j − 20k)× (−3i + 4k)× 10−4N

= 10−5 (−30 · −k + 40i + 60j + 0)N

Effects of magnetic field is usually quite small.

~F = q~v × ~B

|~F | = qvB sin θ, where θ is the angle between ~v and ~B

∴ Magnetic force is maximum when θ = 90 (i.e. ~v ⊥ ~B)

Magnetic force is minimum (0) when θ = 0, 180 (i.e. ~v ‖ ~B)

Graphical representation of B-field: Magnetic field linesCompared with Electric field lines:

Similar characteristics :

(1) Direction of E-field/B-field indicated by tangent of the field lines.

(2) Magnitude of E-field/B-field indicated by density of the field lines.

Differeces :

(1) ~FE ‖ E-field lines; ~FB ⊥ B-field lines

(2) E-field line begins at positive charge and ends at negative charge; B-field line forms a closed loop.

Example : Chap35, Pg803 Halliday

Note: Isolated magnetic monopoles do not exist.

6.2. MOTION OF A POINT CHARGE IN MAGNETIC FIELD 75

6.2 Motion of A Point Charge in Magnetic Field

Since ~FB ⊥ ~v, therefore B-field only changes the direction of the velocity but notits magnitude.

Generally, ~FB = q~v × ~B = q v⊥B ,∴ We only need to consider the motioncomponent ⊥ to B-field.

We have circular motion. Magneticforce provides the centripetal force on themoving charge particles.

∴ FB = mv2

r

|q| vB = mv2

r

∴ r =mv

|q|Bwhere r is radius of circular motion.

Time for moving around one orbit:

T =2πr

v=

2πm

qBCyclotron Period

(1) Independent of v (non-relativistic)

(2) Use it to measure m/q

Generally, charged particles with con-stant velocity moves in helix in the pres-ence of constant B-field.

6.3. HALL EFFECT 76

Note :

(1) B-field does NO work on particles.

(2) B-field does NOT change K.E. of particles.

Particle Motion in Presence of E-field & B-field:

~F = q ~E + q~v × ~B Lorentz Force

Special Case : ~E ⊥ ~B

When |~FE| = |~FB|qE = qvB

⇒ v =E

B

∴ For charged particles moving at v = E/B, they will pass through thecrossed E and B fields without vertical displacement.⇒ velocity selector

Applications :

• Cyclotron (Lawrence & Livingston 1934)

• Measuring e/m for electrons (Thomson 1897)

• Mass Spectrometer (Aston 1919)

6.3 Hall Effect

Charges travelling in a conducting wire will be pushed to one side of the wire bythe external magnetic field. This separation of charge in the wire is called theHall Effect.

6.3. HALL EFFECT 77

The separation will stop when FB experienced by the current carrier is balancedby the force ~FH caused by the E-field set up by the separated charges.

Define :

∆VH = Hall Voltage

= Potential difference across the conducting strip

∴ E-field from separated charges: EH =∆VH

Wwhere W = width of conducting strip

In equilibrium: q ~EH + q~vd × ~B = 0, where ~vd is drift velocity

∴∆VH

W= vdB

Recall from Chapter 5,i = nqAvd

where n is density of charge carrier,A is cross-sectional area = width × thickness = W · t

∴∆VH

W=

i

nqWtB

⇒ n =iB

qt∆VH

To determine densityof charge carriers

Suppose we determine n for a particular metal (∴ q = e), then we can measureB-field strength by measuring the Hall voltage:

B =net

i∆VH

6.4. MAGNETIC FORCE ON CURRENTS 78

6.4 Magnetic Force on Currents

Current = many charges moving together

Consider a wire segment, length L,carrying current i in a magnetic field.

Total magnetic force = ( q~vd × ~B︸︷︷︸force on onecharge carrier

) · n A L︸︷︷︸Total number ofcharge carrier

Recall i = nqvdA

∴ Magnetic force on current ~F = i~L× ~B

where ~L = Vector of which: |~L| = length of current segment; direction =direction of current

For an infinitesimal wire segment d~l

d ~F = i d~l × ~B

Example 1: Force on a semicircle current loop

d~l = Infinitesimal

arc length element ⊥ ~B

∴ dl = R dθ

∴ dF = iRB dθ

By symmetry argument, we only need to consider vertical forces, dF · sin θ

∴ Net force F =

ˆ π

0

dF sin θ

= iRB

ˆ π

0

sin θ dθ

F = 2iRB (downward)


Method 2: Write d~l in i, j components

d~l = −dl sin θ i + dl cos θ j

= R dθ (− sin θ i + cos θ j)

~B = −B k (into the page)

∴ d~F = i d~l × ~B

= −iRB sin θ dθ j − iRB cos θ i

∴ ~F =

ˆ π

0

d~F

= −iRB

[ˆ π

0

sin θ dθ j +

ˆ π

0

cos θ dθ i

]

= −2iRB j

Example 2: Current loop in B-field

For segment2:

F2 = ibB sin(90 + θ) = ibB cos θ (pointing downward)

For segment4:

F2 = ibB sin(90 − θ) = ibB cos θ (pointing upward)


For segment1: F1 = iaBFor segment3: F3 = iaB∴ Net force on the current loop = 0But, net torque on the loop about O

= τ1 + τ3

= iaB · b

2sin θ + iaB · b

2sin θ

= i ab︸︷︷︸A = area of loop

B sin θ

Suppose the loop is a coil with N turns of wires:

Total torque τ = NiAB sin θ

Define: Unit vector n to represent the area-vector (using right hand rule)

Then we can rewrite the torque equation as

~τ = NiA n× ~B

Define: NiA n = ~µ = Magnetic dipole moment of loop

~τ = ~µ× ~B

Chapter 7

Magnetic Field

7.1 Magnetic Field

A moving charge

experiences magnetic force in B-field.

can generate B-field.

Magnetic field ~B due to moving point charge:

~B =µ0

4π· q~v × r

r2=

µ0

4π· q~v × ~r

r3

where µ0 = 4π × 10−7 Tm/A (N/A2)

Permeability of free space (Magnetic constant)

| ~B| = µ0

4π· qv sin θ

r2

maximum when θ = 90

minimum when θ = 0/180

~B at P0 = 0 = ~B at P1~B at P2 < ~B at P3

However, a single moving charge will NOT generate a steady magnetic field.stationary charges generate steady E-field.steady currents generate steady B-field.


Magnetic field at point P can beobtained by integrating the contribu-tion from individual current segments.(Principle of Superposition)

∴ d ~B =µ0

4π· dq ~v × r

r2

Notice: dq ~v = dq · d~s

dt= i d~s

d ~B =µ0

4π· i d~s× r

r2Biot-Savart Law

For current around a whole circuit:

~B =

ˆ

entirecircuit

d ~B =

ˆ

entirecircuit

µ0

4π· i d~s× r

r2

Biot-Savart Law is to magnetic field asCoulomb’s Law is to electric field.

Basic element of E-field: Electric charges dqBasic element of B-field: Current element i d~s

Example 1 : Magnetic field due to straight current segment


∴ |d~s× r| = dz sin φ

= dz sin(π − φ) (Trigonometry Identity)

= dz · d

r=

d · dz√d2 + z2

dB =µ0

4π· i dz

r2· d

r=

µ0i

4π· d

(d2 + z2)3/2dz

∴ B =

ˆ L/2

−L/2

dB =µ0id

4π

ˆ +L/2

−L/2

dz

(d2 + z2)3/2

B =µ0i

4πd· z

(z2 + d2)1/2

∣∣∣∣+L/2

−L/2

B =µ0i

4πd· L

(L2

4+ d2)1/2

Limiting Cases : When L À d (B-field due to long wire)

(L2

4+ d2

)−1/2 ≈(L2

4

)−1/2=

2

L

∴ B =µ0i

2πd;

direction of B-field determinedfrom right-hand screw rule

Recall : E =λ

2πε0dfor an infinite long line of charge.

Example 2 : A circular current loop


Notice that for every current element id~s1, generating a magnetic field d ~B1

at point P , there is an opposite current element id~s2, generating B-fieldd ~B2 so that

d ~B1 sin α = −d ~B2 sin α

∴ Only vertical component of B-field needs to be considered at point P .

dB =µ0

4π· i ds sin

∵d~s⊥r︷︸︸︷90

r2

∴ B-field at point P :

B =

ˆ

aroundcircuit

dB cos α︸︷︷︸consider verticalcomponent

∴ B =

2πˆ

0

µ0i cos α

4πr2· ds︸︷︷︸

Rdθ

=µ0i

4π· R

r3

ˆ 2π

0

ds

︸︷︷︸Integrate around circum-ference of circle = 2πR

∴ B =µ0iR

2

2r3

B =µ0iR

2

2(R2 + z2)3/2;

direction of B-field determinedfrom right-hand screw rule

Limiting Cases :

(1) B-field at center of loop:

z = 0 ⇒ B =µ0i

2R

(2) For z À R,

B =µ0iR

2

2z3(1 + R2

z2

)3/2≈ µ0iR

2

2z3∝ 1

z3

Recall E-field for an electric dipole: E =p

4πε0x3

∴ A circular current loop is also called a magnetic dipole.


(3) A current arc:

B =

ˆ

aroundcircuit

dB cos α︸︷︷︸z = 0 ⇒α = 0 here.

=µ0i

4π· R

r3︸︷︷︸R = r

when α = 0

·

Rθ = length of arc︷︸︸︷ˆ θ

0

ds︸︷︷︸Rdθ

B =µ0i θ

4πR

Example 3 : Magnetic field of a solenoidSolenoid is used to produce a strong and uniform magnetic field inside itscoils.

Consider a solenoid of length L consisting of N turns of wire.

Define: n = Number of turns per unit length =N

L

Consider B-field at distance d from thecenter of the solenoid:For a segment of length dz, number ofcurrent turns = ndz

∴ Total current = ni dz

7.2. PARALLEL CURRENTS 86

Using the result from one coil in Example 2, we get B-field from coils oflength dz at distance z from center:

dB =µ0(ni dz)R2

2r3

However r =√

R2 + (z − d)2

∴ B =

ˆ +L/2

−L/2

dB(Integrating over theentire solenoid)

=µ0niR2

2

ˆ +L/2

−L/2

dz

[R2 + (z − d)2]3/2

B =µ0ni

2

L2

+ d√R2 + (L

2+ d)2

+L2− d√

R2 + (L2− d)2

along negative z direction

Ideal Solenoid :

L À R

then B =µ0ni

2[1 + 1]

∴ B = µ0ni ;direction of B-field determinedfrom right-hand screw rule

Question : What is the B-field at the end of an ideal solenoid? B=µ0ni

2

7.2 Parallel Currents

Magnetic field at point P ~B due to twocurrents i1 and i2 is the vector sum ofthe ~B fields ~B1, ~B2 due to individual cur-rents. (Principle of Superposition)

7.2. PARALLEL CURRENTS 87

Force Between Parallel Currents :

Consider a segment of length L on i2 :

~B1 =µ0i12πd

(pointing down) ~B2 =µ0i12πd

(pointing up)

Force on i2 coming from i1:

|~F21| = i2~L× ~B1 =µ0Li1i2

2πd= |~F12| (Def’n of ampere, A)

∴ Parallel currents attract, anti-parallel currents repel.

Example : Sheet of current

Consider an infinitesimal wire of width dx at position x, there exists anotherelement at −x so that vertical ~B-field components of ~B+x and ~B−x cancel.∴ Magnetic field due to dx wire:

dB =µ0 · di

2πrwhere di = i

(dx

a

)

∴ Total B-field (pointing along −x axis) at point P :

B =

+a/2ˆ

−a/2

dB cos θ =

+a/2ˆ

−a/2

µ0i

2πa· dx

r· cos θ

7.3. AMPERE’S LAW 88

Variable transformation (Goal: change r, x to d, θ, then integrate over θ):

d = r cos θ ⇒ r = d sec θx = d tan θ ⇒ dx = d sec2 θ dθ

Limits of integration: −θ0 to θ0, where tan θ0 =a

2d

∴ B =µ0i

2πa

ˆ θ0

−θ0

d sec2 θ dθ

d sec θ· cos θ

=µ0i

2πa

ˆ θ0

−θ0

dθ

B =µ0iθ0

πa=

µ0i

πatan−1

( a

2d

)

Limiting Cases :

(1) d À a

tan θ =a

2d⇒ θ ≈ a

2d

∴ B =µ0i

2πaB-field due toinfinite long wire

(2) d ¿ a (Infinite sheet of current)

tan θ =a

2d→ ∞ ⇒ θ =

π

2

∴ B =µ0i

2aConstant!

Question : Large sheet of opposite flowing currents.

What’s the B-field between & outside the sheets?

7.3 Ampere’s Law

In our study of electricity, we notice that the inverse square force law leadsto Gauss’ Law, which is useful for finding E-field for systems with high level ofsymmetry.For magnetism, Gauss’ Law is simple


‹

S

˛

S

~B · d ~A = 0∵ There is no mag-netic monopole.

A more useful law for calculating B-field for highly symmetric situations is theAmpere’s Law:

˛

C

˛

C

~B · d~s = µ0i

˛

C

= Line intefral evaluated around a closed loop C (Amperian curve)

i = Net current that penetrates the area bounded by curve C∗ (topological property)

Convention : Use the right-hand screw rule to determine the sign of current.

˛

C

~B · d~s = µ0(i1 − i3 + i4 − i4)

= µ0(i1 − i3)

Applications of the Ampere’s Law :

(1) Long-straight wire

Construct an Amperiancurve of radius d:

By symmetry argument, we know ~B-field only has tangential compo-nent

∴˛

C

~B · d~s = µ0i


Take d~s to be the tangential vector around the circular path:

∴ ~B · d~s = B ds

B

˛

C

ds

︸︷︷︸Circumferenceof circle = 2πd

= µ0i

∴ B(2πd) = µ0i

B-field due to long,straight current B =

µ0i

2πd(Compare with 7.1 Example 1)

(2) Inside a current-carrying wire

Again, symmetry argumentimplies that ~B is tangentialto the Amperian curve and~B → B(r)θ

Consider an Amperian curve of radius r(< R)˛

C

~B · d~s = B

˛ds = B(2πr) = µ0iincluded

But iincluded ∝ cross-sectional area of C

∴iincluded

i=

πr2

πR2

∴ iincluded =r2

R2i

∴ B =µ0i

2πR2· r ∝ r

Recall: Uniformly charged infinite long rod

(3) Solenoid (Ideal)

Consider the rectangularAmperian curve 1234.


˛

C

~B · d~s =

ˆ

1

~B · d~s +

½½

½½½

ˆ

2

~B · d~s +

½½

½½½

ˆ

3

~B · d~s +

½½

½½½

ˆ

4

~B · d~s

ˆ

2

=

ˆ

4

= 0 ∵

~B · d~s = 0 inside solenoid~B = 0 outside solenoid

ˆ

3

= 0 ∵ ~B = 0 outside solenoid

∴˛

C

~B · d~s =

ˆ

1

~B · d~s = Bl = µ0itot

But itot = nl︸︷︷︸Number of coils included

·i

∴ B = µ0ni

Note :

(i) The assumption that ~B = 0 outside the ideal solenoid is onlyapproximate. (Halliday, Pg.763)

(ii) B-field everywhere inside the solenoid is a constant. (for idealsolenoid)

(4) Toroid (A circular solenoid)

By symmetry argument, the B-field lines form concentric circles insidethe toroid.Take Amperian curve C to be a circle of radius r inside the toroid.

˛

C

~B · d~s = B

˛

C

ds = B · 2πr = µ0(Ni)

∴ B =µ0Ni

2πrinside toroid

7.4. MAGNETIC DIPOLE 92

Note :

(i) B 6= constant inside toroid

(ii) Outside toroid:Take Amperian curve to be circle of radius r > R.

˛

C

~B · d~s = B

˛

C

ds = B · 2πr = µ0 · iincl = 0

∴ B = 0

Similarly, in the central cavity B = 0

7.4 Magnetic Dipole

Recall from §6.4, we define the magnetic dipole moment of a rectangularcurrent loop

~µ = NiAn

where n = area unit vector with directiondetermined by the right-hand rule

N = Number of turns in current loop

A = Area of current loop

This is actually a general definition of a magnetic dipole, i.e. we use it forcurrent loops of all shapes.

A common and symmetric example: circular current.

Recall from §7.1 Example 2, magneticfield at point P (height z above the ring)

~B =µ0iR

2n

2(R2 + z2)3/2=

µ0~µ

2π(R2 + z2)3/2

7.5. MAGNETIC DIPOLE IN A CONSTANT B-FIELD 93

At distance z À R,

~B =µ0~µ

2πz3~E =

~p

4πε0z3

due to magnetic dipole due to electric dipole(for z À R) (for z À d)

Also, notice ~µ = magnetic dipole moment

[Unit: Am2

J/T

]

µ0 = Permeability of free space

= 4π × 10−7Tm/A

7.5 Magnetic Dipole in A Constant B-field

In the presence of a constant magnetic field, we have shown for a rectangular

current loop, it experiences a torque ~τ = ~µ× ~B . It applies to any magneticdipole in general.

7.6. MAGNETIC PROPERTIES OF MATERIALS 94

∴ External magnetic field aligns the magneticdipoles.

Similar to electric dipole in a E-field, we can con-sider the work done in rotating the magnetic di-pole. (refer to Chapter 2)

dW = −dU, where U is potential energy of dipole

U = −~µ · ~B

Note :

(1) We cannot define the potential energy of a magnetic field in general.However, we can define the potential energy of a magnetic dipole in aconstant magnetic field.

(2) In a non-uniform external B-field, the magnetic dipole will experiencea net force (not only net torque)

7.6 Magnetic Properties of Materials

Recall intrinsic electric dipole in molecules:

Intrinsic dipole (magnetic) in atoms:

In our classical model of atoms, electronsrevolve around a positive nuclear.

∴ ”Current” i =e

P, where P is period of one orbit around nucleus


P =2πr

v, where v is velocity of electron

∴ Orbit magnetic dipole of atom:

µ = iA =( ev

2πr

)(πr2) =

erv

2

Recall: angular momentum of rotation l = mrv

∴ µ =e

2m· l

In quantum mechanics, we know that

l is quantized, i.e. l = N · h

2π

where N = Any positive integer (1,2,3, ... )

h = Planck’s constant (6.63× 10−34J · s)∴ Orbital magnetic dipole moment

µl =eh

4mπ︸︷︷︸Bohr’s magneton µB=9.27×10−24J/T

·N

There is another source of intrinsic magnetic dipole moment inside an atom:

Spin dipole moment: coming from the intrinsic ”spin” of electrons.

Quantum mechanics suggests that e− are always spinning and it’s either an ”up”spin or a ”down” spin

µe = 9.65× 10−27 ≈ µB

So can there be induced magnetic dipole?


Recall: for electric field

Edielectric = KeEvacuum ; Ke ≥ 1

For magnetic field in a material:

~Bnet = ~B0 + ~BM

↑ ↑appliedB-field

B-field producedby induced dipoles

In many materials (except ferromagnets),

~BM ∝ ~B0

Define :~BM = χm

~B0

χm is a number called magnetic susceptibility.

∴ ~Bnet = ~B0 + χm~B0

= (1 + χm) ~B0

~Bnet = κm~B0 ; κm = 1 + χm

Define : κm is a number called relative permeability.

One more term ......

Define : the Magnetization of a material:

~M =d~µ

dVwhere ~µ is magnetic dipolemoment, V is volume

(or, the net magnetic dipole moment per unit volume)

In most materials (except ferromagnets),

~BM = µ0~M

Three types of magnetic materials:

(1) Paramagnetic:

κm ≥ 1(χm ≥ 0)

,induced magnetic dipoles alignedwith the applied B-field.


e.g. Al (χm.= 2.2× 10−5), Mg (1.2× 10−5), O2 (2.0× 10−6)

(2) Diamagnetic:

κm ≤ 1(χm ≤ 0)

,induced magnetic dipoles alignedopposite with the applied B-field.

e.g. Cu (χm ≈ −1× 10−5), Ag (−2.6× 10−5), N2 (−5× 10−9)

(3) Ferromagnetic:

e.g. Fe, Co, NiMagnetization not linearly proportionalto applied field.

⇒ Bnet

Bapp

not a constant (can be as

big as ∼ 5000− 100, 000)

Interesting Case : Superconductors

χm = −1

A perfect diamagnetic.NO magnetic field inside.

Chapter 8

Faraday’s Law of Induction

8.1 Faraday’s Law

In the previous chapter, we have shown that steady electric current can givesteady magnetic field because of the symmetry between electricity & magnetism.We can ask: Steady magnetic field can give steady electric current. ×

OR Changing magnetic field can give steady electric current. X

Define :

(1) Magnetic flux through surface S:

Φm =

ˆ

S

~B · d ~A

Unit of Φm : Weber (Wb)1Wb = 1Tm2

(2) Graphical:

Φm = Number of magnetic field lines passing through surface S

Faraday’s law of induction:

Induced emf |E| = N

∣∣∣∣∣dΦm

dt

∣∣∣∣∣

where N = Number of coils in the circuit.

8.2. LENZ’ LAW 99

~B = Constant ~B = Constant B = Constant ~B = Constant~A = Constant A = Constant dB/dt 6= 0 A = Constant

dA/dt 6= 0 ~A = Constant dA/dt 6= 0E = 0 ∴ |E| > 0 ∴ |E| > 0 ∴ |E| > 0

Note : The induced emf drives a current throughout the circuit, similar to thefunction of a battery. However, the difference here is that the induced emfis distributed throughout the circuit. The consequence is that we cannotdefine a potential difference between any two points in the circuit.

Suppose there is an induced current in the loop, can wedefine ∆VAB?

Recall:

∆VAB = VA − VB = iR > 0

⇒ VA > VB

Going anti-clockwise (same as i),

If we start from A, going to B, then we get VA > VB.If we start from B, going to A, then we get VB > VA.

∴ We cannot define ∆VAB !!

This situation is like when we study the interior of a battery.

A battery

The loop

provides the energy needed to drive thecharge carriers around the circuit by

chemical reactions.

changing magnetic flux.

sources of emf non-electric means

8.2 Lenz’ Law

(1) The flux of the magnetic field due to induced current opposes the changein flux that causes the induced current.

8.3. MOTIONAL EMF 100

(2) The induced current is in such a direction as to oppose the changes thatproduces it.

(3) Incorporating Lentz’ Law into Faraday’s Law:

E = −NdΦm

dt

IfdΦm

dt> 0, Φm ↑ ⇒ E appears ⇒ Induced current

appears.

⇒ ~B-field due toinduced current

⇒ change in Φmso that=⇒ Φm ↓

(4) Lenz’ Law is a consequence from the principle of conservation of energy.

8.3 Motional EMF

Let’s try to look at a special case when the changing magnetic flux is carried bymotion in the circuit wires.

Consider a conductor of length L movingwith a velocity v in a magnetic field ~B.


Hall Effect for the charge carriers in the rod:

~FE + ~FB = 0

⇒ q ~E + q~v × ~B = 0 (where ~E is Hall electric field)

⇒ ~E = −~v × ~B

Hall Voltage inside rod:

∆V = −ˆ L

0

~E · d~s∆V = −EL

∴ Hall Voltage : ∆V = vBL

Now, suppose the moving wire slides withoutfriction on a stationary U-shape conductor.The motional emf can drive an electric cur-rent i in the U-shape conductor.⇒ Power is dissipated in the circuit.⇒ Pout = V i (Joule’s heating)(see Lecture Notes Chapter 4)

What is the source of this power?Look at the forces acting on the conducting rod:

• Magnetic force:

~Fm = i~L× ~B

Fm = iLB (pointing left)

• For the rod to continue to move at constant velocity v, we need to applyan external force:

~Fext = −~Fm = iLB (pointing right)

∴ Power required to keep the rod moving:

Pin = ~Fext · ~v= iBLv

= iBLdx

dt

= iBd(xL)

dt( xL = A, areaenclosed by circuit)

= id(BA)

dt( BA = Φm, magnetic flux)


Since energy is not being stored in the system,

∴ Pin + Pout = 0

iV + idΦm

dt= 0

We ”prove” Faraday’s Law ⇒ V = −dΦm

dt

Applications :

(1) Eddy current: moving conductors in presence of magnetic field

Induced current

⇒ Power lost in Joule’s heating(E2

R

)

⇒ Extra power input to keep moving

To reduce Eddy currents:

(2) Generators and Motors:Assume that the circuit loop is rotating at a constant angular velocityω, (Source of rotation, e.g. steam produced by burner, water fallingfrom a dam)


Magnetic flux through the loop

Number of coils↓

ΦB = N´

loop

~B · d ~A = NBA cos θ↓

changes with time! θ = ωt

∴ ΦB = NBA cos ωt

Induced emf: E = −dΦB

dt= −NBA

d

dt(cos ωt)

= NBAω sin ωt

Induced current: i =ER

=NBAω

Rsin ωt

Alternating current (AC) voltage generator

Power has to be provided by the source of rotation to overcome thetorque acting on a current loop in a magnetic field.

~τ =

~µ︷︸︸︷Ni ~A× ~B

∴ τ = NiAB sin θ

8.4. INDUCED ELECTRIC FIELD 104

The net effect of the torque is to oppose the rotation of the coil.

An electric motor is simply a generatoroperating in reverse.⇒ Replace the load resistance R witha battery of emf E .

With the battery, there is a current in the coil, and it experiences atorque in the B-field.⇒ Rotation of the coil leads to an induced emf, Eind, inthe direction opposite of that of the battery. (Lenz’ Law)

∴ i =E − Eind

R

⇒ As motor speeds up, Eind ↑, ∴ i ↓∴ mechanical power delivered = torque delivered = NiAB sin θ ↓In conclusion, we can show that

Pelectric = i2R + Pmechanical

Electric power input Mechanical power delivered

8.4 Induced Electric Field

So far we have discussed that a change in mag-netic flux will lead in an induced emf distributedin the loop, resulting from an induced E-field.

However, even in the absence of the loop (so that there is no induced current),the induced E-field will still accompany a change in magnetic flux.


∴ Consider a circular path in a regionwith changing magnetic field.

The induced E-field only has tangential components. (i.e. radial E-field = 0)Why?

Imagine a point charge q0 travelling around the circular path.Work done by induced E-field = q0Eind︸︷︷︸

force

· 2πr︸︷︷︸distance

Recall work done also equals to q0E , where E is induced emf

∴ E = Eind2πr

Generally,

E =

˛~Eind · d~s

where¸

is line integral around a closed loop, ~Eind is induced E-field, ~s istangential vector of path.∴ Faraday’s Law becomes

˛

C

~Eind · d~s = − d

dt

ˆ

S

~B · d ~A

L.H.S. = Integral around a closed loop CR.H.S. = Integral over a surface bounded by C

Direction of d ~A determined by direction of line integration C (Right-Hand Rule)


”Regular” E-field Induced E-field

created by charges created by changing B-field

E-field lines start from +ve and endon −ve charge

E-field lines form closed loops

can define electric potential so thatwe can discuss potential differencebetween two points

Electric potential cannot be defined(or, potential has no meaning)

⇓ ⇓Conservative force field Non-conservative force field

The classification of electric and magnetic effects depend on the frame of referenceof the observer. e.g. For motional emf, observer in the reference frame of themoving loop, will NOT see an induced E-field, just a ”regular” E-field.(Read: Halliday Chap.33-6, 34-7)

Chapter 9

Inductance

9.1 Inductance

An inductor stores energy in the magnetic field just as a capacitor stores energyin the electric field.

We have shown earlier that a changing B-field will lead to an induced emf ina circuit.

Question : If a circuit generates a changing magnetic field, does it lead to aninduced emf in the same circuit? YES! Self-Inductance

The inductance L of any current element is

EL = ∆VL = −Ldi

dtThe negative signcomes from Lenz Law.

Unit of L: Henry(H) 1H=1·VsA

• All circuit elements (including resistors) have some inductance.

• Commonly used inductors: solenoids, toroids

• circuit symbol:

Example : Solenoid

EL = VB − VA = −Ldi

dt< 0 EL = VB − VA = −L

di

dt> 0

∴ VB < VA VB > VA

9.1. INDUCTANCE 108

Recall Faraday’s Law,

EL = −NdΦB

dt= − d

dt(NΦB)

where ΦB is magnetic flux, NΦB is flux linkage.∴ Alternative definition of Inductance:

− d

dt(NΦB) = −L

di

dt⇒ L =

NΦB

i

∴ Inductance is also flux linkage per unit current.

Calculating Inductance:

(1) Solenoid:To first order approximation,

B = µ0ni

where n = N/L = Number ofcoils per unit length.

Consider a subsection of length l of the solenoid:

Flux linkage = N ΦB

= nl ·BA where A iscross-sectional area

∴L =

NΦB

i= µ0n

2lA

L

l= µ0n

2A = Inductance per unit length

Notice :

(i) L ∝ n2

(ii) The inductance, like the capacitance, depends only on geometricfactors, not on i.

9.1. INDUCTANCE 109

(2) Toroid:

Recall: B-field lines are concentric cir-cles.Inside the toroid:

B =µ0iN

2πr(NOT a constant)

where r is the distance from center.Outside the toroid:

B = 0

Flux linkage through the toroid

NΦB = N

ˆ~B · d~a

Notice ~B ‖ d~aWrite da = h dr

⟩KEY

=µ0iN

2

2π

ˆ b

a

h dr

r

=µ0iN

2h

2πln

( b

a

)

∴ Inductance L =NΦB

i=

µ0N2h

2πln

( b

a

)

Again, L ∝ N2

Inductance with magnetic materials :We showed earlier that for capacitors:

~E → ~E/κe

C → κeC(after insertion ofdielectric κe > 1)

For inductors, we first know that

~B → κm~B

(after insertion ofmagnetic material)

Inductance L =NΦB

i

However ΦB =

ˆ~B · d ~A → κmΦB

9.2. LR CIRCUITS 110

∴ L → κmL(after insertion ofmagnetic material)

∴ To increase inductance, fill the interior of inductor with ferromagneticmaterials. (×103 − 104)

9.2 LR Circuits

(A) ”Charging” an inductor

When the switch is adjusted to position a,By loop rule (clockwise) :

E0 − ∆VR + ∆VL = 0↓ ↓

E0 − iR − Ldi

dt= 0

∴di

dt+

R

Li =

E0

LFirst Order Differ-ential Equation

Similar to the equation for charging a capacitor! (Chap5)

Solution: i(t) =E0

R

(1− e−t/τL

)

where τL = Inductive time constant = L/R

∴ |∆VR| = iR = E0(1− e−t/τL)

|∆VL| = Ldi

dt= L · E0

R· 1

τL

· e−t/τL = E0 e−t/τL

9.2. LR CIRCUITS 111

(B) ”Discharging” an inductorWhen the switch is adjusted at position b after the inductor has been”charged” (i.e. current i = E0/R is flowing in the circuit.).

By loop rule:

∆VL − ∆VR = 0↓ ↓

−Ldi

dt− iR = 0

(Treat inductor as source of emf)

∴di

dt+

R

Li = 0

Discharging a capacitor(Chap5)

i(t) = i0 e−t/τL

where i0 = i(t = 0) = Current when the circuit just switch to position b.

Summary : During charging of inductor,

1. At t = 0, inductor acts like open circuit when current flowing is zero.

2. At t → ∞, inductor acts like short circuit when current flowing isstablized at maximum.

3. Inductors are used everyday in switches for safety concerns.

9.3. ENERGY STORED IN INDUCTORS 112

9.3 Energy Stored in Inductors

Inductors stored magnetic energy through the magnetic field stored in the circuit.Recall the equation for charging inductors:

E0 − iR− Ldi

dt= 0

Multiply both sides by i :

E0i︸︷︷︸Power input by emf(Energy supplied toone charge = qE0)

= i2R︸︷︷︸Joule’s heating(Power dissipatedby resistor)

+ Lidi

dt︸︷︷︸Power storedin inductor

∴ Power stored in inductor =dUB

dt= Li

di

dtIntegrating both sides and use initial condition

At t = 0, i(t = 0) = UB(t = 0) = 0

∴ Energy stored in inductor: UB =1

2Li2

Energy Density Stored in Inductors :Consider an infinitely long solenoid of cross-sectional area A.For a portion l of the solenoid, we know from §8.1,

L = µ0n2lA

∴ Energy stored in inductor:

UB =1

2Li2 =

1

2µ0n

2i2 lA︸︷︷︸Volume ofsolenoid

∴ Energy density (= Energy stored per unit volume) inside inductor:

uB =UB

lA=

1

2µ0n

2i2

Recall magnetic field inside solenoid (Chap7)

B = µ0ni

∴ uB =B2

2µ0

This is a general result of the energy stored in a magnetic field.

9.4. LC CIRCUIT (ELECTROMAGNETIC OSCILLATOR) 113

9.4 LC Circuit (Electromagnetic Oscillator)

Initial charge on capacitor = QInitial current = 0No battery.

Assume current i to be in the direction that increases charge on the positivecapacitor plate.

⇒ i =dQ

dt(9.1)

By Lenz Law, we also know the ”poles” of the inductor.

Loop rule: VC + VL = 0

−Q

C− L

di

dt= 0 (9.2)

Combining equations (9.1) and (9.2), we get

d2Q

dt2+

1

LCQ = 0

This is similar to the equation of motionof a simple harmonic oscillator:

d2x

dt2+

k

mx = 0

Another approach (conservation of energy)Total energy stored in circuit:

U = UE + UB

↓ ↓U =

Q2

2C+

1

2Li2

Since the resistance in the circuit is zero, no energy is dissipated in the circuit.∴ Energy contained in the circuit is conserved.

∴dU

dt= 0

⇒ Q

C·¶

¶¶dQ

dt+ L¢i

di

dt= 0 (∵ i =

dQ

dt)

9.4. LC CIRCUIT (ELECTROMAGNETIC OSCILLATOR) 114

⇒ Ldi

dt+

Q

C= 0

⇒ d2Q

dt2+

1

LCQ = 0

The solution to this differential equation is in the form

Q(t) = Q0 cos(ωt + φ)

∴dQ

dt= −ωQ0 sin(ωt + φ)

d2Q

dt2= −ω2Q0 cos(ωt + φ)

= −ω2Q

∴d2Q

dt2+ ω2Q = 0

∴ ω2 =1

LCAngular frequencyof the LC oscillator

Also, Q0, φ are constants derived from the initial conditions. (Two initial condi-

tions, e.g. Q(t = 0), and i(t = 0) = dQdt

∣∣∣t=0

are required.)

Energy stored in C =Q2

2C=

Q20

2Ccos2(ωt + φ)

Energy stored in L =1

2Li2 =

1

2Lω2Q2

0 sin2(ωt + φ)

∵ Lω2 =1

C=

Q20

2Csin2(ωt + φ)

∴ Total energy stored =Q2

0

2C= Initial energy stored in capacitor

9.5. RLC CIRCUIT (DAMPED OSCILLATOR) 115

9.5 RLC Circuit (Damped Oscillator)

In real life circuit, there’s always resistance.In this case, energy stored in the LC oscillator isNOT conserved,

anddU

dt= Power dissipated in the resistor = −i2R (Joule’s heating)

Negative sign shows that energy U is decreasing.

∴ Lidi

dt+

Q

C·

i︷︸︸︷dQ

dt= −i2R

⇒ d2Q

dt2+

R

L· dQ

dt+

1

LCQ = 0

This is similar to the equation of motion of a damped harmonic oscillator (e.g.

if a mass-spring system faces a frictional force ~F = −b~v).Solution to the equation is in the form Q(t) = eλt

If damping is not too big (i.e. R not too big), solution would become

Q(t) = Q0 e−R2L

t︸︷︷︸

exponentialdecay term

cos(ω1t + φ)︸︷︷︸oscillatingterm

where ω21 =

1

LC−

( R

2L

)2

ω21 = ω2 −

( R

2L

)2

Damped oscillator always oscillatesat a lower frequency than thenatural frequency of the oscillator.(Refer to Halliday, Vol1, Chap17 formore details.)

Check this at home: What is UE(t) + UB(t) for the case when damping is small?(i.e. R ¿ ω)

Chapter 10

AC Circuits

10.1 Alternating Current (AC) Voltage

Recall that an AC generator described in Chapter 9 generates a sinusoidal emf.

i.e. E = Em sin(ωt + δ)

Note :

This circuit is the RLC circuit with oneadditional element : the time varying ACpower supply. This is similar to a driven(damped) oscillator.

Ld2Q

dt2+ R

dQ

dt+

1

CQ = Em sin(ωt + δ)

The general solution consists of two parts:

transient : rapidly dies away in a few cycles (not interesting)

steady state : Q(t), i(t) varies sinusoidally with the same frequency as input

Note : Current does NOT vary at frequency ω21 =

1

LC−

( R

2L

)2

Since we only concern about the steady state solution, therefore we can take anytime as starting reference time = 0For convenience, we can write

E = Em sin ωt

And we can write

i = im sin(ωt− φ)

where im is current amplitude, φ is phase constant.Our goal is to determine im and φ.

10.2. PHASE RELATION BETWEEN I, V FOR R,L AND C 117

10.2 Phase Relation Between i, V for R,L and C

(A) Resistive Element

∆VR = VA − VB = iR

∴ ∆VR = imR sin(ωt− φ)

∆VR and i are in phase, i.e. what’sinside the ”sine bracket” (phase) is thesame for ∆VR and i.

Graphically, we introduce phasor diagrams properties of phasors:

(1) Length of a phasor is proportional to the maximum value.

(2) Projection of a phasor onto the vertical axis gives instantaneous value.

(3) Convention: Phasors rotate anti-clockwise in a uniform circular mo-tion with angular velocity.

∴ ∆VR = (∆VR)m sin(ωt− φ)

(∆VR)m = imR ”Ohm’s Law like” rela-tion for AC resistor

10.2. PHASE RELATION BETWEEN I, V FOR R,L AND C 118

(B) The Inductive Element

Potential drop across inductor

∆VL = VA − VB = −EL = Ldi

dt

∴ ∆VL = Limω cos(ωt− φ)

= Limω sin(ωt− φ +π

2) [∵ cos θ = sin(θ +

π

2)]

= imXL sin(ωt− φ +π

2)

(∆VL)m = imXL”Ohm’s Law like” rela-tion for AC inductor

where XL = Inductive Reactance

XL = ωL

As i ↑, VA > VB ∴ ∆VL > 0i ↓, VA < VB ∴ ∆VL < 0

∆VL leads i byπ

2

i lags ∆VL byπ

2

(C) Capacitive Element

∆VC = VA − VB =Q

C

10.3. SINGLE LOOP RLC AC CIRCUIT 119

where Q = charge on the positive plate of the capacitor.

∴ i =dQ

dt⇒ Q =

ˆi dt

=

ˆim sin(ωt− φ) dt

= −imω

cos(ωt− φ)

∴ ∆VC = − imωC

cos(ωt− φ)

= imXC sin(ωt− φ− π

2) [∵ − cos θ = sin(θ − π

2)]

∴ (∆VC)m = imXC”Ohm’s Law like” rela-tion for AC capacitor

where XC =1

ωC= Capacitive Reactance

∆VC lags i byπ

2

i leads ∆VC byπ

2

10.3 Single Loop RLC AC Circuit

Given that E = Em sin ωt, we want tofind im and φ so that we can write i =im sin(ωt− φ)

Loop rule: E −∆VR −∆VL −∆VC = 0

⇒ E = ∆VR + ∆VL + ∆VC

10.3. SINGLE LOOP RLC AC CIRCUIT 120

Using results from the previous section, we can write

Em sin ωt = imR sin(ωt− φ)

+imXL cos(ωt− φ)− imXC cos(ωt− φ)

Em sin ωt = im[R sin(ωt− φ) + (XL −XC) cos(ωt− φ)

]

Answer :

1. Take tan φ =XL −XC

R

2. Define Z =√

R2 + (XL −XC)2

as the impedance of the circuit.

3. Then

im =Em

Zor Em = imZ ”Ohm’s Law like” relation

for AC RLC circuits

Check :

R.H.S. = imZ[R

Zsin(ωt− φ) +

XL −XC

Zcos(ωt− φ)

]

= imZ[cos φ sin(ωt− φ) + sin φ cos(ωt− φ)

]

Use the relation:sin(A + B) = sin A cos B + cos A sin BHere: A = ωt− φ, B = φ

= imZ sin(ωt− φ + φ)

= imz sin ωt

= L.H.S. if Em = imZ QED.

Phasor Approach :

10.4. RESONANCE 121

10.4 Resonance

im =Em

Zis at maximum for an AC circuit of fixed input frequency ω when Z

is at minimum.

Z =√

R2 + (XL −XC)2 =

√R2 +

(ωL− 1

ωC

)2

is at a minimum for a fixed ω when

XL −XC = ωL− 1

ωC= 0

⇒ ωL =1

ωC

⇒ ω2 =1

LCsame as that fora RLC circuit

In Hong Kong, the AC power input is 50Hz.(In US, as mentioned in Halliday, is 60Hz.)

∴ ω = 2πf = 314.2s−1

10.5 Power in AC Circuits

Consider the Power dissipated by R in an AC circuit:

P = i2R = i2mR sin2(ωt− φ)

The average power dissipated in each cycle:

Pave =

´ 2π/ω

0P dt

2π/ω(2π

ωis period of each cycle)

ˆ 2π/ω

0

P dt = i2mR

ˆ 2π/ω

0

sin2(ωt− φ) dt

= i2mR

ˆ 2π/ω

0

1

2

[1− cos 2(ωt− φ)

]dt

= i2mR ·[ t

2−

©©©©©©©sin2(ωt− φ)

4ω

]∣∣∣∣2π/ω

0

= i2mR · 1

2· 2π

ω

10.5. POWER IN AC CIRCUITS 122

∴ Pave =i2m2

R = i2rmsR

where irms = root-mean-square current

irms =im√

2∵ Current is asinusoidal func.

Symbol : 〈P 〉 = Pave = Average of P over time

For sine and cosine functions of time:

Average : 〈sin ωt〉 = 〈cos ωt〉 = 0

Amplitude : Peak value, e.g. Em, im, (∆VR)m, · · ·Root-Mean-Square(RMS) : It’s a measure of the ”time-averaged” deviation

from zero.xrms =

√〈x2〉

For sines and cosines, for whatever quantity x:

xrms =xm√

2(xm is amplitude)

For an AC resistor circuit:

〈P 〉 = i2rmsR =E2

rms

R

Laws for DC circuits can be used to describe AC circuits if we use rms valuesfor i and E .

For general AC circuits:

P = Ei =

E︷︸︸︷Em sin ωt ·

i︷︸︸︷im sin(ωt− φ)

= Emim sin ωt [sin ωt cos φ− cos ωt sin φ]

P = Emim [ sin2 ωt︸︷︷︸12

cos φ− sin ωt cos ωt︸︷︷︸0

(check this!)

sin φ ]

〈P 〉 =Emim

2cos φ

〈P 〉 = Ermsirms cos φ︸︷︷︸power factor

10.6. THE TRANSFORMER 123

Recall tan φ =XL −XC

R

∴ cos φ =R

Z

Maximum power dissipated in circuit when

cos φ = 1

Two possibilities:

(1) XL = XC = 0

(2) XL −XC = 0 ⇒ XL = XC ⇒ ωL =1

ωC⇒ ω2 =

1

LC(Resonance Condition)

10.6 The Transformer

Power dissipated in resistor〈P 〉 = i2rmsR

∴ For power transmission, we’d like to keep irms at minimum.⇒ HIGH potential difference across transmission wires. (So that total powertransmitted P = irmsErms is constant.)However, for home safety, we would like LOW emf supply.

Solution : Transformers

Primary : Number of winding = NP

10.6. THE TRANSFORMER 124

Secondary : Number of winding = NS

In primary circuit, RP ≈ CP ≈ 0

∴ Pure inductive

Power factor : cos φ =R

Z≈ 0

∴ No power delivered from emf to transformer.

The varying current (∵ AC!) in the primary produces an induced emf in thesecondary coils. Assuming perfect magnetic flux linkage:

emf per turn in primary

= emf per turn in secondary

= −dΦB

dt

emf per turn in primary =∆VP

NP

(∆VP is P.D.across primary)

emf per turn in secondary =∆VS

NS

⇒ ∆VP

∆VS

=NP

NS

If NP > NS , then ∆VP > ∆VS Step-DownIf NP < NS , then ∆VP < ∆VS Step-Up

Consider power in circuit:iP ∆VP = iS∆VS

In the secondary, we have∆VS = iSR

Combining the 3 equations, we have

∆VP =(NP

NS

)2R · iP

”Equivalence Resistor” =(NP

NS

)2R

Chapter 11

Displacement Current andMaxwell’s Equations

11.1 Displacement Current

We saw in Chap.7 that we can useAmpere’s law to calculate magneticfields due to currents.We know that the integral

¸C

~B · d~saround any close loop C is equal toµ0iincl, where iincl = current passing anarea bounded by the closed curve C.

e.g.

= Flat surface bounded by loop C

= Curved surface bounded by loop C

If Ampere’s law is true all the time, then the iincl determined should be inde-pendent of the surface chosen.

11.1. DISPLACEMENT CURRENT 126

Let’s consider a simple case: charging acapacitor.From Chap.5, we know there is a currentflowing i(t) = E′

Re−t/RC , which leads

to a magnetic field observed ~B. WithAmpere’s law,

¸C

~B · d~s = µ0iincl.BUT WHAT IS iincl?

If we look at , iincl = i(t)

If we look at , iincl = 0

(∵ There is no charge flow between thecapacitor plates.)∴ Ampere’s law is either WRONG orINCOMPLETE.

Two observations:

1. While there is no current between the capacitor’s plates, there is a time-varying electric field between the plates of the capacitor.

2. We know Ampere’s law is mostly correct from measurements of B-fieldaround circuits.

⇓

Can we revise Ampere’s law to fix it?

Electric field between capacitor’s plates: E =σ

ε0

=Q

ε0A, where Q = charge on

capacitor’s plates, A = Area of capacitor’s plates.

∴ Q = ε0E · A︸︷︷︸Electric flux

= ε0ΦE

∴ We can definedQ

dt= ε0

dΦE

dt= idisp

where idisp is called Displacement Current (first proposed by Maxwell).Maxwell first proposed that this is the missing term for the Ampere’s law:

˛

C

~B · d~s = µ0(iincl + ε0dΦE

dt) Ampere-Maxwell law

11.2. INDUCED MAGNETIC FIELD 127

Where iincl = current through any surface bounded by C,ΦE = electric flux through that same surface bounded by curve C, ΦE =

´S

~E ·d~a.

11.2 Induced Magnetic Field

We learn earlier that electric field can be generated bychargeschanging magnetic flux

.

We see from Ampere-Maxwell law that a magnetic field can be generated bymoving charges (current)changing electric flux

.

That is, a change in electric flux through a surface bounded by C can lead to aninduced magnetic field along the loop C.

Notes The induced magnetic field is along the same direction as caused by thechanging electric flux.

Example What is the magnetic field strength inside a circular plate capacitorof radius R with a current I(t) charging it?

Answer Electric field of capacitor

E =Q

ε0A=

Q

ε0πR2

11.3. MAXWELL’S EQUATIONS 128

Electric flux inside capacitor through aloop C of radius r:

ΦE = E · πr2 =Qr2

ε0R2

Ampere-Maxwell Law inside capacitor:

˛

C

~B · d~s︸︷︷︸

∵ ~Binduced‖d~s

= µ0(©©©iincl + ε0dΦE

dt)

2πr︸︷︷︸Length of loop C

Binduced = µ0ε0d

dt

( Qr2

ε0R2

)

= µ0r2

R2

dQ

dt︸︷︷︸I(t)

∴ Binduced =µ0r

2πR2I(t) for r < R

Outside the capacitor plate:Electric flux through loop C: ΦE = E ·πR2 =

Q

ε0

˛

C

~B · d~s = µ0(iincl + ε0dΦE

dt)

2πrBinduced = µ0ε0

( 1

ε0

· dQ

dt

)

∴ Binduced =µ0I(t)

2πr

11.3 Maxwell’s Equations

The four equations that completely describe the behaviors of electric and magneticfields.

11.3. MAXWELL’S EQUATIONS 129

˛

S

~E · d~a =Qincl

ε0˛

S

~B · d~a = 0

˛

C

~E · d~s = − d

dt

ˆ

S

~B · d~a˛

C

~B · d~s = µ0iincl + µ0ε0d

dt

ˆ

S

~E · d~a

The one equation that describes how matter reacts to electric and magnetic fields.

~F = q( ~E + ~v × ~B)

Features of Maxwell’s equations:

(1) There is a high level of symmetry in the equations. That’s why the studyof electricity and magnetism is also called electromagnetism.There are small asymmetries though:

i) There is NO point ”charge” of magnetism / NO magnetic monopole.

ii) Direction of induced E-field opposes to B-flux change.Direction of induced B-filed enhances E-flux change.

(2) Maxwell’s equations predicted the existence of propagating waves of E-fieldand B-field, known as electromagnetic waves (EM waves).

Examples of EM waves: visible light, radio, TV signals, mobile phonesignals, X-rays, UV, Infrared, gamma-ray, microwaves...

(3) Maxwell’s equations are entirely consistent with the special theory of rela-tivity. This is not true for Newton’s laws!

phys1112 - electricity and magnetism lecture notes · 2.2. the electric field 9 (2) if q1, q2 are...

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