Chapter 21

Electric FieldsLecture 2

Dr. Armen Kocharian

Electric Field – IntroductionThe electric force is a field forceField forces can act through space

The effect is produced even with no physical contact between objects

Faraday developed the concept of a field in terms of electric fields

Electric Field – Definition

An electric field exists in the region of space around a charged object

This charged object is the source chargeWhen another charged object, the test charge, enters this electric field, an electric force acts on it

Electric Field – Definition, contThe electric field is defined as the electric force on the test charge per unit chargeThe electric field vector, E, at a point in space is defined as the electric force Facting on a positive test charge, qoplaced at that point divided by the test charge: E = Fe / qo

Electric Field, NotesE is the field produced by some charge or charge distribution, separate from the test chargeThe existence of an electric field is a property of the source charge

The presence of the test charge is not necessary for the field to exist

The test charge serves as a detector of the field

Relationship Between F and EFe = qE

This is valid for a point charge onlyOne of zero sizeFor larger objects, the field may vary over the size of the object

If q is positive, F and E are in the same directionIf q is negative, F and E are in opposite directions

Electric Field Notes, FinalThe direction of E is that of the force on a positive test chargeThe SI units of E are N/CWe can also say that an electric field exists at a point if a test charge at that point experiences an electric force

Electric Field, Vector FormRemember Coulomb’s law, between the source and test charges, can be expressed as

Then, the electric field will be

2ˆo

e eqqkr

=F r

2ˆe

eo

qkq r

= =FE r→= = =

0

2

0 2

/ ˆ ˆlim lime e o eq

o o

k qq r k qq q rFE r r

More About ElectricField Directiona) q is positive, F is directed away from qb) The direction of E is also away from the positive source chargec) q is negative, F is directed toward qd) E is also toward the negative source charge

Superposition with Electric Fields

At any point P, the total electric field due to a group of source charges equals the vector sum of electric fields of all the charges

2ˆi

e ii i

qkr

= ∑E r

Superposition ExampleFind the electric field due to q1, E1Find the electric field due to q2, E2E = E1 + E2

Remember, the fields add as vectorsThe direction of the individual fields is the direction of the force on a positive test charge

Superposition Example, cont

Find E1

Find E2,

E = E1 + E2

cos ,

sin

=

=

3545

θ

θ

.

. .

=

= −

51

5 52

39 10

11 10 1 4 10

NE x jC

N NE x i x jC C

( )( )

( )( )

.. . ,

.

.. . ,

.

−

−

⎛ ⎞= = =⎜ ⎟

⎝ ⎠

⎛ ⎞= = =⎜ ⎟

⎝ ⎠

621 9 5

1 22 21

622 9 5

2 22 22

70 10 CNm N899 10 39 10C C040m

50 10 CNm N899 10 18 10C C050m

xqE k x x

r

xqE k x x

r

. .= + = +5 51 2 11 10 25 10N NE E E x i x j

C C

Electric Field – Continuous Charge Distribution

The distances between charges in a group of charges may be much smaller than the distance between the group and a point of interestIn this situation, the system of charges can be modeled as continuousThe system of closely spaced charges is equivalent to a total charge that is continuously distributed along some line, over some surface, or throughout some volume

Electric Field – Continuous Charge Distribution, contProcedure:

Divide the charge distribution into small elements, each of which contains ΔqCalculate the electric field due to one of these elements at point PEvaluate the total field by summing the contributions of all the charge elements

Electric Field – Continuous Charge Distribution, equations

For the individual charge elements

Because the charge distribution is continuous

2ˆe

qkrΔ

Δ =E r

2 20ˆ ˆlim

i

ie i eq i i

q dqk kr rΔ →

Δ= =∑ ∫E r r

Charge DensitiesVolume charge density: when a charge is distributed evenly throughout a volume

ρ = Q / VSurface charge density: when a charge is distributed evenly over a surface area

σ = Q / ALinear charge density: when a charge is distributed along a line

λ = Q / ℓ

Amount of Charge in a Small Volume

For the volume: dq = ρ dVFor the surface: dq = σ dAFor the length element: dq = λ dℓ

Problem Solving HintsUnits: when using the Coulomb constant, ke, the charges must be in C and the distances in mCalculating the electric field of point charges: use the superposition principle, find the fields due to the individual charges at the point of interest and then add them as vectors to find the resultant field

Problem Solving Hints, cont.Continuous charge distributions: the vector sums for evaluating the total electric field at some point must be replaced with vector integrals

Divide the charge distribution into infinitesimal pieces, calculate the vector sum by integrating over the entire charge distribution

Symmetry: take advantage of any symmetry to simplify calculations

Example 2 (field of a ring of charge)

Uniformly charged ring, total charge Q, radius aWhat is the electic field at a point P, a distance x, on the axis of the ring.How to solve

Consider one little piece of the ringFind the electric field due to this pieceSum over all the pieces of the ring (VECTOR SUM!!)

Px

dE = electric field due to a small piece of the ring of length ds

dQ = charge of the small piece of the ringSince the circumference is 2πa, and the total charge is Q:

dQ = Q (ds/2πa)

The next step is to look at the componentsBefore we do that, let’s think!

We are on the axis of the ringThere cannot be any net y or z components

A net y or z component would break the azimuthal symmetry of the problem

Let’s just add up the x-components and forget about the rest!

z

What is going on with the y and z components?

The y (or z) component of the electric field caused by the element ds is always exactly cancelled by the electric field caused by the element ds' on the other side of the ring

z dS’dQ

Now we sum over the whole ring, i.e. we take the integral:

Time to think about the integral now.The integration is "over the ring"

• k is a constant of nature• a is the ring-radius, a constant for a given ring• x is the distance from the center of the ring

of the point at which we want the E-field,x is also a constant

= Q

Sanity check: do limiting cases make sense?

What do we expect for x=0 and x ∞?At x=0 expect E=0

Again, because of symmetry Our formula gives E=0 for x=0 ☺

As x ∞, ring should look like a point.Then, should get E kQ/x2

As x ∞, (x2+a2) x2

Then E kxQ/x3 = kQ/x2 ☺

Electrical Field of Charged Ring (cont)

( )=

+3

2 2 24 o

xQ

πε x RE

Example – Charged DiskThe ring has a radius R and a uniform charge density σChoose dq as a ring of radius rThe ring has a surface area 2πr dr

Electric Field LinesField lines give us a means of representing the electric field pictoriallyThe electric field vector E is tangent to the electric field line at each point

The line has a direction that is the same as that of the electric field vector

The number of lines per unit area through a surface perpendicular to the lines is proportional to the magnitude of the electric field in that region

Another electric field example

Electric field on axis of a uniformly charged disk?As we did last time:

Calculate the field due to a small piece of the diskAdd-up the contributions from all the small pieces

Key question: What is the most convenient way of breaking up the disk into small pieces?Exploit result from last time: field due to a ring

Break up the disk into a bunch of concentric rings

Back to the disk

Strategy:Consider ring, at radius r, small width drCompute electric field dE due to this ringAdd up the electric fields of all the rings that make up the disk

Electric field due to ring, using prev. result:

dQ = charge of ring

dQ = charge of ring

Next question: what is dQ?• Total charge of the disk = Q• dQ/Q = [Area of the ring] / [Area of the disk]• Area of the disk = πR2

• Area of ring = 2πr dr

• dQ = Q(2r/R2)dr

width of the ring

circumference of the ring

Surface charge density:σ = Q/Area = Q/(πR2)

Q = πσR2

Now use k=1/(4πε0)

Now we sum over the rings we integrate:

What are we integrating over?Integrate over r, from r=0 to r=R

A curious result (counterintuitive ?)

Consider infinitely large disk, i.e., R ∞x2+R2 ∞, the 2nd term in parenthesis 0E σ /(2ε0)Constant, independent of x !!

The electric field of an infinitely large, uniformly charged plane is perpendicular to the plane, and constant in magnitude E=σ/(2ε0) In the limit, this holds also for a finite plane provided the distance from the plane is small compared to the size of the plane

Another example (Problem 21.102)Find electric field at any point on x-axis

Brute force approach:- find the field due to a small piece of the disk- sum over all of the pieces (integrate)There is a better way:- use previous results + a trick !

Previous result:

How can we apply it to:

Imagine two rings with no holesRadius R2, charge density +σRadius R1, charge density -σ

The sum of the fields due to these two rings will be the same as the field of the ring with the hole!

Field due to ring of radius R2, surface charge density +σ:

Field due to ring of radius R1, surface charge density -σ:

Total field is the sum of the two:

Total electrical field at the center of semicircle

( ) cos 0 N/Cx i i i yxi i

E E E Eθ= = =∑ ∑

2 20 04 4ii

q qEr Rπε πε

Δ Δ= =

•The linear charge density on the rod is λ = Q/L,where L is the rod’s length

Δq = λ Δs = (Q/L)Δs

( )2 2

0 0

/cos cos

4 4x i ii i

Q L s QE sR LR

θ θπε πε

Δ= = Δ∑ ∑

0

cos4x i

i

QELR

θ θπε

= Δ∑

• We note that the arc length Δs is related to the small angle Δθ by

Δs = RΔθ,

• With Δθ → dθ, the sum becomes an integral over all angles forming the rod. θ varies from Δθ = −π/2 toθ = +π/2. So we finally arrive at

/ 2 / 2

/ 2/ 20 0 0

2cos sin4 4 4xQ Q QE dLR LR LR

π π

ππθ θ θ

πε πε πε−−= = =∫

Total electrical field at the center of semicircle

Electric Field Lines, GeneralThe density of lines through surface A is greater than through surface BThe magnitude of the electric field is greater on surface A than BThe lines at different locations point in different directions

This indicates the field is non-uniform

Electric Field Lines, Positive Point Charge

The field lines radiate outward in all directions

In three dimensions, the distribution is spherical

The lines are directed away from the source charge

A positive test charge would be repelled away from the positive source charge

Electric Field Lines, Negative Point ChargeThe field lines radiate inward in all directionsThe lines are directed toward the source charge

A positive test charge would be attracted toward the negative source charge

Electric Field Lines – Dipole The charges are equal and oppositeThe number of field lines leaving the positive charge equals the number of lines terminating on the negative charge

Electric Field Lines – Like ChargesThe charges are equal and positiveThe same number of lines leave each charge since they are equal in magnitudeAt a great distance, the field is approximately equal to that of a single charge of 2q

Electric Field Lines, Unequal ChargesThe positive charge is twice the magnitude of the negative chargeTwo lines leave the positive charge for each line that terminates on the negative chargeAt a great distance, the field would be approximately the same as that due to a single charge of +q

Electric Field Lines – Rules for DrawingThe lines must begin on a positive charge and terminate on a negative charge

In the case of an excess of one type of charge, some lines will begin or end infinitely far away

The number of lines drawn leaving a positive charge or approaching a negative charge is proportional to the magnitude of the chargeNo two field lines can cross

Motion of Charged ParticlesWhen a charged particle is placed in an electric field, it experiences an electrical forceIf this is the only force on the particle, it must be the net forceThe net force will cause the particle to accelerate according to Newton’s second law

Motion of Particles, contFe = qE = maIf E is uniform, then a is constantIf the particle has a positive charge, its acceleration is in the direction of the fieldIf the particle has a negative charge, its acceleration is in the direction opposite the electric fieldSince the acceleration is constant, the kinematic equations can be used

Accelerating Positive ChargeUniform field –constant forceUse kinematics eqs.

For displacement

For velocity

( )

= + +

= +

= + −

2

2 2

1 ,2

,2

f i i

f i f i

x x v t at

vf vi atv v a x x

Positive charge in a uniform electric field

Accelerating Positive Charge, cont

Positive charge in a uniform electric field

= =

= =

= =

= =

= = =

= Δ =

22

2

2

,

0 ,

,2 2

22 ,

22 2

.

i i

f

f f f

ff f

f

F q Eam m

v x

a t q Ex tm

q Ev a x xm

m v m q EK E x q E xm

W K q E x

Work and Kinetic Energy

Electron in a Uniform Field, Example

The electron is projected horizontally into a uniform electric fieldThe electron undergoes a downward acceleration

It is negative, so the acceleration is opposite E

Its motion is parabolic while between the plates

Electron in an Uniform Field, Example, cont

Projectile motion of electron in a uniform electric field

= = −

= =

= = −

,

const,

t

y

fx ix

fy y

F eEam m

v veEv a tm

=

= = −2

2

,

.2 2

fx ix

f

x v t

at qEy tm

= = − ⇒ = −

=

222

2

2

2 2

2

fxf f fx

ix

ix

xat qEy y Axm v

qEAmv

Motion in Nonuniform Field

The Cathode Ray Tube (CRT)A CRT is commonly used to obtain a visual display of electronic information in oscilloscopes, radar systems, televisions, etc.The CRT is a vacuum tube in which a beam of electrons is accelerated and deflected under the influence of electric or magnetic fields

Cathode Ray Tube (CRT), cont

The electrons are deflected in various directions by two sets of platesThe placing of charge on the plates creates the electric field between the plates and allows the beam to be steered

Electric Field Lines (last time)Visualization of the electric fieldLines drawn parallel to the E-direction

With arrows pointing in the direction of EStart on +ve charge, end on –ve chargeHigh density of lines strong fieldUniqueness of E-field lines never cross

Otherwise would have two directions at crossing point

Are these field-lines pattern correct?from

http://www.glenbrook.k12.il.us/gbssci/phys/Class/estatics/u8l4c.html

A. OKB. OKC. No

Direction of arrows is wrongD. No

Density of lines suggests that field is stronger on one side of the charge

E. NoDirection of arrows is wrong

Are these OK field lines?No. Lines cannot cross

What are the signs of charges A & B?A: -ve B: +ve

(lines start on +ve charge, end on –ve)

Several locations are labeled. Rank them in order of electricfield strength, from smallestto largest DAECB or perhaps DAEBChigh density of lines high E What are the signs of charges A through I?

A. +B. -C. +D. -

E. -F. +G. +H. +I. +

Rank magnitude of charges in each sketchA > B

Density of lines to the left of A > density to right of BD > CF > E > GI > H > J

Chapter 23Quiz questions

1. Left.2. Down.3. Right.4. Up.5. The electric field is zero.

At the position of the dot, the electric field points

1. Left.2. Down.3. Right.4. Up.5. The electric field is zero.

At the position of the dot, the electric field points

A piece of plastic is uniformly charged with surface charge density 1. The plastic is then broken into a large piece with surface charge density 2 and a small piece with surface charge density 3. Rank in order, from largest to smallest, the surface charge densities 1 to 3.

1P > η2 > η32. η1 > η2 = η33. η1 = η2 = η34. η2 = η3 > η15. η3 > η2 > η1

A piece of plastic is uniformly charged with surface charge density 1. The plastic is then broken into a large piece with surface charge density 2 and a small piece with surface charge density 3. Rank in order, from largest to smallest, the surface charge densities 1 to 3.

1. η1 > η2 > η32. η1 > η2 = η33. η1 = η2 =η34. η2 = η3 > η15. η3 > η2 > η1

Which of the following actions will increase the electric field strength at the position of the dot?

1. Make the rod longer without changing the charge. 2. Make the rod shorter without changing the charge. 3. Make the rod fatter without changing the charge. 4. Make the rod narrower without changing the charge. 5. Remove charge from the rod.

Which of the following actions will increase the electric field strength at the position of the dot?

1. Make the rod longer without changing the charge. 2. Make the rod shorter without changing the charge.3. Make the rod fatter without changing the charge. 4. Make the rod narrower without changing the charge. 5. Remove charge from the rod.

Which electric field is responsible for the trajectory of the proton?

(1) (2) (3) (4) (5)

Which electric field is responsible for the trajectory of the proton?

(1) (2) (4) (5)(3)