norah ali almoneef 1 25.1 potential difference and electric potential 25.2 potential differences in...

Norah Ali Almoneef1

25.1 Potential Difference and Electric Potential25.2 Potential Differences in a Uniform Electric Field25.3 Electric Potential and Potential Energy Due to Point charges

Norah Ali Almoneef2

25.1 Potential Difference and Electric Potential

When a test charge q0 is placed in an electric field E created by some source charge distribution, the electric force acting on the test charge is q0 E.

When the test charge is moved in the field by some external agent, the workdone by the field on the charge is equal to the negative of the work done by the external agent causing the displacement

This is analogous to the situation of lifting an object with mass in a gravitational field—the work done by the external agent is mgh and the work done by the gravitational force is -mgh.

the work done by the electric field on the charge is

work =F. ds =q0E“.dsds is the displacement of a charge

The potential energy of the charge–field system is changed by an amount dU = q0E“.ds

For a finite displacement of the charge from point A to point B, the change in potential energy of the system ∆U =UB - UA is

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Dividing the potential energy by the test charge gives a physical quantity that depends only on the source charge distribution. The potential energy per unit charge U/q0 is

independent of the value of q0 and has a value at every point in an electric field. This quantity U/q0 is called the electric potential (or simply the potential) V.

The potential difference ∆V =VB -VA between two points A and B in an electric field is defined as the change in potential energy of the system when a test charge is moved between the points divided by the test charge q0:

The potential difference between A and B depends only on the source charge distribution (consider points A and B without the presence of the test charge)

the work done by an external agent in moving a charge q through an electricfield at constant velocity is

W= q ∆V

Norah Ali Almoneef4 4

Definition of the Electric PotentialThe electric potential energy of a charged

particle in an electric field depends not only on the electric field but on the charge of the particle.

We want to define a quantity to probe the electric field that is independent of the charge of the probe.

We define the electric potential as

• Unlike the electric field, which is a vector, the electric potential is a scalar.The electric potential has a value everywhere in

space but has no direction.

qU

V

Units: [V] = J / C, by definition, volt

“potential energy per unit charge of a test particle”

Norah Ali Almoneef5 5

Electric Potential EnergyThe electric force, like the gravitational

force, is a conservative force. When an electrostatic force acts between

two or more charges within a system, we can define an electric potential energy, U, in terms of the work done by the electric field, We, when the system changes its configuration from some initial configuration to some final configuration.

Change in electric potential energy = -Work done by electric field

U U f Ui We

Ui is the initial electric potential energy

U f is the final electric potential energy

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Like gravitational or mechanical potential energy, we must define a reference point from which to define the electric potential energy.

We define the electric potential energy to be zero when all charges are infinitely far apart.

We can then write a simpler definition of the electric potential taking the initial potential energy to be zero,

The negative sign on the work:If E does positive work then U < 0If E does negative work then U > 0

WUUU f 0

Norah Ali Almoneef77

Constant Electric Field - Special CasesDisplacement is in the same direction as

the electric field

A positive charge loses potential energy when it moves in the direction of the electric

field.Displacement is in the direction opposite

to the electric field

A positive charge gains potential energy when it moves in the direction opposite to the electric field.

qEdUqEdW so

qEdUqEdW so

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Potential Difference and Potential Difference and Electric PotentialElectric Potential

The potential difference between points A and B, VB – VA, is defined as the change in potential energy (final value minus initial value) of a charge q moved from A to B divided by the size of the charge.

V VB – VA =

Uq

Final point Initial point

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WorkWork

WAB = -U = - q(VB – VA)

Initial point Final point Final point Initial point

Equation is true if the only force is the conservative electrostatic force. That is, there are no non conservative forces acting on the system.


Change in Potential Energy• As the electric field accelerates the charge, the

charge gains kinetic energy.• As the charged particle gains kinetic energy, it loses

an equal amount of potential energy. K = - U

• By definition, the work done by a conservative force equals the negative change in potential energy, U. U = - WAB = - qEd

• This equation is valid only for a uniform electric field.

Norah Ali Almoneef

Potential Energy

U = Ub – Ua = qVba

Final point Initial point

1111


25-2 Potential Differences in a Uniform Electric Field (Constant Electric Field )Let’s look at the electric potential energy

when we move a charge q by a distance d in a constant electric field.

The definition of work is

For a constant electric field theforce is F = qE

the work done by the electric field on the charge is

dFW

cosqEddEqW

= angle between E and d.


Work and Potential Energy

There is a uniform field between the two plates

As the positive charge moves from A to B, work is done

WAB=F d=q E d

ΔPE =-W AB=-q E donly for a uniform

field


The positive side is always the

“high” potential side, regardless

of the sign of the charge.

The definition of the “high” side is

done for a positive test charge.

Usually take “low” side as V = 0.

Important!


Conservative ForcesConservative ForcesA force is conservative if the work it does on

a particle moving between any two points is independent of the path taken by the particle.

The work done by a conservative force exerted on a particle moving through any closed path is zero.

Gravitational (Newton’s law of gravity) and Electrical (Coulomb’s law of electrical force) are both conservative forces.

Since electrostatic force is conservative, electrostatic phenomena can be described in terms of electrical potential energy.

Potential Diff. In Uniform E field

Show that the potential diff. between path (1) and (2) are the same as expected for a conservative force field.

Charged particle moves from A to B in uniform E field.

V = E . ds = = E . d

B

A

U = qo V = - qo E . d = qo E d cos

16 Norah Ali Almoneef

Potential Diff. In Uniform E field (Path independence)

Show that the potential difference between path (1) and (2) are the same as expected for a conservative force field.

V = - E . ds = Es cos

B

A

B

c

V = -c

A

E . ds + - E . ds

d

V =

c

A

E . ds = Ed= E s cos

= 0 since E ds

Same

path (1)

path (2)

path (2)

path (1)

Conservative force: The work is path-independent.17 Norah Ali Almoneef


Electric Potential

The SI units of electric potential are joules per coulomb.

The unit of potential is the volt.A VOLT is defined:

1 V = 1 J/CThe Electron VoltThe electron volt is defined as the energy acquired by a particle carrying a charge equal to that on the electron (q = e) as the result of moving through a potential difference of 1 V.

1eV = (1.6 x 10-19 C)(1.0 V) = 1.6 x 10-

19 J

• Unlike the electric field, which is a vector, the electric potential is a scalar.The electric potential has a value everywhere in

space but has no direction.

Units: [V] = J / C, by definition, volt


Equipotential

V = E . ds = 0

B

c

VC = VB ( same potential)

In fact, points along this line has the same potential. We have an equipotential line.

If s is perpendicular to E (path C-B), the electric potential does not change.


Equipotential Surfaces

•The name equipotential surface is given to any surface consisting of a continuous distribution of points having the same electric potential.

•No work is done in moving a test charge between any two points on an equipotential surface.

•The equipotential surfaces of a uniform electric field consist of a family of planes that are all perpendicular to the field.

Equipotential surfaces are always perpendicular to electric field lines.


Equipotential Surface

Equipotential Surfaces (dashed blue lines) and electric field lines (orange lines) for (a) a uniform electric field produced by infinite sheet of charge, (b) a point charge, and (c) an electric dipole. In all cases, the equipotential surfaces are perpendicular to the electric field lines at every point.

Norah Ali Almoneef22 Norah Ali Almoneef22

Equipotential Surfaces

•The name equipotential surface is given to any surface consisting of a continuous distribution of points having the same electric potential.

•No work is done in moving a test charge between any two points on an equipotential surface.

•The equipotential surfaces of a uniform electric field consist of a family of planes that are all perpendicular to the field. Equipotential surfaces are always perpendicular to electric field lines.

Lines of constant ELines of constant V(perpendicular to E)


Equipotential Surfaces and the Electric Field

An ideal conductor is an equipotential surface. Therefore, if two conductors are at the same potential, the one that is more curved will have a larger electric field around it. This is also true for different parts of the same conductor.

All points on the surface of a charged conductor in electrostatic equilibrium are at the same potentialTherefore, the electric potential is a constant everywhere on the surface of a charged conductor in equilibrium


A point particle of mass m = 1.8x10-5 kg and charge q = +3.0x10-5 C is released from rest at point A and accelerates until it reaches point B. The only force acting on the particle is the electric force and the electric potential at A is 25 V greater than at B. What is the speed of the particle when it reaches B?

What happens if q is negative?

BBAA EPEKEEPEKE

0

2

2

1Bmv

BqVAqV

BAB VVqmv 2

2

1 V25

solve for vB

smvB /13.9

Example


Example

An ion accelerated through a potential difference of 115 V experiences an increase in kinetic energy of 7.37 x 10 –17 J. Calculate the charge on the ion.

qV= 7.37x10-17 J , V=115 V q = 6.41x10-19 C


In a hydrogen atom the e- revolves around the p+ at a distance of 5.3 x 10-11 m. Find the electric potential at the e- due to the p+, and the electrostatic potential energy between them.Electric potential due to proton:

Electrostatic p.E. is given by:

p+

e -

r

V 27

10 x 5.310 x 1.610 x 9

11-

19-9

rq

krV

J10 x 4.3 2710 x 1.6 18-19-

12

2112

pVe

rqq

kU

Example

A proton is released from rest in a uniform E field that has a magnitude of 8 x 104 V/m and is directed along the positive x-axis. The proton undergoes a displacement of 0.50 m in the direction of E.

(a)Find the change in electric potential between points A and B.

(b)Find the change in potential energy of the proton for this displacement.

(a) V = Ed = (8.0x104 V/m) (0.50m) = 4.0x104 V

(b) U = q V = (1.6 x 10-19 C) (4 .0x104 V) = 6.4 x 10-15 J

Example


Example

Suppose an electron is released from rest in a uniform electric field whose magnitude is 5.90 x 103 V/m. (a) Through what potential difference will it have passed after moving 1.00 cm? (b) How fast will the electron be moving after it has traveled 1.00 cm?

(a) V| = Ed = (5.90 x 103 V/m)(0.0100 m) = 59.0 V

(b) q V| = mv2/2 v = 4.55x106 m/s


29

Calculate the electrostatic potential energy between 2 protons in a Uranium nucleus separated by 2 x 10-15 m.

J10~

10 x 210 x 1.6

10 x 9.0

13

15-

219-9

21

rqq

krU

Norah Ali Almoneef

Example


Forces on Charged Forces on Charged ParticlesParticlesIn a CRT an electron moves 0.2 m in a straight

line (from rest) driven by an electric field of 8 x 103 V/m. Find:(a) The force on the electron.(b) The work done on it by the E-field.(c) Its potential difference from start to finish.(d) Its change in potential energy.(e) Its final speed.

Example


Example(a) Force is in opposite direction to the E-field,

magnitude:(b) Work done by force:

(c) Potential difference is defined as work/unit charge:

Alternatively (e- opposite to p+):

N 10 x 3.110 x 810 x 1.6 15-319- qEF

J 10 x 6.20.2 10 x 3.1 16-15- FsWork

V 10 x 1.6 10 x 1.610 x 6.2 3

19-

16-

qW

V

V 10 x 1.6 2.010 x 8

33

0

EddxEsdEVdb

a


(d) Change in potential energy:

(e) Loss of PE = gain in KE = ½mv2

donework

J 10 x 2.6-

10 x 1.610 x 1.6-16-

319-

0

0

Vq

sdEqUb

a

1-7

31-

16-

ms 10 x 2.4

10 x 9.110 x 2.62

2

mKE

v


A proton is accelerated across a potential difference of 600 V. Find its change in K.E. and its final velocity.

By definition, 1 eV = 1.6 x 10-19 J.Acceleration across 600 VProton gains 600 eV.

K.E. = 600(1.6 x 10-19) = 9.6 x 10-17 J

Final velocity is:

• If it started from rest

1-5

27-

17-

ms 10 x 3.4

10 x 1.710 x 9.62

v

Example


Two parallel metal plates have an area A = 225 cm2 and are d =0.5 cm apart, with a p.d. of 0.25 V between them. Calculate the electric field.0V 0.25V

0.1V 0.2V

x =0 x =0.5m

ds

1-Vm 505.0

25.0

d

VE

Ed

dsE

dsE

sdE

VVV

d

d

b

a

rightleft

0

0

Example

Electric Potential and Potential Energy due to point charges

Consider isolated positive point charge q. (i.e. E directed radially outward from the charge)

To find electric potential at a point located at a distance r from the charge, start with the general expression for potential difference:

VB VA = E . ds

B

A

Where A and B are two arbitrary points as shown.

E = kq/r2 r, where r is a unit vector directed from the charge toward the field point.



We can express E . ds as

E . ds = kq/r2 r . ds

The magnitude of r is 1, dot product r . ds = ds cos , where is the angle between r and ds .

ds cos is the projection of ds onto r , thus

ds cos = dr.

VB-VA = - E . ds

B

A

= -

B

A

kq/r2 dr



VB-VA = -

B

A

Er dr

VB-VA =

rB

rA

kq/r2 dr= -

kqr

rA

rB

VB-VA = kq 1rB

1rA

Depends only on the coordinates and not on the path.



V =kqr

rA = infinity (and VA = 0), we have electric potential created by a point charge at a distance r from the charge given by

Points at same distance r from q have the same potential V, i.e. the equipotential surfaces are spherical and centered on the charge.


Potential due to two or more charges: Superposition

where potential is taken to be zero at infinity and ri is the distance from the point P to the charge qi.

Note that this is a scalar sum rather than a vector sum.

V = qi

rik i P

q1

q2

q3

q5

q4

r1

r2

r3r4

r5

The potential is positive if the charge is positive and negative if the charge is negative



Finding the Electric Potential at Point P

5.0 C -2.0 C

V1060.3)m0.4()m0.3(

)C100.2()C/Nm1099.8(

,V1012.1m0.4

C100.5)C/Nm1099.8(

3

22

6229

2

46

2291

V

V

Superposition: Vp=V1+V2

Vp=1.12104 V+(-3.60103 V)=7.6103 V

Example


Example:How many electrons should be removed from

an initially uncharged spherical conductor of radius 0.300 m to produce a potential of 7.5 kV at the surface?

= C 1050.2 7q

N = 1.56 x 1012 electrons

r

qkV e = 7.50 x 103 V

)m300.0(

)/CNm10x99.8( 229 qV

A charge +q is at the origin. A charge –2q is at x = 2.00 m on the x axis. For what finite value(s) of x is (a) the electric field zero ? (b) the electric potential zero ?

x2 + 4.00x – 4.00 = 0 (x+4.83)(x0.83)=0

x = - 4.83 m (other root is not physically valid)

0)00.2(

2

x

q

x

qV

x = 0.667 m and x= -2.00 m

0)00.2(

222

x

q

x

qkE


Example

Potential Energy of a system of two charges

If two point charges are separated by a distance r12, the potential energy of the pair of charges is given by

U12 = k

q1 q2

r12

k =

1

4

V1 = potential at a point P due to q1, external agent must do work to bring a second charge q2 from infinity to P and this work = q2V1.

Definition: This work done is equal to the potential energy U of the two-particle system.

P


Potential Energy

Three point charges are fixed at the positions shown. The potential energy of this system of charges is given by

U = k

q1 q2

r12

q1 q3

r13

q2 q3

r23

+ +



Notes About Electric Potential Energy of Two ChargesIf the charges have the same sign, PE is

positivePositive work must be done to force the two

charges near one anotherThe like charges would repel

If the charges have opposite signs, PE is negativeThe force would be attractiveWork must be done to hold back the unlike

charges from accelerating as they are brought close together

ExampleWhat is the amount of work required to assemble four identical point charges of magnitude Q at the corners of a square of side s?

U = 0 + U12 + (U13 + U23) + (U14 + U24 + U34)

1

2

111

2

10

222

s

Qk

s

Qk

s

Qk eee

22

42

s

QkU e

s

s

s

s



ExampleA proton is placed in an electric field of E=105

V/m and released. After going 10 cm, what is its speed?

Use conservation of energy.

a b+

E = 105 V/m

d = 10 cm

V = Vb – Va = -Ed U = q V

U + K = 0

K = -U

K = (1/2)mv2

(1/2)mv2 = -q V = +qEd

m

qEdv

2

kg

mCv m

V

23

519

1067.1

110106.12

s

mv 6104.1

48

In a CRT an electron moves 0.2 m in a straight line (from rest) driven by an electric field of 8 x 103 V/m. Find:(a) The force on the electron.(b) The work done on it by the E-field.(c) Its potential difference from start to finish.(d) Its change in potential energy.(e) Its final speed.

Example

Norah Ali Almoneef

49

(a) Force is in opposite direction to the E-field, magnitude:

(b) Work done by force:

(c) Potential difference is defined as work/unit charge:

Alternatively (e- opposite to p+):

N 10 x 3.110 x 810 x 1.6 15-319- qEF

J 10 x 6.20.2 10 x 3.1 16-15- FsWork

V 10 x 1.6 10 x 1.610 x 6.2 3

19-

16-

qW

V

V 10 x 1.6 2.010 x 8

33

0

EddxEsdEVdb

a

Norah Ali Almoneef

50

(d) Change in potential energy:

(e) Loss of PE = gain in KE = ½mv2

donework

J 10 x 2.6-

10 x 1.610 x 1.6-16-

319-

0

0

Vq

sdEqUb

a

1-7

31-

16-

ms 10 x 2.4

10 x 9.110 x 2.62

2

mKE

v

Norah Ali Almoneef

51

Example

Calculate the electrostatic potential energy between 2 protons in a Uranium nucleus separated by 2 x 10-15 m.

J10~

10 x 210 x 1.6

10 x 9.0

13

15-

219-9

21

rqq

krU

Norah Ali Almoneef

Example

The charge distribution as shown is referred to as a linear quadrupole. (a) What is the potential at a point on the axis where x > a? (b) What happens when x >> a?


V = ke Q

axxax121

23

22xax

Qake

=

3

22xQakeV = As x >> a

When a negative charge moves in the direction of the electric field,1. the field does positive work on

it and the potential energy increases

2. the field does positive work on it and the potential energy decreases

3. the field does negative work on it and the potential energy increases

4. the field does negative work on it and the potential energy decreases


Example

The electric potential energy of two point charges approaches zero as the two point charges move farther away from each other.

If the three point charges shown here lie at the vertices of an equilateral triangle, the electric potential energy of the system of three charges is

1. positive

2. negative

3. zero

4. not enough information given to decide


Example

The electric potential due to a point charge approaches zero as you move farther away from the charge.

If the three point charges shown here lie at the vertices of an equilateral triangle, the electric potential at the center of the triangle is 1. positive

2. negative

3. zero

4. not enough information given to decide


Example


E-field between two parallel plates

Assume uniform field, and 3 mm plate separation

E = |VB – VA| / d = 12 / 3.0x10-3 = 4000 V/m

E directed from A (+ve) to B (ve)

A(+ve) plate is at higher potential than –ve plate.

Potential difference between plates = potential difference between battery terminals because all points on a conductor in equilibrium are at the same electric potential; no potential difference exists between a terminal and any portion of the plate to which it is connected.

Example

If a positive charge be moved against the electric field, then what will happen to the energy of the system?

If a positive charge be moved against the electric field, then energy will be used from an outside source.


Example

If 80 J of work is required to transfer 4 C charge from infinity to a point, find the potential at that point

W 80V = = =20 V

Q 4

W =80 J, q = 4 C, V =?



4. Which of the following figures have V=0 and E=0 at red point?

Electric Field and Electric Potential

A

q -q

B

q q

q q

q q

C D

q

E

-q

q -q

-q

q

Example


V = kq/r = 9x109 N m2//C2 x1.6x10-19 C/0.529 x10-10m

V = 27. 2 J/C = 27. 2 Volts

What is the electric potential at a distance of 0.529 A from the proton?

Example

ExampleTwo test charges are brought separately to the vicinity of a positive charge Q

A

qrQ

BQ

2q

2r

Charge +q is brought to pt A, a distance r from QCharge +2q is brought to pt B, a distance 2r from Q

(a) UA < UB (b) UA = UB (c) UA > UB

I) Compare the potential energy of q (UA) to that of 2q (UB)

Therefore, the potential energies UA and UB are EQUAL!!!

The potential energy of q is proportional to Qq/r

The potential energy of 2q is proportional to Q(2q)/(2r) = Qq/r


The principle at work here is CONSERVATION OF ENERGY. Initially:

The charge has no kinetic energy since it is at rest. The charge does have potential energy (electric) = UB.

Finally: The charge has no potential energy (U 1/R) The charge does have kinetic energy = KE

(a) (b) (c)

II) Suppose charge 2q has mass m and is released from rest from the above position (a distance 2r from Q). What is its velocity vf as it approaches r = ∞ ?

mr

Qqv f

04

1

mr

Qqv f

02

1

0fv

Example

KEUB 2

0 2

1

2

)2(

4

1fmv

r

qQ

mr

Qqv f

0

2

2

1



: The electron in the Bohr model of the atom can exist at only certain orbits. The smallest has a radius of .0529nm, and the next level has a radius of .212m.

a)What is the potential difference between the two levels?

b)Which level has a higher potential?+e

r1 r

2

r

qkV

11 r

ekV

VV 2.27100529.

106.1)109(

9

199

1

VV 79.6100212.

106.1)109(

9

199

2

VVdiffpotential 4.2079.62.27

r1 is at a higher potential.

Example


Example


VVV BA 240

What is the electric potential difference between A and B?

Example


Ex: Given that q1 = +2.40 nC and q2 = -6.50 nC.(a) what is the electric potential at points A and

B?

AVElectric Potential at A:

Electric Potential at B:

m

Ck

050.

1040.2 9 m

Ck

050.

105.6 9 V737

BV

m

Ck

080.

1040.2 9 m

Ck

060.

105.6 9 V704

The potential is a scalar (not vector) sum of the electric potentials produced by the individual charges:

Example


Ex: Given that q1 = +2.40 nC and q2 = -6.50 nC.(b) what is the work done by the electric field on a

point charge of 2.50 nC that travels from A to B?

V = EPE/q0 WAB = EPEA – EPEB

BAAB VqVqW 00 VnCVnC 70450.273750.2

nJ5.82

Example


What is the total energy of the electron in a hydrogen atom? In the ground state the electron orbit about the proton has a radius equal to one Bohr radius rB = 5.29x10-11 m.

p

B

pe

kq

r

kqqEPE

2

2

1eevmKE

VqEPE e J181035.4

qe = -qp = -1.60x10-19 C, me = 9.11x10-31 kg and ve = 2.2x106 m/s

JKE 181020.2

PEKEE Energy Total

But the electron in a hydrogen atom also has kinetic energy:

Total energy of e- in hydrogen atom is EPE + KE = -2.15x10-18 JThis is the electron’s binding energy, i.e., how much energy is required to rip off an electron!

Example


Three identical point charges (q = +2.0 μC each) are brought from infinity and fixed to a straight line so that the spacing between adjacent charges is d = 0.40 m. Determine the electric potential energy of this group.

q q q

d d

EPE = 0.23 J

Example:

U = kq1 q2

r12

q1 q3

r13

q2 q3

r23

+ +

U = k0.4 0.4 0.8

+ +2x10-6 x2x10-

6

x2x10-

6

2x10-6 x2x10-

6

2x10-6


Example


What is the potential difference between points A and B?ΔVAB = VB - VA

a) ΔVAB > 0 b) ΔVAB = 0c) ΔVAB < 0

E

A

BC

Example

Points A, B, and C lie in a uniform electric field.

Since points A and B are in the same relative horizontal location in the electric field there is on potential difference between them

The electric field, E, points in the direction of decreasing potential

E

A

BC

Point C is at a higher potential than point A. True

False

Example


As stated previously the electric field points in the direction of decreasing potential

Since point C is further to the right in the electric field and the electric field is pointing to the right, point C is at a lower potential

The statement is therefore false


If a negative charge is moved from point A to point B, its electric potential energy

a) Increases. b) decreases. c) doesn’t change.

E

A

BC

Example


The potential energy of a charge at a location in an electric field is given by the product of the charge and the potential at the locationAs shown in Example, the potential at points A and B are the same

Therefore the electric potential energy also doesn’t change73 Norah Ali Almoneef

Compare the potential differences between points A and C and points B and C. a) VAC > VBC b) VAC = VBC

c) VAC < VBC

E

A

BCPoints A, B, and C

lie in a uniform electric field.

In Example 4 we showed that the the potential at points A and B were the same

Therefore the potential difference between A and C and the potential difference between points B and C are the sameAlso remember that potential and potential energy are scalars and directions do not come into play

Example



A positive charge is released from rest in a region of electric field. The charge moves:a) towards a region of smaller

electric potential

b) along a path of constant electric

potential

c) towards a region of greater

electric potential

Example

A positive charge placed in an electric field will experience a force given byEqF

But E is also given by

Therefore

Since q is positive, the force F points in the direction opposite to increasing potential or in the direction of decreasing potential

d

VE

d

VqEqF


If you want to move in a region of electric field without changing your electric potential energy. You would movea) Parallel to the electric field

b) Perpendicular to the electric field

The work done by the electric field when a charge moves from one point to another is given by

b

a

b

aba ldEqldFW

0

The way no work is done by the electric field is if the integration path is perpendicular to the electric field giving a zero for the dot product

Example


Electric force moves a charge of 2x10-10 C from point A to point B and does 5x10-6 J of work.

What is the difference in potential energies of A and B (EPEA – EPEB)?EPEA – EPEB = 5x10-6 J

What is the potential difference between A and B (VA – VB)?V = 25000 V Point A is higher potential

WAB = - (EPEB - EPEA )∆PE: = EPEA - EPEB = 5x10-6 J

∆V: VA – VB = (EPEA – EPEB ) / q= 5x10-6J/2x10-10C = 25000 V

Example

What is the electric potential at the center of the square?

45º

45ºr

r r

r

mr

r

rr

071.

01.2

10.02

222

C

JV

V

r

qkV

total

total

i

i

6

5566

1027.1

1034.61034.6)1027.1(1027.1

Vr

qkV 5

69 1034.6

071.

105109

Vr

qkV 6

69 1027.1

071.

1010109


Example

A proton is moved from the negative plate to the positive plate of a parallel-plate arrangement. The plates are 1.5cm apart, and the electric field is uniform with a magnitude of 1500N/C.

a)How much work would be required to move a proton from the negative to the positive plate?

b)What is the potential difference between the plates?

c) If the proton is released from rest at the positive plate, what speed will it have just before it hits the negative plate?

cosxFW qEFE

JW

mC

NCW

xqEW

18

19

106.3

)015)(.1500)(106.1(

1

Example



b) What is the potential difference between the plates?

C

JV

mC

NV

EdV

5.22

)015)(.1500(

Example



c) If the proton is released from rest at the positive plate, what speed will it have just before it hits the negative plate?

s

mv

v

m

qVv

mvqV

KU E

4

27

19

2

1057.6

1067.1

)5.22)(106.1(2

2

2

1

Example


Calculate the electric potential energy between each pair of charges and add them together.

JU

JJJU

total

total

72.

)72.()72.(72.

Jr

qqkU 72.0

2.

)104)(104()109(

66932

23

Jr

qqkU 72.0

2.

)104)(104()109(

66931

13

Jr

qqkU 72.0

2.

)104)(104()109(

66921

12

Example


Charges +Q and –Q are arranged at the corners of a square as shown. When the magnitude of the electric field E and the electric potential V are determined at P, the center of the square, we find that

A. E ≠ 0 and V > 0. B. E = 0 and V = 0. C. E = 0 and V > 0. D. E ≠ 0 and V < 0. E. None of these is correct.


EXAMPLE

Q

200V 00V a b c

Q

Two equal positive charges are placed in an external electric field. The equipotential lines shown are at 100 V intervals. The potential for line c is

A.100 V.B.100 V.C.200 V..200 V.E.zero


EXAMPLE

Two equal positive charges are placed in an external electric field. The equipotential lines shown are at 100 V intervals. The work required to move a third charge, q = e, from the 100 V line to b is

A.100 eV.B.100 eV.C.200 eV..200 eV.E.zero

Q

200V 00V a b c

Q


EXAMPLE

The potential at a point due to a unit positive point charge is found to be V. If the distance between the charge and the point is tripled, the potential becomes

A. V/3.B. 3V.C. V/9.D. 9V. E. 1/V 2 .


EXAMPLE

EXAMPLEA proton is released from rest in a uniform electric field that has a magnitude of 8.0 104 V/m and is directed along the positive x axis. The proton undergoes a displacement of 0.50 m in the direction of E. (a) Find the change in electric potential between points A and B.(b) Find the change in potential energy of the proton forthis displacement.

The negative sign means the potential energy of the proton decreases as it moves in the direction of the electric field. As the proton accelerates in the direction of the field, it gains kinetic energy and at the same time loses electric potential energy (because energy is conserved).

V

mmVEdV4

4

10*0.4

)50.0)(/10*0.8(

J

VC

eVVqU

15

419

0

10*4.6

)10*0.4)(10*6.1(

(b)

(a)



A proton is released from rest at point B, where the potential is 0 V. Afterward, the proton

1. moves toward A with an increasing speed.2. moves toward A with a steady speed. 3. remains at rest at B. 4. moves toward C with a steady speed.5. moves toward C with an increasing speed.

EXAMPLE


EXAMPLE

Potential DifferenceThe potential difference between two points A and B is the work per unit positive charge done by electric forces in moving a small test charge from the point of higher potential to the point of lower potential.

Potential Difference: VAB = VA - VB

WorkAB = q(VA – VB) Work BY E-field

The positive and negative signs of the charges may be used mathematically to give appropriate signs.


Summary

Parallel PlatesConsider Two parallel plates of equal and opposite charge, a distance d apart.

Constant E field: F = qE

Work = Fd = (qE)d

Also, Work = q(VA – VB)

So that: qVAB = qEd and VAB = Ed

The potential difference between two oppositely charged parallel plates is the product of E and d.


Summary

Summary of Formulas

; kQq U

U Vr q

; kQq U

U Vr q

kQV

r

kQV

r

WorkAB = q(VA – VB) Work BY E-field

; V

V Ed Ed

; V

V Ed Ed

Electric Potential Energy and Potential

Electric Potential Near Multiple charges:

Oppositely Charged Parallel Plates:


94

SummaryElectric potential differenceSince only differences in potential matter,

the location of the zero of the potential can be chosen as we wish

Electric potential of a point chargeIf we chose the zero to be infinitely far from

a point charge, we can write its potential as kq

Vr

V = E . ds = = E . d

B

A

EquipotentialsThese are surfaces of equal potential difference.

The surface of a conductor in equilibrium is an equipotential.

norah ali almoneef 1 25.1 potential difference and electric potential 25.2 potential differences in...

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