chapter 25 electric potential 25-1 potential difference and electric potential 25-2 potential...

Chapter 25 electric potential25-1 Potential difference and electric Potential25-2 Potential Difference and electric field25-3 Electric Potential and Potential energy due to point charges

04/19/23 1Norah Ali Al-moneef king saud university

25-1 Potential difference and electric Potential

dW dU


Work and Potential Energy

q 0

FE lim F qE

q

Electric Field Definition:

Work Energy Theorem

a

b

E


Electric Potential Difference

Definition:

a

b

E


Conventions for the potential “zero point”

b a baba b a

U U WV V V

q q

Choice 1: Va=0b a

b a

U UV V

q

0 0

bb

UV

q

Choice 2:

“Potential”

V 0

00


25-2 Potential Difference and electric fieldWhen a force is “conservative” ie gravitational and the electrostatic force a potential energy can be defined

Change in electric potential energy is negative of work done by electric force:


∆ V = -∫ E ds = -Ed

Units of Potential Difference

Joules JVolt V

Coulomb C

Because of this, potential difference is often referred to as “voltage”

b a baba b a

U U WV V V

q q

So what is an electron Volt (eV)?

In addition, 1 N/C = 1 V/m - we can interpret the electric field as a measure of the rate of change with position of the electric potential.


•The change in potential energy is directly related to the change in voltage.

U = qVV = U/q

• U: change in electrical potential energy (J)• q: charge moved (C)• V: potential difference (V)•All charges will spontaneously go to lower potential energies if they are allowed to move.

Electron-Volts• Another unit of energy that is commonly used in atomic and nuclear

physics is the electron-volt• One electron-volt is defined as the energy a charge-field system

gains or loses when a charge of magnitude e (an electron or a proton) is moved through a potential difference of 1 volt1 eV = 1.60 x 10-19 J


• Since all charges try to decrease UE, and UE = qV, this means that spontaneous movement of charges result in negative U.

• V = U / q• Positive charges like to DECREASE their potential (V < 0)• Negative charges like to INCREASE their potential. (V > 0)

04/19/23 Norah Ali Al-moneef king saud university 9

. . . cosB

B

B A AA

V V V E ds E ds E d E s ��

VB – VA = VC - VA

A uniform electric field directed along the positive x axis. Point B is at a lower electric potential than point A. Points B and C are at the same electric potential.

VB = VC

If a 9 V battery has a charge of 46 C how much chemical energy does the battery have?

E = V x Q = 9 V x 46C = 414 Joules


Example

ExampleA pair of oppositely charged, parallel plates are separated by 5.33 mm. A

potential difference of 600 V exists between the plates. (a) What is the magnitude of the electric field strength between the plates? (b) What is the magnitude of the force on an electron between the plates?

Cxq

CNEE

EVV

EdVmd

e

19106.1

/55.207,113?

)0053.0(600600

00533.0

N 1081.1

106.114

19

e

ee

F

Cx

F

q

FE


ExampleCalculate the speed of a proton that is accelerated from rest

through a potential difference of 120 V

?

120

1067.1

106.1

27

19

v

VV

kgxm

Cxq

p

p

m/s 1052.11067.1

)120)(106.1(22

21

527

19

2

x

x

m

Vqv

qmv

q

K

q

WV


ds for a point charge

25-3 Electric Potential and Potential energy due to point charges

+Q2

kqˆE r

r


Recall the convention for the potential “zero point”

V 0

ba b ab a

1 1V V V kq

r r

b bb

1 1V V V kq

r

kqV r

r

Equipotential surfaces are concentric spheres


Superposition of potentials

0 1 2 3V V V V ...

+Q3

+Q2

+Q110r

20r

30r0

31 20

10 20 30

kQkQ kQV ...

r r r

Ni

0i 1 i0

kQV

r

04/19/23 15Norah Ali Al-moneef

king saud university

E and V for a Point Charge• The equipotential lines are the

dashed blue lines• The electric field lines are the brown

lines• The equipotential lines are

everywhere perpendicular to the field lines

An equipotential surface is a surface on which the electric potential is the same everywhere.



Figure 25.4 (Quick Quiz 25.3) Four equipotential surfaces

Equipotential surfaces (the dashed blue lines are intersections of these surfaces with the page) and electric field lines (red- rown lines) for (a) a uniform electric field produced by an infinite sheet of charge, (b) a point charge, In all cases, the equipotential surfaces are perpendicular to the electric field lines at every point


Example (25.1) A 12-V battery connected to two parallel plates. The electric field between the plates has a magnitude given by the potential difference V divided by the plate separation d =0.3 cm

Example (25.2)


Example: (a) In figure a, 12 electrons are equally spaced and fixed around a circle of radius R. Relative to V=0 at infinity, what are the electric potential and electric field at the center C of the circle due to these electrons? (b) If the electrons are moved along the circle until they are nonuniformly spaced over a 120 are (figure b), what then is the potential at C? Solution:

012

:)( ER

eKVa

R

eKVb

12:)(

Potential due to a group of point charges

1 0

1

4

n ni

ii i i

qV V

r


Example (25.3) (a) The electric potential at P due to the two charges q1 and q2 is the algebraic sum of the potentials due to the individual charges. (b) A third charge q3 = 3.00 C is brought from infinity to a position near the other charges.

ExampleAn electric dipole consists of two charges q1 = +12nC and q2 = -12nC,

placed 10 cm apart as shown in the figure. Compute the potential at points a,b, and c.

V 899

)04.0

1012

06.0

1012(1099.8

)(

999

21

a

a

aaa

V

xxxV

r

q

r

qkV

V 0

4.1926

)14.0

1012

04.0

1012(1099.8

)(

999

21

c

b

b

bbb

V

VV

xxxV

r

q

r

qkV


V 240

m 60.0

C100.8CmN1099.8

m 20.0

C100.8CmN1099.8 82298229

AV

V 0

m 40.0

C100.8CmN1099.8

m 40.0

C100.8CmN1099.8 82298229

BV

Example The Total Electric Potential

At locations A and B, find the total electric potential.


1

1 2

12e

U V q

q qU k

r

(a) If two point charges are separated by a distance r12, the potential energy of the pair of charges is given by keq1q2/r 12 . (b) If charge q1 is removed, a potential keq2/r

12 exists at point P due to charge q 2 .

Potential energy due to multiple point charges

+Q1

21r

kqV r

r 1

12

kqV

r+Q2

1 22

12

kq qU q V

r

+Q3

+Q1+Q2

1 2

13 23

kq kqV

r r

21r

13r23r1 3 2 31 2

12 13 23

kq q kq qkq qU

r r r


Example 1. What is the potential energy if a +2 nC charge moves from to point A, 8 cm away

from a +6 C charge?

+6 C

+Q

A

+2 nC

kQqU

r

Potential Energy:Potential Energy:

2

2

9 -6 -9NmC

(9 x 10 )( 6 x 10 C)(+2 x 10 C)

(0.08 m)U

The P.E. will be positive at point A, because the field can do + work if q is released.

U = 1.35 mJU = 1.35 mJ Positive potential energy

8 cm


Signs for Potential Energy

+6 C

+Q

A

8 cm

B

C

12 cm

4 cm

Consider Points A, B, and C.Consider Points A, B, and C.

For +2 nC at A: For +2 nC at A: U = +1.35 mJU = +1.35 mJ

If +2 nC moves from A to B, does field E do + or – work? Does P.E. increase or decrease?

Questions:

+2 nC

Moving positive q

The field E does positive work, the P.E. decreases.The field E does positive work, the P.E. decreases.

If +2 nC moves from A to C (closer to +Q), the field E does negative work and P.E. increases.


Example. What is the change in potential energy if a +2 nC charge moves from to B?

kQqU

r

Potential Energy:Potential Energy:

2

2

9 -6 -9NmC

(9 x 19 )( 6 x 10 C)(+2 x 10 C)0.900 mJ

(0.12 m)BU

U = -0.450 mJU = -0.450 mJ

Note that P.E. has decreased as work is done by E.

+6 C

+Q

A

8 cm

B

12 cm

From Ex-1: UA = + 1.35 mJ

U = UB – UA = 0.9 mJ – 1.35 mJ


Example What is the change in potential energy if a -2 nC charge moves from to B?

kQqU

rPotential Energy:

2

2

9 -6 -9NmC

(9 x 19 )(6 x 10 C)(-2 x 10 C)0.900 mJ

(0.12 m)BU

+6 C

+Q

A

8 cm

B

12 cm

From Ex-1: UA = -1.35 mJ

(Negative due to – charge)

UB – UA = -0.9 mJ – (-1.35 mJ) U = +0.450 mJU = +0.450 mJ

A – charge moved away from a + charge gains P.E.A – charge moved away from a + charge gains P.E.


Example :Find the potential at a distance of 6 cm from a –5 nC charge.

Q = -5 nC

--

--

-

---

Q

.

r

P

6 cm

2

29 -9Nm

C9 x 10 ( 5 x 10 C)

(0.06 m)

kQV

r

VP = -750 VVP = -750 VNegative V at Point P :

What would be the P.E. of a –4 What would be the P.E. of a –4 C charge placed C charge placed at this point P?at this point P?

U = qVU = qV = (-4 x 10 = (-4 x 10-6-6 C)(-750 V);C)(-750 V); U = 3.00 mJU = 3.00 mJ

Since P.E. is positive, E will do + work if q is released.Since P.E. is positive, E will do + work if q is released.

q = –4 C


Example : Two charges Q1= +3 nC and Q2 = -5 nC are separated by 8 cm. Calculate the electric

potential at point A.

+

Q2 = -5 nC-

Q1 +3 nC

6 cm

2 cm

2 cm

A

B1 2

1 2A

kQ kQV

r r

2

29 -9Nm

C1

1

9 x 10 ( 3 x 10 C)450 V

(0.06 m)

kQ

r

2

29 -9Nm

C2

2

9 x 10 ( 5 x 10 C)2250 V

(0.02 m)

kQ

r

VA = 450 V – 2250 V; VA = -1800 VVA = -1800 V


Example Calculate the electric potential at point B for same charges.

+

Q2 = -5 nC-

Q1 +3 nC

6 cm

2 cm

2 cm

A

B1 2

1 2B

kQ kQV

r r

2

29 -9Nm

C1

1

9 x 10 ( 3 x 10 C)1350 V

(0.02 m)

kQ

r

2

29 -9Nm

C2

2

9 x 10 ( 5 x 10 C)450 V

(0.10 m)

kQ

r

VB = 1350 V – 450 V; VB = +900 VVB = +900 V


Example : What is the potential difference between points A and B. What work is done by the E-field if a +2 C

charge is moved from A to B?

VB = +900 VVB = +900 VVA = -1800 VVA = -1800 V

VVABAB= V= VAA – V – VBB = -1800 V – 900 V = -1800 V – 900 V

VAB = -2700 VVAB = -2700 VNote point B is at higher Note point B is at higher potential.potential.

WorkWorkABAB = = q(Vq(VAA – V – VBB) = ) = (2 x 10(2 x 10-6-6 C )(-2700 V) C )(-2700 V)

Work = -5.40 mJWork = -5.40 mJ

Thus, an external force was required to move the charge.

+

-5 nC-

Q1 +3 nC

6 cm

2 cm

2 cm

A

B

Q2

E-field does negative work.E-field does negative work.


Example 6 (Cont.): Now suppose the +2 C charge is moved from back from B to A?

VB = +900 VVB = +900 VVA = -1800 VVA = -1800 V

VVBABA= V= VBB – V – VAA = 900 V – (-1800 V) = 900 V – (-1800 V)

VBA = +2700 VVBA = +2700 VThis path is from high to low potential.

WorkWorkBABA = = q(Vq(VBB – V – VAA) = ) = (2 x 10(2 x 10-6-6 C )(+2700 V) C )(+2700 V)

Work = +5.40 mJWork = +5.40 mJ

The work is done BY the E-field this time !

+

-5 nC-

Q1 +3 nC

6 cm

2 cm

2 cm

A

B

Q2

E-field does positive work.


ExampleAn electron is accelerated in a TV tube through a

potential difference of 5000 V. a) What is the change in PE of the electron?V = PE/qPE = qV = (-1.60 X 10-19 C)(+5000 V)= -8.0 X 10-16 JWhat is the final speed of the electron (m = 9.1 X 10-31 kg)PE + KE = 0 (Law of conservation of energy)PE = -KEPE = - ½ mv2

v2 = (-2)(PE) = (-2)(-8.0 X 10-16 J) m 9.1 X 10-31 kg

v = 4.2 X 107 m/s04/19/23 36Norah Ali Al-moneef

king saud university

Summary• Electric potential energy:

• Electric potential difference: work done to move charge from one point to another

• Relationship between potential difference and field:

• Equipotential: line or surface along which potential is the same

• Electric potential of a point charge:



1:

2:


3:

4:


The electrons in a particle beam each have a kinetic energy of 1.60 x 10-17 J. What are the magnitude and direction of the electric field that stops these electrons in a distance of 10.0 cm?

An electron and a proton are each placed at rest in an electric field of 520 N/C. Calculate the speed of each particle 48.0 ns after being released.

5:

6:

d

d

d

d P

q1

q2

q3

q4

What is the potential at point P, located at the center of the square of point charges. Assume that d = 1.3m and the charges are

q1 = +12 n C, q2= -24 n Cq3 = +31 n C, q4= +17 n C


7:

1- The electric field has a magnitude of 3.0 N/m at a distance of 60 cm from a point charge. What is the charge?(a)1.4 nC(b)120 pC(c) 36 mC(d)12 C(e)3.0 nC


1- A conducting sphere has a net charge of −4.8 × 10−17 C. What is the approximate number of excess electrons on the sphere?(a) 100 (b) 200 (c) 300 (d) 400 (e) 500


2- Two point charges, 8x10-9 C and -2x10-9 C are separated by 4 m. The electric field magnitude (in units of V/m) midway between them is:A)9x109 B) 13,500 C) 135,000 D) 36x10-9 E) 22.5

N= (-4.8x10-17 C/-1.6x10-19 C=300 electrons)e = ×

19 C1.60 1

0

Q = Ne (N =1 、 2 、 3…)

Electric charge always occurs in multiples of e


2

9

2

9

222

2

99

2

99

22

21

21

)4(102108

/ 5.224

90

2)8 (2

9

2

29

2

892

102109

2108109

22

xx

CNE

E

E

kqkqEEE


3 - If 10000 electrons are removed from a neutral ball, its charge is;(a)+1.6×10-15 C (b) +1.6×10-23 C (c) -1.6×10-15 C (d) -1.6×10-23 C

4 - A charge of 10-6 C is in a field of 9000 N/C, directed upwards. The magnitude and direction of the force it experiences are;(a) 9×10-3 N, downwards (b) 3×10-3 N, downwards(c) 9×10-3 N, upwards (d) 3×10-3 N, upwards

Q = Ne =10000 x -1.6×10-19 Q = -1.6×10-15 C

F= q E = 9000 x 10-6

F = 9 x 10 -3 N

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