1

Chapter 23--Examples

2

Problem

In the figure below, point P is at the center of the rectangle. With V=0 at infinity, what is the net electric potential at P due to the six charged particles?

-2q+3q

+5q -2q -3q

+5q

P

d d

d

dd

d

3

Find distance from corners to P

-2q+3q

+5q -2q -3q

+5q

P

d d

d

dd

d

s

d

d/2

4

Potentials (Voltage) is a scalar

654321 VVVVVVVVi

i

2

22 d

qkV

s

qkV

53

-2q+3q

+5q -2q -3q

+5q

P

d d

d

dd

d

s

d

d/2

123

4 5 6

s

qkV

31

2

25 d

qkV

s

qkV

56 s

qkV

34

5

654321 VVVVVVVVi

i

d

kqV

d

kqV

s

qdq

kV

s

qdq

s

q

s

qdq

s

qkV

VVVVVVV

94.0

854

10

2

4

5

2

235

2

23

654321

6

Problem

A charge q is distributed uniformly throughout a spherical volume of radius, R.

a) Setting V=0 at infinity shown that the potential at a distance r from the center, where r<R is given by

b) What is the potential difference between a point on the surface and the sphere’s center?

3

0

22

8

3

R

rRqV

7

First, use Gauss’s law to find the E-field inside and outside the sphere

rr

qE

qrE

qAdE

outside

outside

enclosedoutside

ˆ1

4

4

20

0

2

0

rrR

qrrE

rrE

R

q

VrE

qAdE

inside

inside

inside

enclosedinside

ˆ4

ˆ3

3

414

34

14

300

3

0

2

3

0

2

0

8

Outside is simpler

R

qV

r

qVV

drr

qrdrEsdEVV

rr

qE

R

RR

RRf

i

if

outside

1

4

|1

4

1

4ˆ

ˆ1

4

0

0

20

20

9

Inside

R

qr

R

qVV

RrR

qVV

rR

qVV

rdrR

qsdEVV

rrR

qrrE

Rinr

Rinr

rRRinr

r

R

r

R

Rinr

inside

0

23

0

223

0

23

0

30

300

88

8

|8

4

ˆ4

ˆ3

10

Voltage=Outside+Inside

)3(8

88

3

88

1

4

88

1

4

223

0

23

0

23

0

0

23

00

0

23

00

rRR

qV

rR

qR

R

qV

R

qr

R

q

R

qV

R

qr

R

q

R

qVVVV RinrR

11

Part b) What is the potential difference between a point on the surface and the sphere’s center?

R

q

R

q

R

qVVV

VR

qV

rat

rRR

qV

CenterR

Center

000

0

223

0

88

3

4

8

3

0

)3(8

12

Okay, that is if V=0 at infinity what if V=0 at the center of the sphere?

30

2

0

30

2

0

03

00

03

00

30

8

0

84

4

ˆ4

R

qrV

VBut

R

qrVrdr

R

qVV

rdrR

qVV

rrR

qrE

r

r

r

r

r

inside

R

qV

Rrat

Rin08

Same as previous

Same as previous

13

Problem

The electric potential at points in a space are given by

V=2x2-3y2+5z3

What is the magnitude and direction of the electric field at the point (3,2,-1)?

14

E=-grad(V)

CNE

zyxE

zyxE

zz

Vy

y

Vx

x

V

zyxV

VE

/6.22151212

ˆ15ˆ12ˆ12)1,2,3(

ˆ)1(15ˆ)2(6ˆ)3(4)1,2,3(

1564

532

222

2

2

322

15

Directions

01

222

01

4.1386.22

15cos

6.22cos

135)1(tan

112

12tan

ˆ15ˆ12ˆ12)1,2,3(

zyxrwherer

z

x

y

zyxE

Direction w.r.t +x axis

Direction w.r.t +z axis

16

Problem

Three +0.12 C charges form an equilateral triangle, 1.7 m on a side. Using energy that is supplied at a rate of 0.83 kW, how many days would be required to move one of the charges to the midpoint of the line joining the other two charges?

17

Draw It

0.12C

0.12C

0.12C

1.7 m1.7 m

1.7 m

Initially, this charge is 1.7 m from the other two charges

Finally, this charge is 0.85 m from the other two charges

0.85 m 0.85 m

18

Potential Difference

CqmL

whereL

kqV

L

kq

L

kqVVV

L

kqLkq

Lkq

V

L

kq

L

kq

L

kqV

VVV

if

f

i

if

12.07.1

2

24

4

22

2

19

Potential Energy

CqmL

where

Jxx

U

L

kqVqU

12.07.1

101527.1

)12.0)(109(2

2

629

2

20

Power=Work per unit time

P=W/tW=-USo 0.83 kW= 830 J/sAnd t= U/P=152x106/830t=183,699 s or 51 hours or 2.12 days