PHY2061
9-29-05
Name:_______________________ ___
+2q

7q +4q

5q

3q
+3q

2q

3q d

3q

5q
+4q

7q +2q
1.
[6 points] A central particle of charge

2 q is surrounded by a square array of charged particles. The length of the square side is d , and charges are placed at the square corners, at the midpoint of the sides, and midway between the corner and the midpoint for two sides. What are the magnitude and direction of the net electrostatic force on the central particle due to the other particles?
Note that most charges come in pairs with the same sign and at equal and opposite distances from the central charge. Thus the net force on the central charge cancels.
The only charges which do not balance lie along a horizontal line crossing the central charge.
Left side:
F
 
K
  
 d / 2

2
ˆ
attractive, force to left
Right side:
F
 
K
  
 d / 2

2
ˆ
repulsive, force to left
Total:
F
 
48 K q
2 d
2
ˆ
Page 1 of 11

PHY2061
9-29-05
Name:_______________________ ___
2.
[6 points] Particles 1, with a charge q
1
, and 2, with a charge q
2
, are on the x -axis with particle 1 at x = 2a and particle 2 at x =
 a. For the net force on a third charged particle placed at the origin to be zero, what must be q
2
in terms of q
1
? q
2 q
1 x=
 a x=2a
The only way for the forces to balance is if q
1
and q
2
have the same sign charge, and the charge placed at the origin has the opposite sign. Let’s assume that q
1
and q
2
are both positive, and q
3
is negative. Then the forces on the third particle are:
F
32
 
K
| q q
3 2 a
2
F
31

K
| q q
3 1
4 a
2
Requiring these to balance, add to zero, implies that q
2
 q
1
4
Page 2 of 11

PHY2061
9-29-05
Name:_______________________ ___ y
E
 x
3.
Consider electric charge distributed along a one-dimensional path in the form shown as ¾ of a circle. The circle is centered at the origin with a radius of
R , and the linear charge density

is positive.
(a) [6 points] Find the component of the electric field along the x -axis (E x
) at the origin (0,0).
(b) [6 points] Find the component of the electric field along the y -axis (E y
) at the origin (0,0).
Note that the ¾ ring of positive charge is equivalent to a full ring of positive charge plus a ¼ ring of negative charge in the first quadrant. For the full ring, by symmetry, the electric field must cancel. That leaves the field from the negatively charged segment, which would point at an angle of 45

(that is E x
and E y
are equal and both positive in the shown coordinate system).
Let’s solve for each component explicitly for a ¼ ring with charge density
   
: dq
  ds
   dE x

K
E x
 
0

/ 2 dq cos

R
2

R
K cos
 
R
2 dE y

K
E y
 
0
 dq
/ 2
K sin

R
2

R sin
R
2

K
 sin

R 0

/ 2 
K

R
 
K
 cos

R 0

/ 2 
K

R
Indeed the components are equal, and the angle is 45

. We could also have integrated the ¾ positive ring from 
/ 2

2

and obtained the same answer.
Page 3 of 11

PHY2061
9-29-05
Name:_______________________ ___
4.
[8 points] In the figure shown, a small, non-conducting ball of mass and charge q
   m

6
(distributed uniformly through its volume) hangs from an insulating thread that makes an angle

with a vertical, uniformly charged nonconducting sheet (shown in cross section). The sheet has a surface charge density of
   
11 C/m 2
. Considering the gravitational force of the ball and assuming that the sheet extends far vertically and into and out of the page, calculate the angle

.
By the relation of forces, tan
 
F
E
 qE
F mg g
Now for an infinite sheet of charge, the magnitude of the electric field is E


2

0
So: tan
  q

2

0 mg





12
 
6

 
11
C/m
2
2
C / N m
2

 
6


2
 tan
 
0.277
15.5
Page 4 of 11

PHY2061
9-29-05 y
Name:_______________________ ___ x z
5.
Consider a cube with side length s = 0.5 m and one corner at the origin (0,0,0) as shown.
(a) [6 points] What is the total charge enclosed by the cube if the electric field is
E


2
ˆ 
4 y
3 ˆ  z z

N/C ?
By Gauss’ Law,
 
E

S
Ε Α  q enc

0
 
E i
6 

1
 i
(sum over flux through all 6 sides)
Only the y and z components of the electric field vary and thus give rise to a net flux out of the cube (the constant x component has zero net flux). Moreover, The face at z=0 and the face at y=0 has E=0, so zero flux as well. That leaves just the faces at z=s and y=s:
E A top
  front
 

2 x

4 y
3 ˆ 
6
ˆ

 s
2 ˆ 

2
ˆ 
4 y
3 ˆ 
6 z

 s
2 z
 
E
4
3 2 s s

6 ss
2 
4 s
5 
6 s
3 
0.875
 
E q enc

0
 q enc



12


7.75 10

12
C
(b) [6 points] What is the electric charge density (C/m
3
) at the center of the top face at y = 0.5 m if the electric field is the same as in part (a)?


Gauss' Law in differential form
0


E x x


E y
 y



E z z
  y
2  
9 at y

0.5
 
E
 
11
8.0 10 C/m
3
Page 5 of 11

PHY2061
9-29-05
Name:_______________________ ___
6.
(a) [6 points] A solid non-conducting sphere of radius R

0.05 m density
 
0.01 C/m
3 has a uniform charge
constant throughout the volume of the sphere. What is the ratio of the magnitude of the electric field at r

R / 2 to that at r

R ?
Apply Gauss’ Law to a spherical surface with radius r:
 
E

S
Ε A  q enc

0
4

2 r E
 q enc

0
   
K q enc r 2
Now the total charge enclosed is obtained by integration: q enc
 
V
 dV

 q enc
 r

R
 
3 r
3
 q enc
 r

R / 2
 
1
2
3 q enc
 r

R
 
So the ratio of the electric field is




R / 2
R



Kq enc


R r

/ 2

R
2
/ 2


R
2
Kq enc r

R


1 2
2

1
3
2 1 2
(b) [6 points] Suppose the solid non-conducting sphere in part (a) has a non-uniform charge density
    
0 r , where

0

0.267 C/m
4
(i.e. varies with radius). What is the ratio of the magnitude of the electric field at r

R / 2 to that at r

R ?
The difference with part (a) is in calculating the enclosed charge: q enc
 r

R
  
V
 q enc
 r

R
 
 dV
 
0
R 
0
4

2 r r dr

4

0
R
4
4
 
0
R
4
 q enc
 r

R / 2
 
1 q enc
 r

R
 
2
4
So the ratio of the electric field is




R / 2
R



Kq enc r

R


R / 2

2
/ 2


R
2
Kq enc r

R


1 2
2

1
4
2 1 4
Page 6 of 11

PHY2061
9-29-05
R
3
R
2
R
1
Name:_______________________ ___
7.
Three charged, concentric, spherical conducting shells have radii
R
1

10 cm, R
2

15 cm, R
3

20 cm . The thickness of the conducting shells, while not zero, is considered negligible. The charge on the innermost shell is

15

C, the charge on the middle shell is

20

C, and the charge on the outermost shell is +25

C.
(a) [6 points] What is the charge on the inner surface of the outermost conducting shell?
The electric field must be zero inside an isolated conductor. So a spherical Gaussian surface within each conductive shell must have zero flux since E=0, which implies that by
Gauss’ Law the enclosed charge must be zero. Thus, the inside surface of each shell must have an electric charge that exactly balances the charge from the inside conductors.
Therefore, the inside surface of the outermost shell must balance a charge of

35

C, which is thus +35

C.
(b) [6 points] What is the direction and magnitude of the electric field at radius r

17 cm ?
The two inner shells have a total electric charge of

35
 C. By Gauss’ Law, the flux through a spherical surface with radius 17 cm enclosing these two shells must equal the enclosed charge /

0
:

S
Ε A 

E

 
4

0 r
2 q enc

0

9
   r
2
  7
1.09 10 N/C
The direction of the electric field in this region is inward toward the center of the sphere(and toward the negative charge)
Page 7 of 11

PHY2061
9-29-05
Name:_______________________ ___
E
+
+
+
+ p v
+
+
+
8.
[8 points] A proton is injected with a velocity of v
  6
2 10 m/s into a region of uniform electric field between two large plates separated by 1m and maintained with an electric potential difference of 30,000 V. The proton travels from lower electric potential to higher on a path perpendicular to the plates. Does the proton reach the far plate, and if not, what is the distance of closest approach to it? The charge of the proton is

19 e 1.6 10 C , and the proton mass is m p

1.67 10

27
kg .
As the proton climbs from lower electric potential to higher, it will gain potential energy at the expense of kinetic energy. The total potential energy to climb to reach the other plate is:
U q V

1.6 10

19
     
15
J
The initial kinetic energy is
K

1
2 mv 2 

 
27 kg

 6

2   
15 J which is not enough energy to reach the far plate, as the energy is short by
1.46 10

15
J


 
19
   
9125 eV
For parallel plates, the electric field is uniform and the equipotential lines equally spaced. So the distance of closest approach to the far plate is: dca

9125
30000

1.0 m
 
0.3 m a little less than 1/3 of the plate separation.
Page 8 of 11

PHY2061
9-29-05
Name:_______________________ ___
9.
[6 points] The electric potential along the x -axis (in kV=10
3
V ) is plotted versus the value of x (in meters). Evaluate the x -component of the electrical force (in Newtons) on a charge of 4.70

C located on the x-axis at x=

1.2 m.
The electric field is given by the negative rate of change of the electric potential per unit length:
E
 

V
 
6000
 
 
7500 V/m (by estimation from graph)
F
 qE
 x


6
1.2 m
4.7 10 C
  
7500 V/m
 so F
  x
Force is directed in negative x direction.
Page 9 of 11

PHY2061
9-29-05
Name:_______________________ ___
A
B
10.
(a) [6 points] Consider the shown arrangements of capacitors above. Calculate the effective capacitance, C , of the network between the terminals A and B given that eff each of the shown capacitors has a capacitance C = 4

F.

1
The rightmost two capacitors are in series, with capacitance C
 
1

1
C C

C
2
This is in parallel with the left capacitor, so the effective capacitance is:
C eff
C C
 
3
2
C

A
  
B
(b) [6 points] Consider the above infinite chain of capacitors. Calculate the effective capacitance, C , of the network between the terminals A and B given that each of eff the shown capacitors has a capacitance C = 4

F.
You could solve this problem by successive approximations. Suppose just two capacitors were added to the end of the chain in part (a). The last two capacitors would be in series, so 2

F capacitance. Adding this in parallel with 4

F gives 6

F, in series with 4

F again gives 2.4

F in parallel with 4

F. So the two loop capacitor case gives 6.4

F. Three loops give 6.46

F, and so on.
However, this semi-infinite ladder can be solved inductively. Let C be the eff capacitance of the circuit. Then the effective capacitance of the circuit is capacitor C in parallel with a circuit containing a capacitor C in series with C ! eff
Page 10 of 11

PHY2061
9-29-05
Name:_______________________ ___
C eff
C


1
C

C
1 eff



1

C eff
C


 
C eff

C
CC eff



1
C eff

C

C eff

C
 
CC eff

C eff
2 
CC eff

C
2 
0


C eff
C eff


C
2
C

C
2 
4 C
2
by quadratic formula
2

1

5


1.62
C
 
11.
[6 points] A parallel plate capacitor with a capacitance of 12

F is connected to a battery that maintains an electric potential difference of 6V. If the plates are squeezed to one-third of the original separation, how much work must be done by the battery to maintain the same potential difference across the plates?
The stored potential energy of a capacitor is U

1
2
CV
2
. The capacitance of a parallel-plate capacitor satisfies: C
 
0
A
, so if the plates get 3 times closer, the d capacitance is 3 times larger

36

F. Thus the potential energy increases since the voltage remains the same across the capacitor but the capacitance increases. The net work done by the battery to change this potential energy is:
W U U f

U i

1
2
V
2

C f

C i

1
2

6 V
       
4
Page 11 of 11