ECE 341 Test 2 Spring 2016
20 points total
1. (a) Two charges + Q and

Q are apart by a distance d , as shown in Fig. (a). Assume
Q is positive. What is the field E due to + Q at the position of

Q ? What is the force F exerted by + Q on

Q ? Indicate the directions of the filed E and the force F by arrows on the figure. You need to label each arrow with vectors E and F .
(b) A charge

Q is a distance d /2 above an infinitely large metal sheet, to be considered as an ideal conductor, as shown in Fig. (b). What is the force on

Q ?
Indicate the direction of the force by an arrow on the figure.
(c) Two charges + Q and

Q are a distance d apart and are both d /2 above an infinitely large metal sheet, to be considered as an ideal conductor, as shown in Fig. (c). What is the force on

Q ? Indicate the direction of the force by an arrow on the figure.
E

Q
E

Q
F d
F d /2
+Q d
F
2 F
F
1
 3
Q d /2
8
4 d /2
+Q
(a)
Solution:
(b)

Q
(c)
+Q
2 points for each magnitude,
2 points for each direction
Directions of E and F shown in figure.
(b) The image of

Q is + Q . Same as (a):
2 points for magnitude, 2 points for direction
(only F is asked for)
8
The direction of F
1
+ F
2
is along the dashed line pointing towards the image charge

Q .
The total force on the real charge

Q is
The direction of the total force F is shown in figure.
4 points for magnitude, 4 points for direction, if student gets the final result right.
Otherwise, give 3 points for working out the image charges, 1 point for getting each of the forces
(F1, F2, F3) right.

ECE 341 Test 2 Spring 2016
18 points total
2. An ideal conductor sphere of radius R is charged with with total charge Q Find the electric field and potential distributions of the entire space. You may use the spherical coordinate, with the origin at the center of the sphere and r being the distance from the origin. Notice that the field is a vector and the potential a scalar. Assume the potential is
0 at r =

. Express your results in terms of r , R , and Q .
Solution:
For r < R , E = 0 since there is no charge.
For r

R , we get by using Gauss’s law
4 points
E

4
Q
 2
. 3 points for magnitude, 3 points for direction. Do not have to be in this form, r
0 but must be correct to earn points.
(Must specify the direction of the E field in some way, if not in this form.)
For r

R , the potential
V ( r )
  
 r
Edr

Q
4

0 r
. 4 points
For r < R , since E = 0, we have
V ( r )

V ( R )

Q
4

0
R
. 4 points
V
Figure just for illustration. Not required.
r
R

ECE 341 Test 2 Spring 2016
Total 32 points
3. Figure (a) below depicts a capacitor consisting of two parallel, conducting plates separated by a distance d . The space between the plates contains two adjacent dielectrics, one with permittivity

1
and surface area A
1
and another with

2
and A
2
. A voltage V is applied between the two plates.
(1) What is the total capacitance of this capacitor? (Ignore any fringe effects when calculating the capacitance.) How much energy does it store when charged to voltage V .
(2) The break-down fields are E b 1
and E b 2
for the two dielectrics, respectively. Given
E b 1
> E b 2
, what is the maximum amount of energy stored in this capacitor?
(3) When the lateral dimensions are sufficiently larger than d , we can ignore the fringe effect to calculate the capacitance, i.e., we assume an electric field distribution at the peripheries of the capacitor as shown in Figure (b). However, the fringe effect is unavoidable. The actual situation is something as shown in Fig. (c). Show that the approximation in Figure (b) is unphysical, i.e. it violates some fundamental physical law.
(4) Find the electric fields E
1
and E
2
in the two dielectrics. Show that your results of E
1 and E
2
are physical and thus there is no fringe effect at the boundary between the two dielectrics, i.e., the kind of field distribution at the peripheries of the capacitor shown in Figure (c) does not occur at this boundary.
E
1
E
2
(b)
8
Solution :
(1)
E
= ½ CV
2
= ½( ) V
2
4 points
4 points
(c)

8
ECE 341 Test 2 Spring 2016
(2)
Student gets full points if final answer right. Partial credit as follows:
3 points
2 points
8 (3)
8 (4)
E
1
= E
2
= V/d . Directions of E
1
and E
2
are shown in Fig. (a).
For a rectangular loop (blue) shown in Fig. (a),
Therefore, E
1
= E
2
is physical. No fringe effect at the boundary.

4
ECE 341 Test 2 Spring 2016
16 points total
4. The ( x , y ) plane separates two dielectric media 1 and 2 of dielectric constants
 r 1
and
 z r 2
, respectively. Medium 1 fills the space where z < 0, and medium 2 fills z > 0. On the
< 0 side of the interface, the electric filed E
1
= ˆ E
0

E
0
.
What is the angle between E
1
and the z axis?
If

 r r 1
2

3 , find the electric field E
2
on the z > 0 side of the interface. What is the magnitude E
2
= | E
2
|? What is the angle between E
2
and the z axis?
Solution :
6
6

ECE 341 Test 2 Spring 2016
Total 14 points + 20 bonus points
5. An infinitely large conductor sheet is at the y = 0 plane, carrying a surface current density J
 
J z , as shown in Fig. (a).
(a) What is the unit of J ? Find the magnetic field ( B ) distribution in the entire space
(for y > 0 and y < 0). Draw the magnetic field lines (indicating directions with arrows) in Fig. (a).
(b) (Optional) Now, let’s place another conductor sheet at y = d , carrying a surface current density J

J z , as shown in Fig. (b). Find the magnetic field ( B ) distribution in the entire space. Draw the magnetic field lines (indicating directions with arrows) in Fig. (b).
(c) (Optional) Do the two sheets in Fig. (b) attract or repel each other? What is the force per unit area between the two sheets?
(a) (b) y y d x x x x x x x x l z x z x
Solution:
14 (a) The unit of J is A/m. 4 points for the unit
Directions of B for y > 0 and y < 0 are shown in Figure (a). 5 points for directions
Draw a rectangular loop with length l and width w as shown. (gets points if student indicates
2 lB =

Jl . directions with vector
Therefore, B =

J /2.
5 points expressions)
(b) For z < 0 and z > d ,
6 bonus points
B = 0.
2 points
For 0 < z < d ,
B =

J
2 points
The direction of B is shown in Figure (b). 2 points
(c) They repel each other. 4 points
The upper sheet is in the field B =

J /2, with the direction shown by red arrows, due to the lower sheet.
For a rectangle in the upper sheet, with a width (perpendicular to J ) w , and a length
(along the direction of J ) l , the force is
F = IlB = JwlB = Jwl

J /2 =

J
2 wl /2.
Therefore, the force per area is
F =

J
2
/2 10 points