PHYS 102 – General Physics II
Summer 2015 – 16, Midterm Exam 1 Solutions
11 June 2016, 13:30
1. Two point charges are placed on 𝑥 axis as shown in
the figure. The first one has a charge +2𝑄 and
coordinate (– 𝑎, 0). The second one has a charge −𝑄
and coordinate (𝑎, 0). Please use vector notation to
answer the following questions, that is, if you have a
vector 𝐴⃗ it should be written as 𝐴⃗ = 𝐴𝑥 𝑖̂ + 𝐴𝑘 𝑗̂ + 𝐴𝑧 𝑘̂ .
(a) (8 Pts.) What is the force on the first particle due to
⃗⃗⃗⃗⃗⃗⃗
⃗⃗⃗⃗⃗⃗⃗
the second particle (𝐹
12 ) and vice versa (𝐹21 )?
Solution:
F12 
(2Q)(Q) ˆ
Q2 ˆ
i

i.
4 0 (2a) 2
8 0 a 2
F21  F12 
Q 2 ˆ
i.
8 0 a 2
y
(b) (9 Pts.) What is the total electric field along the line
connecting the two point charges (as a function of 𝑥)?
2Q
Solution:
2Q
Q
ˆi , E 
ˆi .
E1 
2
2
2
4 0 (a  x)
4 0 (a  x)
E  E1  E2 
𝑎+𝑥
x
𝑎−𝑥


Q  3a 2  3x 2  2ax  ˆ
i.
4 0   a 2  x 2 2 


⃗𝑬⃗1
y
(c) (9 Pts.) What is the total electric field along the
𝑦 axis (as a function of 𝑦)?
Solution:
y
ඥ𝑎2
2Q




2Q
E1 
cos  ˆi  sin  ˆj ,
4 0 (a 2  y 2 )
E2 
-Q
.
x
+
𝑦2
.
𝜃
a
⃗𝑬⃗2
ඥ𝑎2 + 𝑦 2
-Q
𝜃
x
a
Q
a
y
cos  ˆi  sin  ˆj , cos  
, sin  
2
2
2
2
2
4 0 (a  y )
a y
a  y2
E  E1  E2 


Q
3a ˆi  y ˆj .
4 0 (a 2  y 2 )3/2
(d) (8 Pts.) Draw the electric field lines. Comment on the meaning of your drawing and its relationship to
your answers in part (b) and (c).
(d) (8 Pts.) Solution:
The electric field lines along the 𝑥 axis are closer together in the left hand side compared to the right hand
side. This is in accordance with the answer in part (b). The electric field lines along the 𝑦 axis are
pointing towards right with a little contribution in 𝑦 direction. This is in accordance with the answer in
part (c).
2. (a) (10 Pts.) A uniform field E is parallel to the axis of a hollow hemisphere of
radius r. What is the electric flux through the hemispherical surface?
Solution:
From the figure, we see that the flux outward through the hemispherical
surface is the same as the flux inward through the circular surface base of the
hemisphere. On that surface all of the flux is perpendicular to the surface. So, we say that on the circular base, E A.
Thus  E  E A   r 2 E.
(b) (23 Pts.) Two thin concentric spherical shells of radii r1 and r2  r1 < r2  contain uniform surface charge densities
 1 and  2 , respectively. Determine the electric field for 0 < r < r1 , r1 < r < r2 , and
r > r2 . Under what conditions will E  0 for r > r2 ? Neglect the thickness of the
shells.
Solution:
In the region 0  r  r1 , a gaussian surface would enclose no charge. Thus, due to the spherical symmetry, we have
 E dA  E  4 r  
2
Qencl
0
 0  E  0.
In the region r1  r  r2 , only the charge on the inner shell will be enclosed. Therefore
 E dA  E  4 r  
2
Qencl
0
 1 4 r12

0
 1r12
 E
.
 0r 2
In the region r2  r, the charge on both shells will be enclosed. So


2
 E dA  E 4 r 
Qencl
0

 1 4 r12   2 4 r22
0
 E
 1r12   2 r22
.
 0r 2
To make E  0 for r2  r, we must have 1r12   2 r22  0. This implies that the shells are of
opposite charge.
3. The electric potential in a region of space is given by 𝑉(𝑥, 𝑦, 𝑧) = 𝐴(2𝑥 2 − 3𝑦 2 + 𝑧 2 ),
where 𝑉 is in volts, the coordinates are in meters, and 𝐴 is a constant.
a) (11 Pts.) Find the electric field 𝐸⃗⃗ in this region (express the result in terms of unit vectors).
Solution:
V
V
V
E  Ex ˆi  Ey ˆj  Ez kˆ , where Ex  
 4 Ax , E y  
 6 Ay , Ez  
 2 Az .
x
y
z
Hence


E  2 A 2 x ˆi  3 y ˆj  z kˆ .
1
2
b) (11 Pts.) If the work done by the field on a 4-𝜇C test charge when it moves from point ( , 0,1) m to the origin
(0,0,0) is 3.10−5 J, what is 𝐴? (𝑁𝑜𝑡𝑒: 1 𝜇 = 10−6)
Solution:
WE  U  q0 V  q0 (V f  Vi ) . Vi  V ( 12 , 0,1) 
WE  q0 V 
3A 2
m , V f  V (0, 0, 0)  0 . Since q0  4 106 C ,
2
3
q0 A  3 105 J and we have A  5V / m2 .
2
1
2
c) (11 Pts.) Find the magnitude of the electrostatic force acting on the test charge at the point ( , 0,1) m.
Solution:
E( 12 , 0,1)  10(ˆi  kˆ ) V / m  E  10 2 V / m . F  q0 E  4 2 105 N .