Physics 114 Exam 1 Fall 2012
Answer each of the following questions and each of the problems. Points for each
question and problem are indicated in red with the amount being spread equally among
parts (a,b,c etc). Be sure to show all your work.
Questions
1. (10 points) A negatively charged rod is brought close to a small metal ball suspended by
a thin thread, but doesn’t touch it. (a) What does the sphere do, if anything. (b) Why
does it do this or why does it not do anything? You can draw a picture. Now the rod
touches the sphere and is then pulled away. (c) What does the sphere do now when the
rod is brought close again, but doesn’t touch it? (d) Why?
(a) It will move towards the rod.
(b) The charges in the sphere will polarize; the electrons will move away from the sphere
leaving the near-side positively charged. The near-side, being closer to the rod, is
attracted to the rod as opposite charges attract. This attraction is greater than the
repulsion of the far-side of the sphere since the positive, near-side is closer to the rod.
(c) It moves away.
(d) Touching the sphere imparts excess negative charge onto the sphere so now the two
objects have the same charge and they repel.
2. (10 points) Draw the electric field due to (a) a positive point charge, (b) a negative point
charge, (c) a system of two point charges with equal magnitude charge but opposite sign
(positive and negative).
3. (10 points) A charge of 5 C is placed exactly in the center of a cube that has a length of 1
m on each side. (a) What is the total flux through the cube? (b) What is the flux through
each of the six sides of the cube? (c) If the charge is moved to another place in the cube
(not the center), does the total flux through the cube change? How about the flux through
each side? (d) If the charge is now moved outside of the cube, what is the total flux
through the cube?
(a) Gauss law tells us that the total flux through the cube is the charge enclosed over 0. So
  q /  0  5C / (8.85 1012 C2 / Nm 2 )  5.6  1011 Nm 2 / C
(b) Due to the symmetry, the same flux goes through each of the six sides. Thus, we have
  9.5 1010 Nm 2 / C
(c) The total flux through the cube will not change as Gauss’ law tells us that the total flux
only depends on the charge enclosed. However, the flux through each side will change;
more field lines will go through sides that are closer to the charge than those sides further
away from the charge.
(d) The flux would be zero since all the field lines that go in also come out.
4. (10 points) A spherical rubber balloon has a charge uniformly distributed over its surface.
As the balloon is inflated, how does the electric field E vary (a) outside the balloon, at
some point well away from the surface? (b) at the outer surface of the balloon? (c) inside
the balloon? Assume the balloon remains spherical during inflation.
(a) The electric field will not change. This system has spherical symmetry. The electric
field is the same at all radial distances outside the sphere. Drawing a spherical
Gaussian surface at r and applying Gauss law shows E does not change as the charge
inside does not change.
(b) The field decreases. The field at the surface depends on the surface charge density
(more charge density, more field). Gauss’s law shows it is 0. As the balloon
expands, the charge spreads out so the surface charge density decreases. Thus, the
field decreases.
(c) It stays zero. Using a Gaussian sphere inside, we see E = 0 as the charge inside is
zero.
Problems
1. (15 points) Charge q1 = +2 C is placed at the position (x,y) = (3 cm,4 cm), making an
angle of 53.13o above the positive x-axis . Charge q2 = -3 C is placed at the position
(x,y) = (-3 cm,4 cm), making an angle of 53.13o above the negative x-axis. Note that q2
has a negative change. (a) Calculate the electric field at the origin due to q1 and q2. (b) If
a third charge q3 = +5 nC is placed at the origin, what would the force on it be due to the
other two charges.
| q|
. We need to
4  0 r 2
add the vector electric field by resolving the fields and adding the components in the
x and y directions.

 | q1 |
   | q1 |

| q2 |
| q2 |
E tot  
cos(
)
cos(
)
i
sin(
)
sin(
)









1
2
1
2  j
2
2
4 0 r2 2
4 0 r2 2
 4 0 r1
  4 0 r1

Looking at the diagram of the electric fields (obtained by noting E1 points away from
q1 and E2 points towards q2) we see they both have negative x components and E1 has
a negative y component while E2’s is positive. This can be accounted for by using
proper angles measured from the positive x-axis so, in the equation for Etot, 2 is
126.87 and 1 is 233.13.
We have
| q1 |
2  106

 7.19 106 N/C
4 0 r12 4 (8.85 1012 )(0.05) 2
(a) The magnitude of the electric field due to each charge is E 
and
| q2 |
3  106

 1.08  107 N/C
2
12
2
4 0 r2
4 (8.85  10 )(0.05)



E tot   (7.19 106 )(0.6)  (1.08 107 )(0.6)  i   (7.19 106 )(0.8)  (1.08 107 )(0.8)  j


  (4.31106 )  (6.48  106 )  i   5.75  106 )  (8.64  106 )  j


 10.79 106 i  2.89 106 j N/C
This has a magnitude of the square root of the sum of the squares of the components
which is 1.12 x 107 N/C
(b) Since F = qE, we just have to multiply the result from part (a)





Ftot  (5  109 C)(  10.79 106 i  2.89  106 j N/C) =  5.40  102 i  1.45 10 2 j N
This has a magnitude of the square root of the sum of the squares of the components
which is 5.59 x 10-2 N
2. (10 points) An electric field of magnitude 3.79 kN/C is applied along the x axis. Calculate
the electric flux through a rectangular plane 0.350 m wide and 0.700 m long in the
following cases. (a) the plane is parallel to the yz plane. (b) the plane is parallel to the xy
plane
(c) the plane contains the y axis and its normal makes an angle of 40 with the x axis.
(a) Here the normal is in the x-direction, parallel to the field and maximum flux goes
through the surface. The flux is simply EA = (3.79 X103)(0.35)(0.7) = 9.29 x 102
N-m2/C
(b) Now the normal points towards the z-axis and is perpendicular to the field so the
flux is zero.
(c) We have the flux is EAcos(40) = 7.11 x 102 N-m2/C
3. (20 points) A solid insulating sphere of radius a = 5.00 cm carries a net positive charge of Q =
3.00 µC uniformly distributed throughout its volume. Concentric with this sphere is a conducting
spherical shell with inner radius b = 10.0 cm and outer radius c = 15.0 cm as shown in the figure
below, having net charge q = −1.00 µC. Use Gauss’ Law to calculate the electric field at (a) r =
20 cm, (b) r = 12.5 cm, (c) r = 7.5 cm and (d) r = 3 cm.
For all the parts we see that there is spherical symmetry and so we will use spherical
Gaussian surface at a radius that we want to calculate the field for each part. We will get
 
that  E.dA  EA
(a) EA = q/0. A = 4(0.2)2 = 0.5 m2. The charge inside is the total charge 3-1 C = 2
C. Then, solving for E we have E = 2 x 10-6/[(8.85 x 10-12)(0.5)] = 4.52 x 105 N/C.
This points outward radially.
(b) E = 0 inside the conductor.
(c) EA = q/0. A = 4(0.075)2 = 0.071 m2. The charge inside is just the charge of the
inner sphere which is 3 C. Then, solving for E we have E = 3 x 10-6/[(8.85 x 1012
)(0.071)] = 4.77 x 106 N/C, radially outward.
(d) EA = q/0. A = 4(0.03)2 = 0.011 m2. The charge inside is just the charge of the
inner sphere that is contained within the Gaussian surface. This will be q = V
3
3 C
4  (0.03)3  3 C (0.03)

4  (0.05)3 3
(0.05)3
3
=6.48 x 10-7 C
Then, solving for E we have E = 6.48 x 10-7/[(8.85 x 10-12)(0.011)] = 6.6 x 106 N/C,
radially outward.

