Solutions to Exam I
PHY 2049, Summer C, 2007
In one mole there are 6x1023 atoms. Each atom has two electrons. Therefore there are
1.2x1024 electrons, which carry the charge 1.2x1024x 1.6x10-19 C = 1.9 x 105 C.
Suppose that the unknown charge is Q. At the first touch of sphere C with sphere A, half
of the charge is transferred to C, leaving charge Q/2 on A. At the next touch of sphere C,
carrying charge Q/2 with sphere B, which carries charge Q, Q/4 is transferred to C,
leaving sphere A charged to Q/2 and sphere B charged to 3Q/4. If the distances remain
the same, the force F2 = 3F/8.
You now know that the act of touching equates the potential and the capacitance of a
sphere of radius R is C = 4πε0 R. Suppose also that the radius of sphere C is R/2. What
is the magnitude of the final force in terms of F?
Consider the electric field (you can always get force from that by using F = q E), We
want E to be zero.
If q1 and q2 are unlike, then the force between them is never zero. The vector
contributions fro the two charges always add. They must be alike.
In that case
q1
q
= 22
2
a
4a
q 2 = 4q1 .
On the oil drop qE + mg = 0. Here the electric field and the gravity are both pointed
downwards. Therefore q = -mg/E = -6.5 × 10-6 C.
Again we will look at the electric Field. In region II, the electric fields due to the two
unlike charges add. In region III, the larger negative charge dominates since it is also
closer. The electric fields (and the forces on the third charge) can cancel only in region I.
Here the electric field due to the positive charges would be in the IV quadrant while the
one due to the negative charges would be in the III quadrant. Finally all other things
being the same, the cancellation would lead to a field in the –y direction.
A hollow conductor will have all its charge on the outside surface. The uncharged metal
ball will touch the inside surface which has no charge. Hence once the ball is removed, it
will remain uncharged.
Electric field inside a conducting shell is always zero.
This information used with Gauss’ theorem tells us that the charge on the inner surface of
the shell will be –q. On the outer surface the total charge is Q + q.
There are six surfaces in a cube. The flux through each of them should be the same. The
total flux is q/εo. Hence the flux through each one must q/6εo.
−15
⎡ 2q q q ⎤
−3
9 7 × 10
=
10
+
9
×
10
= 2.6 mV
The potential at point P, V = V (∞) + k ⎢ − −
⎥
2 × .02
⎣ d d 2d ⎦
The battery deposits a charge q = C1V = 1.4 ×10-10 ×60 = 8.4 ×10-9 C on the original
capacitor. When this capacitor is disconnected form the battery and the new one of
capacitance C2 connected in parallel, the new combination has an equivalent capacitance
C1 + C2. Therefore:
q = (C1 + C 2 )V
(
8.4 × 10 −9 = 48 1.4 × 10 −10 + C 2
)
C 2 = 35 pF
∂V
= −6 x
∂x
∂V
Ey = −
= 8y
∂y
At (3,3), the electric field is (-18,24) V/m
Ex = −