2049 Spring 2005
Quiz 1
Name:
Period:
Question 1
The figure below shows two charged particles on an axis. The charges are free to move.
However, a third charged particle can be placed at a certain point such that all three particles are then in equilibrium. (a) Is that point to their left, to their right, or between them?
(1 point) (b) Should the third particle be positively or negatively charged? (1 point) (c) Is
the equilibrium stable or unstable? (1 point)
−5q
−q
(a) The only place in which the two negative charges can exert a
force to cancel each other out is in between the two negative charges.
(b) Again, the only charge that can produce a net force of zero in between is a positive
charge.
(c) The equilibrium is unstable. If the positive charge is displaced in either direction, it will
not return to the equilibrium point, much like a a car on the top of a hill will not return to
the top of a hill once nudged off.
Question 2
In the figure below, two identical circular nonconducting rings are centered on the same
line. For three situations, the uniform charges on rings A and B are, respectively, (1) q0 and
q0 , (2) −q0 and −q0 , and (3) −q0 and q0 . Rank the situations according to the magnitude of
the net electric field at (a) point P1 midway between the rings (1 point), (b) point P2 at the
center of ring B (1 point), and (c) point P3 to the right of ring B and equidistant to point
P2 as is P1 (1 point), greatest first.
P1
Ring A
P2
P3
Ring B
There were two interpretations: either you could rank the situations according to the points,
or according to the charge scenarios. The former is obtained by noticing two things: the
electric field due to a circular ring is,
Ering (z) =
kqz
,
(R2 + z 2 )3/2
(1)
and since they are identical, we need only consider how the magnitude behaves with z, the
distance along the axis from the center of the ring (note also that at any ring center, the
electric field due to the ring alone is zero). The direction of the field is dictated by the charge
on the ring. The magnitude decreases as z −2 . The situations therefore breakdown as follows:
1. q0 , q0 : P3 , P2 , P1
2. −q0 , −q0 : same as (a) by symmetry.
3. −q0 , q0 : P1 , P3 , P2 .
and if you are ranking by the charge situations at each point then,
1. P1 : 3, 1 = 2,
2. P2 : 1 = 2 = 3,
3. P3 : 1 = 2, 3.