PHY 2054 Midterm: 2054_Spring14_Exam3-Solutions
PHY2054 Spring 2014
1
Prof. Paul Avery
Prof. Pradeep Kumar
Apr. 26, 2014
Exam 3 Solutions
1. Four charges are placed along a straight line each separated by a distance L from its neighbor.
The order of the charges is +Q, –Q, –Q, +Q. What is the total potential energy of the system
(relative to infinity)?
Answer:
−5kQ2/ 3L
Solution: Let the charges be ordered 1, 2, 3, 4. There are 6 combinations: (12, 23, 24), (13, 24),
14 which give contributions
2−kQ2/L
( )
+ +kQ2/L
( )
+
2−kQ2/ 2 L
( )
+
+kQ2/ 3L
( )
=
−5kQ2/ 3L
for the total potential energy.
2. A wire of length L has a resistance of 60Ω. The wire is uniformly stretched to five times its
original length, maintaining constant volume. If a length L is now cut from the stretched wire,
what is the resistance of the remaining piece?
Answer: 1200Ω
Solution: Stretching the wire at constant volume gives 5 times the length and 1/5 the cross
sectional area, giving a total resistance of
60 ×5×5=1500Ω
. After the piece is cut off, the
resistance of the remaining wire is 4/5 of the new resistance or
1500 ×4
5=1200Ω
.
3. In the circuit shown in the figure, what is the potential difference
Va - Vb between points a and b?
Answer: +12.0 V
Solution: This is two-loop problem. Let the top loop have a
clockwise current i1 and the bottom loop have a clockwise current i2.
Thus the current going through b is i1 + i2 moving to the left. The
equations describing the two loops are
16 −4i
1+12i2−18 =0
(top) and
18 −12i2−8i
1+i2
( )
=0
(bottom). Solving yields
i
1=1.0
and
i2=0.5
, so the current flowing to the left through point b is
1.5 A. Starting from point a the voltage falls by
8×1.5 =12
volts, thus
Va−Vb= +12.0
volts.
There are other ways of defining the loops (e.g., defining i2 as the current flowing through point
b to the left), but the current moving through point b to the left is always 1.5 A.
++
a
b
16 V
18 V
12Ω
4Ω
4Ω
4Ω
PHY2054 Spring 2014
2
4. Positive point charges q and 4q are located at x = 0 and x = 4d, respectively. What is the
magnitude of the electric field at x = d?
Answer:
5kq / 9d2
Solution: For positive q, the charge at x = 0 generates an E field of
kq /d2
to the right while the
other charge gives an E field of
4kq / 9d2
in the opposite direction, so the magnitude of the net
field is
5kq / 9d2
.
5. Three point charges are placed as shown at the corners of a right triangle
(L = 2.0 cm, q = 4.0 µC). What is the y component of the force on the +q
charge?
Answer: +39 N
Solution: The y component of the force from the –q charge is
−kq2/ 4L2
.
The corresponding y force component from the +2q charge is
2kq2/ 5L2
( )
sin
θ
, where
sin
θ
=2 / 5
. The total y force component is thus
kq2/L2
( )
4 5 / 25 −1 / 4
( )
= +39
N.
L
2L
+2q
-q
+q
Document Summary
Exam 3 solutions: four charges are placed along a straight line each separated by a distance l from its neighbor. The order of the charges is +q, q, q, +q. Solution: let the charges be ordered 1, 2, 3, 4. There are 6 combinations: (12, 23, 24), (13, 24), / 3l: a wire of length l has a resistance of 60 . The wire is uniformly stretched to five times its original length, maintaining constant volume. Solution: stretching the wire at constant volume gives 5 times the length and 1/5 the cross sectional area, giving a total resistance of 60 5 5 = 1500 . After the piece is cut off, the resistance of the remaining wire is 4/5 of the new resistance or. = 1200 : in the circuit shown in the figure, what is the potential difference. Let the top loop have a clockwise current i1 and the bottom loop have a clockwise current i2.
