Physics 114 Exam 2 Fall 2015
Name: ___________________________________________________________
For grading purposes (do not write here):
Question
Problem
1.
1.
2.
2.
3.
3.
Answer each of the following questions. Points for each question are indicated in red.
Unless otherwise indicated, the amount being spread among parts (a,b,c etc) are equal.
Be sure to show all your work.
Use the back of the pages if necessary.
Question 1. (10 points) Describe the motion of a proton after it is released from rest in
a uniform electric field: (a) How does its kinetics energy change (increase, decrease or
stay the same)? (b) How does the potential energy of the proton-field change if at all?
(c) Describe the difference in electric potential (Vf – Vi) between the initial and final
positions of the proton (is it zero, positive, or negative)?
How do the answers to all of these questions change if an electron is released instead
of a proton (d) Kinetic energy, (e) potential energy, and (f) electric potential?
Solution
(a) It’s kinetic energy increases due to the acceleration from the force of the electric
field.
(b) The potential energy must decrease due to conservation of mechanical energy.
This is like dropping a mass in the earth’s gravitational field.
(c) The difference is electric potential is negative. The acceleration is in the
direction of the electric field since the force on the proton is in this direction. The
electric field points in direction going from high to low potential. Also,
f
 

V    E .ds   E .s . The angle between the displacement vector and the electric
i
field is zero degrees so Vf – Vi is negative.
(d) The electron just goes in the opposite direction. Kinetic energy still increases.
(e) The potential energy decreases.
(f) Now the displacement is in the opposite direction of the electric field. Vf – Vi is
positive.
Question 2. (10 points) Two capacitors with capacitances C1 = 2 F and C2 = 4 F are
connected in parallel to a 12 V battery.
(a) What is the ratio of the voltages on each capacitor, V1/V2?
(b) What is the ratio of charges on each capacitor, Q1/Q2?
(c) What is the ratio of the energy stored in each capacitor, U1/U2?
Now the capacitors are connected in series to the 12 V battery.
(d) What is the ratio of the voltages on each capacitor, V1/V2?
(e) What is the ratio of charges on each capacitor, Q1/Q2?
(f) What is the ratio of the energy stored in each capacitor, U1/U2?
Solution.
(a) The ratio is 1. Parallel devices have the same voltages across them.
(b) Q1/Q2 = C1V/C2V = C1/C2 = 0.5.
(c) U1/U2 = (½C1V2)/ (½C2V2) = C1/C2 = 0.5.
(d) V1/V2 = (Q/C1)/(Q/C2) = C2/C1 = 2.
(e) Q1/Q2 = 1. Series capacitors have the same charge on them.
(f) U1/U2 = (Q2/2C1)/(Q2/2C2) = C2/C1 = 2.
Question 3. (10 points) Consider the circuit below in which there are 3 identical light
bulbs, labeled A, B, C. In answering each question, include an explanation.
C
A
B
(a) How does the brightness of B compare to that of C?
Brightness goes as the power which we can write as I2R. Both bulbs have the same resistance
and the same current since they are in series, so the brightness is the same.
(b) How does the brightness of A compare to that of B?
We can also express the power as V2/R. The entire emf is across bulb A but this same emf is
split between B and C. Thus, A is brighter than B.
(c) When A is unscrewed, what happens to the brightness of bulbs B, C?
Nothing happens to the brightness of B and C as these are in parallel with bulb A (like in your
house), so bulbs B and C still have the same voltage across them.
(d) When B is unscrewed, what happens to the brightness of bulbs A and C?
Nothing happens to the brightness of A as it is in parallel. Bulb C goes out as the resistance in
that branch of the circuit goes to infinity.
-------(e) If a capacitor were in series with bulb A (see figure below), how does the brightness of A
immediately when the switch is closed compare to when it has been closed for a very
long time?
When the switch is closed, there is no charge yet in the capacitor and current flow through it as if
it were a wire. After a certain amount of time, the capacitor is fully charged and no current flows
through it. Thus, Bulb A is brightest immediately after the switch is closed and then dims to
nothing as time goes on.
(f) In this same circuit, how does the brightness of B immediately when the switch is closed
compare to when it has been closed for a very long time (assume a tungsten filament with
a positive temperature coefficient of resistance, )?
The voltage across B is constant once the switch is closed. Its resistance increases over time due
to heating. Thus, the P=V2/R decreases over time, and it is brightest immediately upon closing
the switch.
C
A
B
Problem 1. (15 points) Consider the system of two charges +q along the y-axis shown
in the Figure below (ignore the presence of the charge –Q for now). Let q = 2 C and let
d = 8 m.
(a) Calculate the electric potential at a point on the x-axis a distance x = 3 m from
the origin, due to the two 2C charges.
(b) Calculate the electric potential at the origin.
(c) Now let the magnitude of the negative charge Q = 3mC which has a mass of 2
Kg be released from rest from the point along the x-axis at x = 3 m. What would
the speed of the charge be at the origin?
Solutions
n
1 n qi
(a) V   Vi 
 r = 9 ×109(2/5 + 2/5) V
4
0 i 1 i
i 1
= 7.2 ×109 V. The distance of 5 m was obtained
using the pythogorean theorem.
n
1 n qi
(b) V   Vi 
 r = 9 ×109(2/4 + 2/4) V
4
0 i 1 i
i 1
= 9 ×109 V
(c) Uf + Kf = Ui +Ki. Ki = 0 since it is released from rest.
So we have
U = -Kf
qV = -½ mv2
(-0.003)(1.8 × 109) = - v2
v = 2.3 ×103 m/s
Problem 2. (15 points) Consider the circuit below,
(a) What is the equivalent capacitance of the four capacitors?
(b) Find the charge on each of the four capacitors.
(c) Find the voltage across each of the four capacitors.
Solution
Problem 3. (15 points) Consider the circuit below.
(a) Determine if there are any resistors in series or in parallel.
(b) Use Kirchoff’s first law applied at point c to obtain one equation relating the three
currents, I1, I2, and I3.
(c) Apply Kirchoff’s loop rule to the top loop to obtain an equation relating I1 and I2.
(d) Apply Kirchoff’s loop rule to the top loop to obtain an equation relating I1 and I3.
(e) Given that I1 = 2A, find the other two currents.
Solutions
(a) None of the resistors here are in series or
parallel.
(b) I1 + I2 = I3.
(c) -14+6*I1-10-4*I2 = 0. -24 +6*I1 – 4*I2 = 0
(d) 10-6*I1-2*I3 = 0.
(e) Taking the last equation we got, we have
10-12= 2*I3
I3 = -1 A
Now taking the junction rule equation, we have
I2 = I3 – I1 = -1-2 = -3A
So I2 and I3 are actually going the opposite
direction as originally guessed.
Possibly Useful Information
1 | q 1 || q 2 |
4  0 r 2
e = 1.6 X 10-19 C
 0  8.85 X 10-12 (C 2 / N  m2 )
| q|
, E = /
4  0 r 2
x = x2 - x1, t = t2 - t1
 
 0    0  E. dA  q enc
s = (total distance) / t
v = dx/dt
a = v / t
a = dv/dt = d2x/dt2
v = vo + at
g = 9.8 m/s2

r = xi + yj + zk
  
 r = r2 - r1
F
E
x-xo = vot + (½)at2
v2 = vo2 + 2a(x-xo)
 
E  Fq
0
v = x / t
x-xo = vt -1/2at2


a = dv / dt

r = (x 2 - x 1 ) i + (y 2 - y 1 ) j + (z 2 - z1 ) k




v = r / t , v = dr / dt


a = v / t
U = Uf - Ui = -W
V = Vf - Vi = -W/q0 = U/q0
f
 
Vf  Vi    E. ds
U=-W∞
V = -W∞/q0
f
 
V    E. ds
x-xo = ½( vo+ v)t
i
1 q
V
4  0 r
Uf + Kf = Ui +Ki
K = ½ mv2
V
Es 
s
V
E
s
Q = CV
C  2 0
l
ln(b / a)
i
n
1
4 0
i 1
1
dq
V

4 0 r
V   Vi 
n
q
 ri
i 1
i
V
V
V
;Ey 
;Ez 
x
y
z
1 q1q2
U  W 
4 0 r12
 0A
C
d
ab
C  4 0
ba
Ex 
C  4  0 R
C eq   C j (parallel)
1
1

(series)
C eq
Cj
Q2 1
U
 2 CV 2
2C
C =  C0
u  12  0 E 2
I= dQ/dt
  1
R
L
P = IV
A
Pemf = I
1
1

(parallel)
R eq
Rj
I = (R)e-t/RC
I = (QRC)e-t/RC, I0 = (QRC)
E = o
V = IR
P = I2R=V2/R
I 
(R  r)
R eq   R j (series)
q(t)= Q(1-e-t/RC)
q(t) = Qet/RC
 = Q/L, = Q/A,  = Q/V