Physics 114 Exam 2 Fall 2014
Solutions
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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.
1. Question 1. (10 points) Consider two charged conducting
spheres of radii, r1 = 0.5 m and r2 = 1 m. The two spheres are
connected by a 10 m long wire and are at equilibrium. The
total charge on the spheres is Q = 5C and each one has charge
q1 and q2 respectively.
(a) How much work would be required to move a 5 nC test
charge along the wire from the surface of the small sphere
to the surface of the large sphere?
(b) How much work would be required to move a 5 nC test
charge from the center of the small sphere to the surface same small
sphere?
Compare [just say which is bigger or are they the same] the
(c) potential at the surfaces of the spheres,
(d) the electric field at the surface of the spheres,
(e) the total charge on the spheres,
(f) surface charge density on the spheres (state which is greater sphere 1 with
radius r1 or sphere 2 with radius r2)?
Solutions
(a) The potential inside a conductor is a constant. Thus, U = 0 and it takes
no work to move the charge inside the sphere. The electric field is also
zero – so there is no force. The answer is zero.
(b) The whole system is at the same potential. . Thus, U = 0 still and the
answer is zero.
(c) The two spheres are at the same potential. Charge will move until
equilibrium is reached and everywhere in a conductor is at the same
potential.
(d) The electric field is greater at the surface of the smaller sphere, E1 > E2.
Firstly, we know E is greatest at places with the largest curvature.
Secondly, we know
V2 = V1 so E2r2 = E1r1 so E2/E1 = r1/r2, so E1 > E2.
(e) Since V2 = V1,
keQ1
r1
(f)  1 

keQ2
Q1
, 2 
2
r2
4 r1
, and
Q1 r1
 , so Q1<Q2.
Q2 r2
Q2
4 r2
, and
2
 1 Q1r2 2 r2

 , so 1>2
 2 Q2 r12 r1
Question 2. (10 points) Consider the circuit below with  = 8 V, R = 4 , and C = 10 F.
Assume the battery is ideal. Immediately after the switch is closed (after having been
open for a long time),
(a) What is the current through each resistor?
(b) What is the current through the battery
(c) What is the potential difference across the
capacitor
(d) What is the capacitance of the capacitor?
After the switch has been closed for a long time,
(e) What is the current through each resistor?
(f) What is the current through the battery
(g) What is the potential difference across the
capacitor
(h) What is the capacitance of the capacitor?
Solutions
(a) Immediately after the switch is closed, current flows freely through the capacitor.
The circuit behaves as if the capacitor is not there. The current through each
resistor is /R = 8/4 = 2 A
(b) The current through the battery must be the sum of these currents = 4A
(c) There is no chage across the capacitor, V = Q/C = 0.
(d) The capacitance does not change. It is always 10 F.
(e) Now there is no current through the resistor on the right. The current through the
one on the left is /R = 8/4 = 2 A.
(f) It is 2 A as this is now a single loop circuit as far as current is concerned.
(g) The potential across the capacitor must be equal to  = 8 V.
(h) Still 10 F!
Question 3. (10 points) You connect an ideal battery to a single resistor,
producing a potential difference V across it and causing a current to flow through
the resistor. Next, the resistor is removed from the circuit and cut in half crosswise
(so its length is halved). One of the halves is placed back into the circuit, with the
battery connected to it. Compared to before the resistor was cut in half
(a) What happens to its resistance?
(b) What happens to the potential difference across the resistor?
(c) What happens to the current across the resistor?
(d) What happens to the power output of the battery?
Solutions
L
(a) We have R 
A , and the length has decreased by a factor of 2, so the resistance is half
of its original value.
(b) The potential across the resistor is the same – it comes from an ideal battery.
(c) Since V = IR, and R is halved, I is doubled.
(d) The power is can be expressed as IV so it doubles. It is also V2/R, so it doubles.
Problem 1. (15 points) A charge of +2 C and mass 20 Kg moves from point A to point B
in an electric field a distance d = 5 m as shown below. The magnitude of the electric
field is 100 V/m. The charge is released from rest (initial velocity is zero).
(a) (3 points) What is the change in potential in going from point A to B?
(b) (4 points) What is the final speed when it gets to point B?
(c) (2 points) Which point is at a higher potential (A or B)?
(d) (3 points) What is the work done by the electric field for this displacement from A
to B?
(e) (3 points) After reaching point B, the charge is moved horizontally, perpendicular
to the electric field, to point C through a displacement of 4 m. What is the work
done by the electric field for this displacement from B to C?
©
Solutions
(a) V = -Ed = -(100V/m)(5m) = -500 V
(b) Potential energy is converted to kinetic energy. U = -K. qV = -½mv2. The
change in potential is negative and we have
v = 2qV
= 2 * 2 *500 = 10 m/s
m
20
(c) Point A is at a higher potential (E always points from high to low potential)
(d) The work done is -U = -(qV) = 1000 J.
(e) Here E is perpendicular to the displacement, so the work is zero.
Problem 2. (15 points) For the system of four capacitors shown below,
(a) find the equivalent capacitance of the system,
(b) the charge on each capacitor,
(c) the potential difference across each capacitor.
Solutions
Problem 3. (15 points) Consider the circuit shown below.
(a) Which resistors are in series and which are in parallel (if any)
(b) Use Kirchoff’s first law to obtain one relation between the three currents
(c) Use Kirchoff’s second law to obtain two additional equations that can be used to solve
for the three currents.
(d) Given that I2 = 1.915 amps, find the other two currents.
Solutions
(a) The two 2  resistors are in series. No other resistors are in series on in parallel.
(b) I1 = I2 + I3
(c) 12 -3 I1 + 6 – 5 I2= 0 → 18 -3 I1 – 5 I2= 0
5 I2 – 6 – 4 I3= 0
(d) Taking the third equation and solving for I2 we get I3 = (5*1.915 - 6)/4 = 0.8938.
Taking the first equation we have I1 = 0.8938 + 1.915 = 2.809
Possibly Useful Information
1 | q 1 || q 2 |
4  0 r 2
e = 1.6 X 10-19 C
F
| q|
, E = /
4  0 r 2
x = x2 - x1, t = t2 - t1
E
 0  8.85 X 10 -12 (C 2 / N  m 2 )
 
E  Fq
0
 
 0    0  E. dA  q enc
v = x / t
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
x-xo = vot + (½)at2
v2 = vo2 + 2a(x-xo)
x-xo = vt -1/2at2


a = dv / dt

 r = (x 2 - x 1 ) i + (y 2 - y 1 ) j + (z 2 - z 1 ) 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
V
1 q
4  0 r
Uf + Kf = Ui +Ki
K = ½ mv2
V
Es 
s
V
E
s
Q = CV
l
ln(b / a)
C  4  0 R
C  2 0
i
n
1
V   Vi 
4 0
i 1
1
dq
V

4  0 r
n
q
 ri
i 1
i
V
V
V
;Ey 
;Ez 
x
y
z
1 q1q 2
U  W 
4 0 r12
 0A
C
d
ab
C  4  0
ba
Ex 
C eq   C j (parallel)
1
1

(series)
C eq
Cj
u  12  0 E 2
I= dQ/dt
1
Q2 1
 2 CV 2
2C
C =  C0
U

R
L
V = IR
A
P = IV
P = I2R=V2/R
I 
(R  r)
Pemf = I
R eq   R j (series)
q(t)= Q(1-e-t/RC)
1
1

(parallel)
R eq
Rj
I = (R)e-t/RC
I = (QRC)e-t/RC, I0 = (QRC)
E = o
q(t) = Qet/RC
 = Q/L, = Q/A,  = Q/V