Physics 114 Exam 2 Fall 2012
Name: ___________________________________________________________
For grading purposes (do not write here):
Question
Problem
1.
1.
2.
2.
3.
3.
Answer each of the following questions and each of the problems. Points for each
question and problem are indicated in red with the amount being spread equally among
parts (a,b,c etc). Be sure to show all your work.
Use the back of the pages if necessary.
Question 1. (10 points) Consider an isolated solid metal sphere of radius r1= 1 m as illustrated
on the left below (case A) that has been given an excess charge of 2 C.
(a) How much work would you have to do to move a point charge of 1 C from the center of
the sphere to a distance of r = 0.5 m from the center of the sphere along a radial path?
Ans: The work is 0, because there is no change of electric potential.
Now also consider an alternative case where you have a larger isolated metal sphere of radius r2=
2 m and charge of 4C illustrated on the right (case B)
(b) Compare the electric potential for each case (A and B) at a distance of r = 0.5 m from the
center of each sphere. Is it greater for the smaller sphere (case A), or for the bigger
sphere, or are they equal?
Ans: The potential is k*Q/R for each sphere and is constant throughout the sphere. We
compare k*2C/1m vs k*4C/2m. They are equal.
(c) Now compare the electric potential for each case (A and B) at a distance of r = 4 m from
the center of each sphere. Is it greater for the smaller sphere (case A), or for the bigger
sphere, or are they equal?
Ans: The bigger sphere B has greater electric potential.
, they have equal
distance, but B has bigger charge 4C.
(d) Say now that you also had a 10 C test charge a distance of r = 4 m from the center of
each sphere. Compare the electric potential energy for each case (A and B).
Ans: For case A, the
; For case B, the
,
obviously, the case B has greater electric potential energy.
A
B
r1
r2
Question 2 (10 points) Consider a capacitor hooked up to a 9 V battery. If we now hook it up to
a 18 V battery,
(a) What happens to the capacitance of the capacitor as we double the voltage across it as
described above? Does it change – if so by how much (increasing or decreasing)?
Ans: It doesn’t change, because capacitance is an attribute of a capacitor and independent
of voltage applied.
(b) What happens to the charge stored on the capacitor as we double the voltage? Does it
change – if so by how much (increasing or decreasing)?
Ans: It will increase, double. Charge is proportional to voltage so that it is doubled.
(c) What about the energy stored in the capacitor? Does it change – if so by how much
(increasing or decreasing) when we double the voltage across the capacitor?
Ans: It will increase by four times. Energy is proportional to square voltage so that it is
four times.
In answering the above questions assume the capacitor to be in steady state both when the 9V
and the 18V battery are connected to it.
Question 3. (10 points) Consider 2 identical light bulbs wired as illustrated below in Circuit I.
(a) How does the voltage across each bulb compare to each other?
Ans: They have the same voltage for each other and the voltage is 12V, because the two
bulbs are connected in parallel with the battery.
(b) How does the current in each bulb compare to each other?
Ans: They have the same current which equal to 12V/R,
where R is the resistance of bulb.
(c) Now answer the same questions if the bulbs are wired as in
Circuit II; how do the voltage and current in each bulb
compare to each other.
Ans: They have the same voltage and current. The voltage equals to 6V, half of supply
voltage12V. The current equals to 12V/2R, where R is the resistance of bulb. They are
connected in series, so the current is the same. Because they are identical and have the
same resistance, the voltage is also the same.
(d) Finally, now compare the 4 identical bulbs, which circuit puts out more light?
Ans: The circuit I puts out more light, which depends the power of every bulb.
If we consider V2/R, we see there is more voltage across each bulb in circuit I. The
voltage 12 V is split in circuit II across each resistor.
In circuit I, the power of bulb can also be evaluated as P=I2R=(12/R)2R=144/R.
Here, thecurrent is ½ of the current in the equivalent circuit which has a resistance of
(1/R + 1/R)-1 = R/2. The current is (0.5)(12/(R/2) = 12/R
In circuit II, the power of bulb P=I2R=(12/2R)2R=36/R.
Problem 1. (15 points) A uniform electric field of magnitude 325 V/m is directed in the negative
y direction as shown in the figure below. The coordinates of point are (-0.200, -0.300) m, and
those of point are (0.400, 0.500)
(a) Calculate the electric potential difference VB − VA using the dashed-line path.
Ans: Assume the point C is (-0.200, 0.500), then
∫
∫
∫
∫
(b) Which point is at a higher potential
Ans: Point B is at a higher potential
(c) If a +2 C m=1kg is accelerated along the vertical direction from rest in this electric
field over a distance of 1 m, what would its final speed be?
Ans:
√
√
Problem 2. (15 points) Four capacitors are connected as shown in the figure below. (Let C = 15
µF.)
(a) Find the equivalent capacitance between points a and b.
Ans:
(b) Calculate the charge on each capacitor, taking ΔVab = 15.0 V.
Ans:
; so
Problem 3. (20 points) Consider the figure below. (Let R1 = 2.0 Ω, and R2 = 1.0 Ω.)
(a) Which resistors are in series (if any)?
Ans: The R1 and the resistor with 4Ω are in series and
Rtot1=6Ω;
The R2 and the resistor with 5Ω are in series and
Rtot2=6Ω;
(b) Which resistors are in parallel?
Ans: There are no resistors in parallel.
(c) Use Kirchoff’s junction rule to obtain an equation relating I1, I2, and I3.
Ans: I1=I2+I3
(d) Apply Kirchoff’s loop rule to the top loop to obtain an equation relating I1 and I3.
Ans: V1-I1R1-I1R4-I3R3=0
24-2I1-4I1-3I3=0 2I1+I3=8
(e) Apply Kirchoff’s loop rule to the bottom loop to obtain an equation relating I2 and I3
Ans: V2+I3R3-I2R2-I2R5=0
12+3I3-1I2-5I2=0
2I2-I3=4
Possibly Useful Information
1 | q 1 || q 2 |
4  0 r 2
e = 1.6 X 10-19 C
F
| q|
, E = /
E
4  0 r 2
x = x2 - x1, t = t2 - t1
 0  8.85 X 10-12 (C 2 / N  m2 )
 
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
x-xo = vot + (½)at2
v2 = vo2 + 2a(x-xo)

r = xi + yj + zk
  
r = r2 - r1
x-xo = vt -1/2at2


a = dv / dt

r = (x 2 - x1 ) i + (y 2 - y1 ) 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
l
ln(b / a)
C  4 0 R
C  2 0
1
1
(series)

C eq
Cj
u  12  0 E 2
I= dQ/dt
1
i
n
1
4 0
i 1
1
dq
V

4 0 r
V   Vi 
n
i 1
Pemf = I
1
1
(parallel)

R eq
Rj
I = (R)e-t/RC
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 eq   C j (parallel)
Q2 1
 2 CV 2
2C
C =  C0
U

L
R
A
P = IV
q
 ri
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
I = (QRC)e-t/RC, I0 = (QRC)
E = o
 = Q/L, = Q/A,  = Q/V