PHY 2049 SPRING 2000 EXAM 2
1. Consider three copper resistors of simple cylindrical shape with resistance R1 , R2 , and
R3 . The first resistor has length L and diameter D, the second has length 2L and diameter
2D, the third one has length 4L and diameter 2D. Which of the following is true? Answer: R3 = R1 > R2
The resistance of a resistor is related to its geometry
R=
ρL
.
A
The cross-sectional area of a cylinder is related to its diameter
A = πR2 = π
D
2
2
=
πD 2
.
4
The expression for the resistance is now
R=
4ρL
.
πD 2
For the 3 resistors
4ρL1
4ρL
=
2
πD1
πD 2
4ρL2
4ρ(2L)
2ρL
=
=
=
2
2
πD2
π(2D)
πD 2
4ρL3
4ρ(4L)
4ρL
=
=
=
.
2
πD3
π(2D)2
πD 2
R1 =
R2
R3
Comparing R1 , R2 , and R3 , it is clear that R3 = R1 > R2 .
2. A charge q is distributed uniformly throughout a spherical volume of radius R. What is
2 −r 2 )
the potential V a distance r from the center if r < R and V = 0 at infinity? Answer: q(3R
8π0 R3
The electric field inside the charged sphere is
E1 =
qr
4π0 R3
E2 =
q
.
4π0 r 2
and outside the sphere it is
The potential difference is given by
Vf − Vi = −
f
i
· ds.
E
· ds =
The integration is in the r direction and E is radially outward. Therefore, E
E dr cos 0 = E dr. In addition, it is given that V = 0 at infinity.
Vf − Vi = −
V (∞) − V (r) = −
0 − V (r) = −
f
i
· ds
E
∞
r ∞
r
E dr
E dr.
It is necessary to break up the integral at r = R since the electric field is different for points
inside and outside the sphere.
−V (r) = −
V (r) =
=
=
=
=
=
∞
R
r
r
R
r
E dr
E1 dr +
E1 dr +
∞
R
∞
R
R
E2 dr
E2 dr
∞
qr
q
dr
+
dr
3
r 4π0 R
R 4π0 r 2
R2 r 2
q
1
q
+
0− −
−
4π0 R3 2
2
4π0
R
q
(R2 − r 2 + 2R2 )
8π0 R3
q(3R2 − r 2 )
.
8π0 R3
3. Two batteries of identical e.m.f. E and three resistors of identical resistance R are
connected as shown. Find the difference of potentials across the central resistor in terms of
E. Answer: (2/3)E
+
+
i1
R
-
i1
R
i1
+
i2
i3
R
+
-
i2
E − i1 R − i3 R = 0
E − i2 R − i3 R = 0.
i3 = i1 + i2 .
+
-
Using Kirchhoff’s loop rule as indicated in the picture:
The junction rule yields
i2
Substituting into the two above equations and performing some reorganizing
E = 2Ri1 + Ri2
E = Ri1 + 2Ri2
Solving the first equation for Ri2
Ri2 = E − 2Ri1
and substituting into the second equation:
E = Ri1 + 2Ri2
= Ri1 + 2(E − 2Ri1 )
= 2E − 3Ri1
E
.
i1 =
3R
The other current is found
Ri2 = E − 2Ri1
E
= E − 2R
3R
E
=
3
E
i2 =
3R
and i1 = i2 as would be expected from the symmetry of the circuit. The current through
the central branch is i3 = i1 + i2 = 2E/3R. So the potential drop in the central R is
∆V = i3 R = 2E/3.
4. A parallel-plate capacitance gap is filled with a slab made of material with dielectric
constant κ. The capacitance (with the slab inserted) is C. The capacitor is then charged
to a potential difference V and the battery is removed. How much work would be needed
to remove the slab from the capacitor gap? Answer: 12 CV 2 (κ − 1)
The work done is equal to the change in the potential energy
W = ∆U = Uf − Ui .
The charge on the capacitor is Q = CV and remains constant even though the dielectric is
removed because the charging battery was removed. The initial energy is
Ui =
1
Q2
= CV 2 .
2C
2
After removing the dielectric, the new capacitance Cf = C/κ. The final energy is
Q2
2Cf
Q2
=
2 Cκ
Q2
= κ
2C
1
= κ CV 2 .
2
Uf =
The work done is
W = Uf − Ui
1
1
= κ CV 2 − CV 2
2
2
1
=
CV 2 (κ − 1).
2
5. Three light bulbs labelled as 60 W / 120 V are connected in series by a curious student
and plugged into a 120 V outlet. How much power would this chain of light bulbs dissipate?
Answer: 20 W
The bulbs are rated as 60 W at 120 V. The resistance is related to the power by
P = V 2 /R. The resistance is
R=
(120 V)2
V2
=
= 240 Ω.
P
60 W
Connecting the light bulbs in series, the total resistance is the sum of the resistances of the
bulbs, Rs = 240 Ω + 240 Ω + 240 Ω = 720 Ω. The power dissipated by the three resistors is
P =
V2
(120 V)2
=
= 20 W.
R
720 Ω
6. Points A and B are separated as shown. What is difference of potentials between points
A and B, VB − VA ? The electric field is uniform and has magnitude E. Answer: xE
The potential difference is given by
VB − VA = −
B
A
· ds.
E
Since the electric force is conservative, the path can be deformed without changing the
potential difference. The new path goes from A to C and then to B.
VB − VA = −
= −
B
A
C
A
· ds
E
· ds −
E
B
C
· ds
E
B
C
c
y
A
x
= 0−
= E
B
C
E ds cos π
ds
= Ex.
7. The current density in a cylindrical wire of radius R at a distance r from its axis is
2
J0 (1 − Rr 2 ) and parallel to the wire axis. The total current is Answer: 12 J0 πR2
To find the total current, integrate the current density.
J dA
I =
=
=
=
=
r2
J0 1 − 2 2πr dr
R
R 3
R
r
r dr −
dr
2πJ0
0
0 R2
1 2
1 R4
R − 2
2πJ0
2
R 4
2
πJ0 R
.
2
8. A resistor R = 15.0 kOhm is connected in series with a capacitor C of unstated value. At
time t = 0 a potential difference E = 12.0 V is applied across this series combination. The
potential difference across the capacitor rises to a measured value of 5.0 V at t = 1.30 µs.
Calculate the value, in picofarads, of the capacitance C. [Hint: use the given data to calculate the RC time constant.] Answer: 160.6 pF
The capacitor is charging since the problem mentions that the potential across the
capacitor rises. In a charging RC circuit, the charge on the capacitor is given by
Q = Q0 (1 − e−t/τ ).
Multiplying by C and replacing CQ0 = E the potential across the capacitor can be written
VC = E(1 − e−t/τ )
VC
E
= 1 − e−t/τ
VC
E
E − VC
E
E
E − VC
E
ln
E − VC
(12 V)
ln
(12 V) − (5 V)
0.539
t
0.539
1.30 × 10−6 s
0.539
2.41 × 10−6 s.
e−t/τ = 1 −
=
et/τ =
t/τ =
=
=
τ =
=
=
The time constant τ = RC can be used to find the capacitance
C=
τ
2.41 × 10−6 s
=
= 1.61 × 10−10 F.
R
15 × 103 Ω
9. Consider two protons heading towards one another, each with kinetic energy E and
charge 1.6 × 10−19 Coulomb. Estimate the minimum energy that would allow the protons to come in contact, assuming that protons can be viewed as solid spheres of diameter
d = 10−15 m. Your estimate is a necessary condition for the nuclear fusion process so it
would be appropriate to express the answer in units of temperature: Use the conversion
factor: one degree Kelvin (K)≈ 2 × 10−23 Joules. Answer: ∼ 6 × 109 K
Energy is conserved in the process. The total kinetic energy is the energy of both moving
charges
U1 + K1 = U2 + K2
k(e)(e)
+0
0+E+E =
d
ke2
E =
2d
(8.99 × 109 Nm2 /C2 )(1.6 × 10−19 C)2
E =
2(10−15 m)
= 1.15 × 10−13 J.
Converting to Kelvins
E = 1.15 × 10−13 J ×
1K
= 6 × 109 K.
2 × 10−23 J
10. Find the total energy stored in the three capacitors assuming that the battery has an
emf of 12 V and that the capacitors all have a capacitance of 4 µF. Answer: 432 µJ
4 µF
12 V
4µF
4 µF
The two capacitors on the right are in series
1
1
1
=
+
Cs
C1 C2
1
1
+
=
4 µF 4 µF
2
=
4 µF
Cs = 2 µF.
This pair of series capacitors is in parallel with the other capacitor
C = C3 + Cs = 4 µF + 2 µF = 6 µF.
The energy is
1
1
U = CV 2 = (6 µF)(12 V)2 = 432 µJ.
2
2