PHY2054 Spring 2009
Prof. Pradeep Kumar
Prof. Paul Avery
Prof. Yoonseok Lee
Feb. 4, 2009
Exam 1 Solutions
1. A proton (+e) originally has a speed of v = 2.0 ×105 m/s as it goes
through a plate as shown in the figure. It shoots through the tiny
holes in the two plates across which 16 V | 100 V | 2 V of electric
potential is applied. Find the speed as it leaves the second plate.
Answer: 1.92 × 105 m/s | 1.44 × 105 m/s | 1.99 × 105 m/s
Solution: The proton is slowed down by the electric field between
the two plates, so the velocity will be reduced. Conservation of energy yields
1 m v2
2 p i
+++
+++
V
−−− −−−
v
− eΔV = 12 m p v 2f , so v f = vi2 − 2eΔV / m p .
2. Five identical charges of +1μC are located on the 5 vertices of a regular hexagon
of 1 m side leaving one vertex empty. A −1μC | −2μC | +2μC point charge is
placed at the center of the hexagon. Find the magnitude and direction of the net
Coulomb force on this charge.
y
Answer: 9.0 × 10−3 N −x-direction | 1.8 × 10−2 N −x-direction | 1.8 × 10−2 N +x-direction
Solution: Only the charge on the far left contributes to the net force on the charge in the center
because its contribution is not cancelled by an opposing charge. If Q represents the charge at
each hexagon vertex, q is the charge at the center and d is the length of a hexagon side (note that
the distance from each hexagon point to the center is also d), the force on q is kQq / d 2 along the
+x direction if Qq is positive and along the −x direction if Qq is negative.
3. Wire A and Wire B are made out of copper and each wire carries the same current. If the diameter of A is twice | half | three times that of Wire B, how does the drift velocity vdA in Wire A
compares to that in Wire B, vdB?
Answer: vdA = vdB / 4 | vdA = 4vdB | vdA = vdB / 9
Solution: The current i is i = Ane evd , where A = π r 2 is the cross sectional area, ne is the free
electron density and vd is the drift velocity. The answers follow by writing vd = i / π r 2ene and
comparing the diameters.
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PHY2054 Spring 2009
4. The free electron density of copper is 8.5 × 1028 m−3. A copper wire is connected to a 1.5 V
battery and carries a current of 0.125 A | 0.015 A | 0.010 A. The electrons inside the wire move
with the drift velocity of 2.5 × 10−4 m/s. What is the cross-sectional area of the wire?
Answer: 3.7 × 10−8 m2 | 4.4 × 10−9 m2 | 2.9 × 10−9 m2
Solution: We write i = Ane evd and solve for A.
5. The earth and a very large cloud (12 km × 12 km | 14 km × 14 km | 16 km × 16 km) 500 | 400 |
300 m above the surface can be considered to be two parallel plates of a very large capacitor. If
an electric field strength E = 3 × 106 V/m causes the air to break down and conduct charge
(lightning), what is the maximum charge the cloud can hold?
Answer: 3820 C | 5200 C | 6800 C
Solution: For a capacitor the electric field strength between the plates is E = σ / ε 0 , where
σ = Q / A is the surface charge density (you can also use Q = CV with C = ε 0 A / d and
V = Ed ). Thus Q = Aσ = ε 0 AE .
6. In the previous problem, the electric field formed by the earth and the cloud is pointing vertically downward and acts on a charged raindrop of mass 0.15 g | 0.20 g | 0.25 g. If the cloud is
charged to its maximum capacity, what charge on the raindrop is needed for it to float at rest in
the air?
Answer: −4.9 × 10−10 C | −6.5 × 10−10 C | −8.2 × 10−10 C
Solution: To float in the air, the electric force must exactly cancel the gravitational force. Thus
qE + mg = 0 or q = − mg / E .
7. A positive charge Q = +2μC | +3μC | +4μC is placed on the y-axis at y = 1
m and a charge −Q is placed on the y-axis at y = −1 m. What is the potential
difference VA − VB between the points A = (0,2 m) and B = (1 m,0)?
Answer: 12 kV | 18 kV | 24 kV
Solution: The potential at B is zero (opposite charges, same distance). Thus
we need only calculate VA. The distance from A to +Q is 1 m and the distance of A to −Q is 3 m. We obtain VA − VB = kQ /1 − kQ / 3 = 2kQ / 3 .
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y
X
A
+Q
B
X
−Q
x
PHY2054 Spring 2009
8. A 30-V battery is connected to a circuit and the current delivered to the circuit falls linearly
from 7 A | 9 A | 11 A to 2 A over 4 | 3 | 2 hours. How much charge (in kilo-Coulombs) was delivered to the circuit by the battery?
Answer: 65 | 59 | 47
Solution: Let i1 and i2 be the initial and final currents, respectively. Then the average current is
( i1 + i2 ) / 2 and the total charge delivered in a time t is Q = ( i1 + i2 ) t / 2 .
9. A 3.0μF | 4.0μF | 5.0μF capacitor is charged by a battery to a voltage of 12 V. It is then disconnected from the circuit and connected across an uncharged 9.0μF | 12.0μF | 15μF capacitor.
What is the final charge on the second capacitor?
Answer: 27μC | 36μC | 45μC
Solution: When the two capacitors are connected, charge flows from the first capacitor to the
second one until the voltages are equal. Let q1 = C1V1 be the initial charge on C1 and q be the
charge that moves from C1 to the initially uncharged C2. Using charge conservation, the condition of equal potential can be written ( q1 − q ) / C1 = q / C2 . Solving for q (which is the charge on
C2) yields q = q1 ( C1 + C2 ) / C1C2 = V1 ( C1 + C2 ) / C2 .
10. Two capacitors with C1 = 5.0μF and C2 = 15.0μF | 20.0μF | 25.0μF are initially uncharged and then connected to a battery as shown in the figure. If neither capacitor can withstand a potential difference greater than 100 volts without breakdown, what is the maximum value of the battery emf that can be sustained by the circuit?
Answer: 133 V | 125 V | 120 V
Solution: When connected to a battery, the initially uncharged capacitors acquire equal charge q
because they are in series. The voltages on the capacitors are V1 = q / C1 and V2 = q / C2 . The first
capacitor is smaller so it will hit 100 V first. When this happens the voltage on C2 is
100 ( C1 / C2 ) and the total emf is 100 + 100C1 / C2 = 100 ( C1 + C2 ) / C2 .
11. Three identical capacitors of capacitance C = 2.5μF | 3.5μF |
4.5μF are connected in series to a battery with emf 130 V | 140 V |
150 V as shown in the figure. What is the combined energy stored
in the capacitors?
Answer: 7.0 × 10−3 J | 1.14 × 10−2 J | 1.7 × 10−2 J
Solution: The equivalence capacitance of the three capacitors in series is C / 3 , so the combined
energy is
1C V2
2 eq
= 16 CV 2 .
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PHY2054 Spring 2009
12. Two small identical conducting balls, insulated from their surroundings and separated by a
large distance, have charges 2.6μC | 3.2μC | 3.8μC and −1.2μC | −1.4μC | −1.6μC, respectively.
The balls are then touched together and then brought back to their original positions. If F1 is the
magnitude of the force between the balls before contact and F2 is the magnitude of the force after
contact, what is the ratio F2 / F1?
Answer: 0.16 | 0.18 | 0.20
Solution: The initial force is F1 = kQ1Q2 / d 2 . When the balls are touched, since they are identi-
cal and conducting, they exchange charge until charges are equal ( Q′ = ( Q1 + Q2 ) / 2 ). Thus
F2 = kQ′2 / d 2 and F2 / F1 = ( Q1 + Q2 ) / 4Q1Q2 .
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13. A parallel-plate capacitor with plate separation d and charge Q is disconnected from a battery. The plates are then moved to a new separation distance of 3d | 4d | 5d. What is the new
charge on the plates?
Answer: Q
Solution: Since the capacitor is isolated the charge cannot go anywhere.
14. Refer to the previous problem. If the initial electric field strength was E, what is the final
electric field strength?
Answer: E
Solution: Electric field strength is independent of the separation.
15. Refer to the previous problem, except that this time when the plate separation changes, the
capacitor remains connected to the battery. If the initial energy stored in the capacitor was U,
what is the final energy stored?
Answer: U / 3 | U / 4 | U / 5
Solution: Since C = ε 0 A / d and V is the same, the energy U = 12 CV 2 will be inversely proportional to the separation distance.
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PHY2054 Spring 2009
16. In the figure, V = 25 volts, C1 = C2 = 6μF and C3 = C4 = 36μF | 42μF |
48μF. Find the charge on C1.
Answer: 129μC | 131μC | 133μC
Solution: The capacitances C1 and C2 can be combined into a single capacitance CA = 2C1. Same with C3 and C4 , which yield CB = 2C3. The net capacitance is Ceq = C ACB / ( C A + CB ) , which allows us to calculate q as
q = Ceq = VC ACB / ( C A + CB ) . This is same as the charge on the C1 + C2 and C3 + C4 pairs since
they are in series with each other. Since C1 = C2, the charge on C1 is q/2 or
q = 12 VCeq = 12 VC ACB / ( C A + CB ) .
17. A special room is set up so that there is a downward pointing electric field of magnitude E. A
ball of mass 4 kg and charge 0.005 C is dropped from a height of 15.0 m | 20.0 m | 25.0 m, striking the ground 1.5 seconds later. What is E? (Don't forget to include gravity, g = 9.8 m/s2.)
Answer: 2830 V/m | 6830 V/m | 9940 V/m
Solution: The total force acting on the ball is mg + qE , so the acceleration is a = g + qE / m . Us-
ing h = 12 at 2 , where h is the height where the ball is dropped, we get h =
1
2
( g + qE / m ) t 2 , from
which we can calculate E.
18. An electron circles a Nitrogen | Oxygen | Fluorine nucleus (7 | 8 | 9 protons, 7 | 8 | 9 neutrons)
at a distance of 0.0302 | 0.0265 | 0.0235 nanometers. What is its velocity in km/s?
Answer: 7650 | 8730 | 9840
Solution: We equate the centripetal force to the electrical force between the electron (charge –e)
and the nucleus (charge +Ze), me v 2 / r = Zke 2 / r 2 . Solving for v yields v = Zke2 / me r .
19. Three charges Q1 = +5μC | −3μC | −3μC, Q2 = −6μC and Q3 = +9μC produce an electric
field of strength E at a point P. All 3 charges are now made 2 | 3 | 4 times larger. The new field at
P is
Answer: 2E | 3E | 4E
Solution: Using vector addition, we see that if we increase each field by the same amount (by
changing its charge), then the net field changes by the same factor.
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PHY2054 Spring 2009
20. A +3μC | +3μC | −1μC point charge is placed at the center of a conducting spherical shell.
The inner radius of the shell is 30 cm and its thickness is 1 cm. The electric field at a point 1 m
away from the point charge is 9000 N/C | 18000 N/C | 18000 N/C and pointing toward the center. What is the net charge on the conducting shell?
Answer: −4.0μC | −5.0μC | −1.0μC
Solution: From Gauss’ law ∑ Ei ⋅ Ai = qenc / ε 0 , where the electric flux on the left is equal to
i
−4π r E (inward directed E field at r = 1 m gives the negative value) where the total “enclosed”
charge qenc is the charge in the center plus the charge on the shell. Gauss’ law then becomes
−4π r 2 E = ( qcenter + qshell ) / ε 0 , which can be rearranged to get qshell = − qcenter − ε 0 4π r 2 E .
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