PHY2054 Spring 2007
Prof. Eugene Dunnam
Prof. Paul Avery
Feb. 6, 2007
Exam 1 Solutions
1. A charge Q1 and a charge Q2 = 1000Q1 are located 25 cm apart. The ratio of the electrostatic
force on Q1 to that on Q2 is:
(1) none of these
(2) 1000
(3) 0.0032
(4) 0.001
(5) 316
The ratio of forces is 1.0, as can be seen from Coulomb’s law or Newton’s third law.
2. The charge on each of two tiny spheres is doubled while their separation distance is tripled.
The ratio of the new electrostatic force to the old force is:
(1) 4/9
(2) 2/3
(3) 9/4
(4) 4/3
(5) 1
Doubling the charge makes the numerator 4 times larger, while tripling the distance makes the
denominator 9 times larger. Together they yield a force that is 4/9 of the original force.
3. An electron circles a carbon nucleus (6 protons, 6 neutrons) at a distance of 0.0353 nanometers. What is its velocity in km/s?
(1) 6560
(2) 9280
(3) 2680
(4) 153
(5) 1950
Here we balance centripetal force with the coulomb force, or mv 2 / r = k ( 6e ) e / r 2 , where 6e is
the charge of the carbon nucleus, r is the orbital radius and m is the mass of the electron. Solving for v gives v = 6ke2 / mr = 6.56 × 106 m / s = 6560 km/s.
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PHY2054 Spring 2007
4. The electric field required to suspend a proton against the force of gravity is:
(1) 102 nanovolts/m directed upward
(2) 102 nanovolts/m directed downward
(3) 557 picovolts/m directed upward
(4) 557 picovolts/m directed downward
(5) none of these
The E field required to balance the proton is found from Ee = m p g , where mp is the mass of the
proton and e is its charge. Solving yields E = 102 nanovolts/m, directed upwards to balance the
gravitational force.
5. Negative charge −Q is located on the y-axis at y = a and charge 2Q is located at y = −a. The
direction of the electric field at the point (a, 0) due to these charges is toward
(1) First quadrant
(2) +y
(3) −x
(4) +x
(5) −y
The net y component of E is positive and the net x component is positive, putting the direction of
E in the first quadrant.
6. A charge q is placed at the origin and a charge −4q is placed on the x-axis at x = 2 cm. At what
value of x on the x axis (in cm) does the electric field equal zero?
(1) −2.00
(2) +4.00
(3) +0.67
(4) −4.00
(5) −0.67
The point where E = 0 must be to the left of the charge q (why?). The x components of E from the
charge q and −4q then point in the negative and positive directions, respectively. They satisfy the
relation − kq / x 2 + 4kq / ( x − d ) = 0 , where d is the distance between the charges. Solving for x
2
(the negative root is the one we want) yields x = −d = −2 cm.
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PHY2054 Spring 2007
7. A thick conducting spherical shell has inner radius r and outer radius R, as shown in the diagram. A point charge of +Q is located at the center of the sphere and a charge of +6Q is placed
on the conducting shell. The charge on the inner surface of the conducting shell is:
(1) −Q
(2) +Q
(3) +7Q
(4) +6Q
(5) +5Q
R
r
+Q
+6Q
From Gauss’ law, as we showed in class, the inside charge on the outer shell must balance the
charge in the center to give E = 0 inside the shell. Thus the charge on the inner shell is –Q (the
charge on the outer shell is +7Q).
8. An alpha particle (nucleus of helium with 2 protons and 2 neutrons) with kinetic energy 20
keV is shot directly at a small conducting sphere carrying charge Q and comes to a complete stop
at a point 20 cm from the sphere's center. The charge Q, in microcoulombs, must be
(1) +0.22
(2) +0.11
(3) +0.044
(4) +0.022
(5) none of these
This is a problem of conservation of energy, where the initial energy is purely kinetic and the
final energy is purely potential. The energy balance condition is then 12 mv 2 = k 2eQ / r , where 2e
is the charge on the alpha particle and r is the minimum distance to the sphere. Solving for Q
(after converting the kinetic energy to joules) yields Q = +0.22 μC (must be positive to repel the
alpha particle charge).
9. The electrostatic potential energy of three 10-microcoulomb charges placed at the corners of
an equilateral triangle of side length 20 cm is:
(1) none of these
(2) 4.5 J
(3) 9.0 J
(4) 135 picoJ
(5) 68 J
The potential energy of the system of three charges is U = 3kQ 2 / r , where r is the distance between any two and Q is the common charge. The factor of 3 accounts for the fact that there are
three pairs of charges. This yields U = 13.5 J.
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PHY2054 Spring 2007
10. An alpha particle (nucleus of helium with 2 protons and 2 neutrons) and an electron are each
accelerated from rest through a potential difference of 2500 volts. Their resultant kinetic energies, in electron volts will be, respectively:
(1) none of these
(2) 2500, 1.36
(3) 10000, 2500
(4) 2500, 1250
(5) 2500, 2500
The kinetic energies of the alpha particle and the electron are 5000 eV and 2500 eV, respectively.
11. Electric field lines and equipotential surfaces
(1) are always perpendicular to one another
(2) make arbitrary angles when they meet
(3) never cross
(4) only exist for certain charge configurations
(5) are always radial and circular, respectively
Answer is (1), as we discussed in class.
12. The movement of charge in an electric field from one point to another at constant speed
without the expenditure of work by or against the field
(1) can only occur along an equipotential
(2) can only occur along a field line
(3) none of these
(4) only holds when the field lines are radial
(5) only occurs in a uniform field
An equipotential is a surface of constant potential energy, so no work is done moving along an
equipotential.
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PHY2054 Spring 2007
13. Two large, parallel, conducting plates are 1.2 mm apart and have equal and opposite charges
on their facing surfaces. An electrostatic force of 6 × 10−13 N acts on an electron placed anywhere between the two plates. What is the potential difference in volts between the plates?
(1) 4500
(2) 720
(3) 3.76 × 106
(4) 2250
(5) 0.00088
The electric field is given by Ee = F, or E = F / e. The potential difference is therefore calculated
from V = Ed = Fd / e = 4500 V.
14. Capacitors C1 = 5μF and C2 = 10μF are connected in series across a battery of V = 100 volts
and remain connected until fully charged. The battery is then carefully removed and C1 and C2
are connected together, + plate to + plate and − plate to − plate. After charge equilibrium is
reached, the potential difference across C1 and C2, respectively, is (in volts):
(1) 44, 44
(2) 67, 67
(3) 50, 50
(4) 100, 100
(5) 0, 0
The charge on each of the capacitors in the initial configuration is found by calculating the
equivalent series capacitance Cs, which is 10/3 μF. The charge on Cs is Q = CsV0 = 1000/3 μC,
which is the same as the charge on each of the capacitors since they are in series. When the capacitors are reconnected, charge flows from one to the other until the voltages across them
equalize. In effect, the capacitors are in parallel, with the positive plates at the same potential
and the negative plates at the same potential. Since the total charge does not change, we can
write the total charge as ( C1 + C2 ) V = 1000 / 3 + 1000 / 3 = 2000 / 3 μC. Solving for the potential
difference yields V = 2000 / 45 = 44.4 V for each capacitor.
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PHY2054 Spring 2007
15. In the figure shown, the battery has a potential difference of 24 V. If C1 = C2 = C3 = C4 = C5
= 18 μF, what is the equivalent capacitance of the circuit in μF?
(1) 28.8
(2) 11.3
(3) 18.0
(4) 12.0
(5) 36.0
C2
C3
C1
+
V
C4
C5
C2 and C3 are in series and are equivalent to C23 = 9 μF. C23 and C4 are in parallel and add,
yielding C234 = 27 μF. C234 and C5 are in series and yield C2345 = 10.8 μF. C2345 and C1 are in
parallel, yielding C12345 = 28.8 μF.
16. A parallel-plate capacitor whose capacitance is 120 pF is charged by a battery to a potential
difference of 6 V between its plates. The battery is now disconnected, and a slab of a dielectric
material, whose dielectric constant is 80, is slipped between the plates. What is the stored energy
of the capacitor-slab device after the slab is inserted, in joules?
(1) 2.7 × 10−11
(2) 5.8 × 10−7
(3) 1.5 × 10−3
(4) 2.2 × 10−9
(5) 7.4 × 10−6
When the dielectric material is inserted, the charge is the same but the electric field (and thus V
= Ed) is reduced by a factor of 80 and the capacitance is increased by a factor of 80. The energy 12 CV 2 is thus reduced by a factor of 80 from the value it would have without the dielectric.
Calculating yields 2.7 × 10−11 J.
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PHY2054 Spring 2007
17. Automobile and similar batteries are sometimes rated in ‘ampere-hours’. What does this represent?
(1) the total charge stored in the battery
(2) the total number of volts in the battery
(3) the maximum current the battery can deliver
(4) the total lifetime of the battery
(5) the total energy stored in the battery
Charge is current times time.
18. In recharging a weak 12-V automobile battery, it is connected to a 14-V electronically controlled charger. The initial charging current of 7.0 A decreases linearly to 3.0 A over a 6.0-hour
period. At this point, how much charge (in Coulombs) has passed through the battery?
(1) 108,000
(2) 36,000
(3) 14,400
(4) 360
(5) 420
The total charge is average current × time. Average current is 5A and total time is 6 hours =
21,600 seconds, yielding a total charge of 108,000 coulombs.
19. The drift speed of electrons in a long copper wire of radius 1 mm cross section and carrying a
current of 1.0 A is 0.1 mm/s. If the current is increased to 20 A, and the radius increased to 2
mm, what would be the drift speed, in mm/s?
(1) 0.5
(2) 1.0
(3) 4.0
(4) 0.1
(5) none of these
Drift speed is proportional to current and inversely proportional to the cross sectional area
(which in turn goes as the square as the diameter). This yields 0.1 × 20 / 22 = 0.5 m/s.
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PHY2054 Spring 2007
20. Three conductors, A, B, and C are connected across a variable voltage source. When the
voltage is doubled, conductor A's current triples, conductor B's current doubles, and conductor
C's current increases by 1.414. Which of the conductors obeys Ohm's Law?
(1) B
(2) A, B, and C
(3) A
(4) A and C
(5) C
Ohm’s law says that the current is proportional to voltage, so only B obeys Ohm’s law.
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