PHY2054 Spring 2013 1 Prof. Paul Avery Prof. Zongan Qiu Feb. 12, 2013 Exam 1 Solutions 1. Three equal charges of value q are placed at 3 corners of a square of side L, as shown. Another charge Q is placed at position 4. What is the value of Q if the charge at position 2 feels no net force? Answer: !2 2 q Solution: The net electric field at point 2 from the other two charges is 2 kq 2 / L 2 along the +45 degree line. To cancel that force the force from Q must satisfy kqQ / 2 L ( ) 2 = ! 2 kq 2 / L 2 . This yields the solution shown. 2. Three charges q 1 , q 2 , and q 3 are placed on the x-axis at x = 0, a, 2a respectively. The electric flux through a sphere of radius 1.5a, centered on the origin, is found to be ! E = 10N " C . On the basis of only this information which of the following must be true? Answer: q 2 = ! q 1 + 10" 0 Solution: The sphere encloses only q 1 and q 2 . From Gauss’ law, ! E = 10 = q 1 + q 2 ( ) / " 0 , from which the solution follows. 3. A uniform electric field of 5,000 V/m is directed along the negative y-axis. A proton is projected upward from the origin at an angle of 60 degrees above the horizontal. The proton’s initial speed is 800,000 m/s. How much time (in μs ) is required for the proton to return to the x-axis? (Ignore gravitational forces.) Answer: 2.9 Solution: This is a standard problem in projectile motion in which the acceleration of gravity is replaced by the proton acceleration a = eE / m p . The time it takes for the proton to return to the x-axis is twice the time it needs to reach its maximum height or t = 2v 0 sin! / a = 2m p v 0 sin! / eE . Plugging in the numbers gives 2.9 μs .
Transcript
PHY2054 Spring 2013
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Prof. Paul Avery Prof. Zongan Qiu Feb. 12, 2013
Exam 1 Solutions 1. Three equal charges of value q are placed at 3 corners of a square of side L, as shown. Another charge Q is placed at position 4. What is the value of Q if the charge at position 2 feels no net force?
Answer: !2 2q
Solution: The net electric field at point 2 from the other two charges is
2kq2 / L2 along the +45 degree line. To cancel that force the force from Q
must satisfy kqQ / 2L( )2 = ! 2kq2 / L2 . This yields the solution shown.
2. Three charges q1, q2, and q3 are placed on the x-axis at x = 0, a, 2a respectively. The electric flux through a sphere of radius 1.5a, centered on the origin, is found to be !E = 10N "C . On the basis of only this information which of the following must be true?
Answer: q2 = !q1 +10"0
Solution: The sphere encloses only q1 and q2. From Gauss’ law, !E = 10 = q1 + q2( ) / "0 , from
which the solution follows.
3. A uniform electric field of 5,000 V/m is directed along the negative y-axis. A proton is projected upward from the origin at an angle of 60 degrees above the horizontal. The proton’s initial speed is 800,000 m/s. How much time (in µs ) is required for the proton to return to the x-axis? (Ignore gravitational forces.) Answer: 2.9
Solution: This is a standard problem in projectile motion in which the acceleration of gravity is replaced by the proton acceleration
a = eE / mp . The time it takes for the proton to return to the
x-axis is twice the time it needs to reach its maximum height or t = 2v0 sin! / a = 2mpv0 sin! / eE .
Plugging in the numbers gives 2.9 µs .
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4. Charges +2Q and +8Q are held in place at positions x = 0 m and x = 4 m, respectively. At what position in x (in m) should a third charge be placed so that it is in equilibrium?
Answer: +1.33 Solution: Both charges are positive so the equilibrium point is somewhere between them at
location 0 < x < L . Balance of electric field from point charges yields k2Q / x2 = k8Q / L! x( )2 . Solving for x then gives x = L / 3= 1.33 m.
5. A special room is set up with a downward pointing electric field of magnitude E = 23,000 V/m. A charged ball of mass 2.1 kg and charge –150 µC is shot vertically from the floor at 15 m/s. How high does the ball reach? g = 9.8 m/s2. Answer: 13.8 m
Solution: The electric field and charge give an additional upward force of qE, thus the net acceleration is a = !g + qE / m = 8.16 m / s2 (q and E are both negative). The maximum height
can be found from simple kinematics, i.e., a h! 0( ) = 1
2 0! v02( ) , which yields 13.8 m. You can
also set up the problem in terms of conservation of energy, where the potential energy includes both gravitational and electrical components. This yields
12 mv0
2 = mgh! qEh which when rearranged gives exactly the same equation as before.
6. Two charged point particles are located at two vertices of an equilateral triangle and the electric field is zero at the third vertex. We conclude
Answer: At least one other charge is present Solution: The two charged particles always give a net electric at the third vertex, because they can never be aligned along a single direction containing the third vertex. Thus another charge must be present somewhere to cancel the net electric field.
7. An electron traveling with velocity 4000 m/s along the +x direction, enters a region of uniform electric field 1!10"8 V/m pointing in the +y direction. After 3 sec, what is the speed of the electron in m/s?
Answer: 6620
Solution: The acceleration of the electron is eE / me in the negative y-direction. Since the initial
motion is along the x-direction, the final velocity is v = vx
2 + eEt / me( )2 = 6620 m/s.
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8. A 2.5-mC charge is on the y-axis at y = 3.0 m and a 6.3-mC charge is on the x-axis at x = 3.0 m. What is the direction of the potential at the origin?
Answer: Potential has no direction Solution: Potential is a scalar quantity, not a vector.
9. Two particles with charges Q and Q are fixed at the vertices of an equilateral triangle with sides of length a. If k = 1/ 4!"0 , the work required to move a particle with a charge q from the other vertex to the center of the line joining the fixed charges is:
Answer: 2kQq / a
Solution: The work done by the electric field is W = !"U = ! 4kQq / a ! 2kQq / a( ) = !2kQq / a . But the work done by an outside force is the negative of this or 2kQq / a . It must be positive because it’s clear that the outside force is pushing against the electric force so requires positive work.
10. The plate areas and plate separations of five parallel plate capacitors are (1) capacitor 1: area A0 , separation d0
(2) capacitor 2: area 2A0 , separation 2d0
(3) capacitor 3: area 2A0 , separation d0 / 2
(4) capacitor 4: area A0 / 2 , separation 2d0
(5) capacitor 5: area A0 , separation d0 / 2
Rank these according to their capacitances, least to greatest. Answer: 4, 1 and 2 tie, then 5, 3
Solution: Capacitance is calculated from C = !0 A / d , where A is the plate area and d is the plate
separation. Defining C0 = !0 A0 / d0 , then C1 = C2 = C0 , C3 = 4C0 , C4 = C0 / 4 and C5 = 2C0 . The ranking is then as shown.
11. A uniform electric field, with a magnitude of 600 V/m, is directed parallel to the positive x-axis. If the potential at x = 3.0 m is 1000 V, what is the change in potential energy of a proton as it moves from x = 3.0 m to x = 1.0 m?
Answer: 1.9!10"16J
Solution: The potential energy in this situation is falling along the +x direction, so the change in potential energy is !U = e!V = "eE!x = "1.6#10"19 # 600# "2( ) = +1.92#10"16 J.
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12. A pair of parallel plates, forming a capacitor, are connected to a battery. While the capacitor is still connected to the battery maintaining a constant voltage, the plates are pulled apart to double their original distance. What is the ratio of the final energy stored to the original energy stored?
Answer: 1/2
Solution: The energy stored in a capacitor is U = 1
2 CV 2 . If the plate separation is doubled and the voltage is held constant by the battery, then the capacitance is halved and the energy is halved.
13. An air-filled 3.0 nF capacitor is charged to 8.0 V. If the plate separation is 100 µm , what is the energy density in the electric field?
Answer: 28 mJ/m3
Solution: The energy in the capacitor is U = 1
2 CV 2 = 12 !0 A / d( )V 2 . The energy density u is
obtained by dividing the energy by the volume Ad, yielding u = 1
2 !0 / d2( )V 2 = 12 !0E2 . Thus
you don’t need to know the plate area A. However, you could solve for the plate area using the capacitance formula C = !0 A / d , then divide the total energy
12 CV 2 by Ad to yield the energy
density.
14. An air-filled parallel-plate capacitor has a capacitance of 2 pF. The plate separation is then tripled and a wax dielectric is inserted, completely filling the space between the plates. As a result, the capacitance becomes 4 pF. The dielectric constant of the wax is:
Answer: 6.0
Solution: The total capacitance is C =! "0 A / d . Since the plate separation is tripled, ! = 6 to cause a doubling of the capacitance.
15. A resistor in the form of a solid cylinder of material is connected across the terminals of an ideal battery and the current is measured to be 6.0A. A second resistor made of the same material and having the same volume but three times the length is put in series with the first resistor. What is the current in the new circuit?
Answer: 0.60 A Solution: Recall that the resistance is proportional to the cross sectional area and inversely proportional to the length. Let the resistance of the first resistor be R. If the volume of the second resistor is constant but the length is tripled, then the cross sectional area is 1/3 the original and its resistance is 9R. Thus the total resistance in the second circuit is 10R and the current is 1/10 the original current or 0.6 A.
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16. A light bulb is basically a wire with resistance that emits light when current goes through it (more current increases brightness). The light bulbs A and B in the circuit shown are identical. When the switch S is closed, what happens to the brightness of the two bulbs?
Answer: B gets brighter, A dimmer Solution: Before the switch is closed, the potential to the left of the switch is 18 V (from the battery) while the potential to the right of the switch is 15 V (30 V is split in half from current voltage drop). Thus A and B both have 15 V across them. When S is closed, the voltage to the right of the switch becomes 18 V, and the voltage across B is increased by 3 V to 18 V, while the voltage across A decreases to 30 – 18 = 12 V. Thus B gets brighter and A dimmer.
17. As shown in the figure, a 12-V battery with a 0.5 Ω internal resistance has initially no load connected across its terminals. Then the switches S1 and S2 are closed successively. The voltages across A and B (the terminals of the real battery) has which set of successive values (in volts)?
Answer: 12.0, 10.9, 10.0
Solution: When S1 is closed, the total resistance is 5.5 Ω and the current is 2.18A. The voltage across the terminals is thus 12! ir = 12! (2.18" 0.5) = 10.9 V. When S2 is closed, the total resistance in the circuit is 0.5 Ω + 2.5 Ω = 3.0 Ω and the total current is 4.0A. Thus the voltage across the terminals is 12! ir = 12! 4" 0.5( ) = 10.0 V.
18. If C1 = 25µF , C2 = 20µF , C3 = 10µF , and !V0 = 22 V,
determine the charge stored by C2 (in µC ).
Answer: 200 Solution: The overall equivalent capacitance is 150/11 µF and thus the charges are
q1 = q23 = !V0Ceq = 22"150 / 11= 300µC . The charge q23 = 300µC on C2 + C3 is divided between them in proportion to their capacitances, thus
19. What is the current through the 8 Ω resistor? Answer: 1.5 A
Solution: This is a two-loop circuit. Let the currents in the three branches (bottom to top) be i1, i2, i3, all directed to the right. From Kirchhof’s first rule, i3 = !i1 ! i2 . The loop equations, using
counterclockwise loops, are !8i1 +12i2 !18 = 0 (bottom loop) and
and i3 = 1.0 . Thus there is 1.5A going through the 8 Ω resistor.
20. What is the equivalent capacitance of the combination shown (in µF )?
Answer: 10 Solution: Combine 12 µF & 24 µF in series to yield 8 µF . That in parallel with 12 µF yields 20 µF which in series with the other 20 µF yields 10 µF .
