Exam 1 Solutions

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. V − − − − − − 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 v the two plates, so the velocity will be reduced. Conservation of en- 11222 ergy yields 22mvp ipf−Δ eV = mv , so vveVmf =−Δip2/.

2. Five identical charges of +1μC are located on the 5 vertices of a regular hexagon y 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 x Coulomb force on this charge.

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 di- ameter 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 2 Solution: The current i is iAnev= ed, where Ar= π is the cross sectional area, ne is the free 2 electron density and vd is the drift velocity. The answers follow by writing virende= /π and comparing the diameters.

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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 iAnev= ed 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 /ε0 , where σ = QA/ is the surface charge density (you can also use QCV= with CAd= ε0 / and VEd= ). Thus QA==σ ε0 AE.

6. In the previous problem, the electric field formed by the earth and the cloud is pointing verti- cally 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 qmgE=− / .

7. A positive charge Q = +2μC | +3μC | +4μC is placed on the y-axis at y = 1 y m and a charge −Q is placed on the y-axis at y = −1 m. What is the potential X A difference VA − VB between the points A = (0,2 m) and B = (1 m,0)? +Q B Answer: 12 kV | 18 kV | 24 kV X x 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 dis- −Q tance of A to −Q is 3 m. We obtain VAB− V=−= kQ /1 kQ / 3 2 kQ / 3.

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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 ()ii12+ /2 and the total charge delivered in a time t is Qiit=+( 12) /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 qCV111= be the initial charge on C1 and q be the charge that moves from C1 to the initially uncharged C2. Using charge conservation, the condi- tion of equal potential can be written (qqCqC112−=) //. Solving for q (which is the charge on

C2) yields qqCC=+11() 2// CCVCC 1211 =+( 2) C 2.

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 nei- ther capacitor can withstand a potential difference greater than 100 volts with- out breakdown, what is the maximum value of the battery emf that can be sus- tained 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 VqC11= / and VqC22= / . The first capacitor is smaller so it will hit 100 V first. When this happens the voltage on C2 is 100()CC12 / and the total emf is 100+=+ 100CC12 / 100( C 1 C 2) / C 2.

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 1122 energy is 26CVeq = CV.

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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 2 Solution: The initial force is FkQQd112= / . When the balls are touched, since they are identical and conducting, they exchange charge until charges are equal (QQQ′ =+( 12) /2). Thus 22 2 FkQd2 = ′ / and F21//4FQQ=+() 1 2 QQ 12.

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 1 2 Solution: Since CAd= ε0 / and V is the same, the energy UCV= 2 will be inversely propor- tional to the separation distance.

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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 CCCCCeq =+A BAB/ (), which allows us to calculate q as

qC==eq VCCCA BAB/ () + C . 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 11 qVCVCCCC==22eq A BAB/ () +.

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, strik- ing 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 agqEm=+ / . Us- 1 2 1 2 ing hat= 2 , where h is the height where the ball is dropped, we get hgqEmt=+2 ()/ , from which we can calculate E.

Answer: 7650 | 8730 | 9840 Solution: We equate the centripetal force to the electrical force between the electron (charge –e) 222 2 and the nucleus (charge +Ze), mve // r= Zke r . Solving for v yields vZkemr= / e .

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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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 ∑EAii⋅=qenc/ε 0 , where the electric flux on the left is equal to i −4π rE2 (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 2 2 −=4/π rE() qcenter + q shellε 0 , which can be rearranged to get qqshell=− center −επ 0 4 rE.