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Chapter 18 Capacitance and Potential
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1. What is the electric potential energy between a proton and an electron separated by a distance of 5.3 x m?
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PE = k q1 q2 / d EPE = (9 x 109)(1. 6 x 10-19)(-1. 6 x 10-19)/(5
PE = k q1 q2 / d EPE = (9 x 109)(1.6 x 10-19)(-1.6 x 10-19)/(5.3 x ) PE = x J The charges are opposite so the energy is negative.
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2. Two positive point charges are separated by a distance of 4 cm
2. Two positive point charges are separated by a distance of 4 cm. If they are moved to a distance of 12 cm apart, by what factor does the electric potential energy between them change?
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PE = k q1 q2 / d Three times as far away, 1/3 the potential energy
PE = k q1 q2 / d Three times as far away, 1/3 the potential energy. This is NOT an inverse-square relationship.
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3. A particle of charge 3 x C moves 5 m in the direction of a uniform electric field of strength 4000 N/C. What is the change in electrical potential energy?
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PEE = qEd PEE = (3 x 10-15)4000(5) PEE = -6 x J (The positive charge is moving in the direction it wants to go, so it is a decrease in potential energy.)
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4. If you move the particle in problem 3 back to its starting position, what is the change in electrical potential energy for this second move?
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It is being moved in the direction it doesn’t want to go, so it is an increase in potential energy. +6 x J
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5. What is the strength of the electric field in problem 3?
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Duh, 4000 N/C. I wanted you to calculate it using Ed = V, but I messed up!
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6. Two point charges, +2. 7 μC and +3. 8 μC, are 0. 55 m apart
6. Two point charges, +2.7 μC and +3.8 μC, are 0.55 m apart. What is the electrical potential energy of this system of two charges?
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PE = k q1 q2 / d EPE = (9 x 109)(+2. 7 x 10-6)(+3. 8 x 10-6)/(0
PE = k q1 q2 / d EPE = (9 x 109)(+2.7 x 10-6)(+3.8 x 10-6)/(0.55 m) PE = J The charges are the same so the energy is positive.
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7. What would be the electrical potential energy of the system in problem 6 if both charges were negative instead of positive?
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The same, PE = J, two negative charges would still produce a positive potential energy by PE = k q1 q2 / d.
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8. A proton (q = 1.6 x 10-19) moves 3 meters along an electric field of 500 N/C. What is the potential difference between the proton’s beginning and final location?
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Ed = V 500 x 3 = V 1500 V = V Moving in the direction of the field is a decrease of potential, so V = V.
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9. A uniform electric field of 1000 N/C is directed downward
9. A uniform electric field of 1000 N/C is directed downward. If the potential at 10 m above the ground is 15,000 V, what is the potential at ground level? At 3 m above the ground?
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Ed = V, so the potential difference between 10 m and 0 m is 1000 x 10 = V. Since the field is directed downward, the potential must be V lower at the ground: = 5000 V. At 3 m above the ground the potential must be 1000 x 3 higher: = 8000 V.
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10. What is the electric potential 2 m from a point charge of 22 μC?
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V = k q / d V = (9 x 109)(22 x 10-6) / 2 V = V Potential is a scalar, so there is no direction associated with this answer.
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11. Two point charges are separated by 3 meters
11. Two point charges are separated by 3 meters. If the charges are 25 μC and 13 μC, what is the potential at the midway point between the two charges?
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V = k q / d V1 = (9 x 109)(25 x 10-6) / V2 = (9 x 109)(13 x 10-6) /1.5 V1 = V V2 = V V = = V
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12. Two point charges are separated by 4 meters
12. Two point charges are separated by 4 meters. If the charges are +36 μC and -24 μC, what is the potential at the midway point between the two charges?
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V = k q / d V+ = (9 x 109)(36 x 10-6) / V- = (9 x 109)(-24 x 10-6) /2 V+ = V V- = V V = = V
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13. At what distance from a point charge of 16 μC is the potential 2 x 105 V?
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V = k q / d 2 x 105 = (9 x 109)(16 x 10-6) / d d = 0.72 m
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14. At what distance from a point charge of 16 μC is the potential zero V?
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The potential is always zero at infinity.
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