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Chapter 18.2 Review
Capacitance and Potential
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1. A 5 μF capacitor is connected to a 12 volt battery. What is the potential difference across the plates of the capacitor when it is fully charged?
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12 volts, a charged capacitor has the same potential difference across the plates as the source.
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2. What is the charge on one plate of the capacitor in problem 1 when it is fully charged? What is the net charge on the capacitor when it is fully charged?
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q = VCq = 12 x 5 = 60 μCor
q = 12(5 x 10-6)= 6 x 10-5 C
But the net charge on a capacitor is always zero because the + and – plate have charges of equal magnitude.
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3. How much electrical potential energy is stored in the capacitor in problem 1?
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PEE = ½ CV2
PEE = ½ 5(12)2
PEE = 360 μJor,PEE = ½ (5x 10-6)(12)2
PEE = 3.6 x 10-4 J
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4. If the plates are separated by a distance of 0.003 m, what is the strength of the electric field across the plates?
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Ed = VE(0.003) = 12E = 4000 N/C
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5. A 9 μF capacitor is connected to a 9 volt battery. What is the potential difference across the plates of the capacitor when it is fully charged?
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9 volts, a charged capacitor has the same potential difference across the plates as the source.
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6. What is the charge on one plate of the capacitor in problem 5 when it is fully charged? What is the net charge on the capacitor when it is fully charged?
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q = VCq = 9 x 9 = 81 μCor
q = 9(9 x 10-6)= 8.1 x 10-5 C
But the net charge on a capacitor is always zero because the + and – plate have charges of equal magnitude.
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7. How much electrical potential energy is stored in the capacitor in problem 5?
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PEE = ½ CV2
PEE = ½ 9(9)2
PEE = 364.5 μJor,PEE = ½ (9 x 10-6)(9)2
PEE = 3.645 x 10-4 J
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8. If the plates are separated by a distance of 0.005 m, what is the strength of the electric field across the plates?
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Ed = VE(0.005) = 9E = 1800 N/C
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9. List four ways to increase the charge on a capacitor.
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1. Larger plates2. Plates closer
together3. Different dielectric4. Increase voltage
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10. What is the difference in electrical potential energy and electric potential?
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Electric potential is electric potential energy divided by charge.
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11. A proton is moved 0.5 m in the direction of an electric field with a strength of 5000 N/C. What is the change in electric potential?
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Ed = V5000 x 0.5 = V2500 V = VMoving in the direction of the field is a decrease of potential, so
-2500 V = V.
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12. What is the change in electrical potential energy of the proton in problem 11?
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PEE = qV
PEE = (1.6 x 10-19)(2500)
PEE = -4 x 10-16 J
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13. An electron is moved 0.8 m in the direction of an electric field with a strength of 3000 N/C. What is the change in electric potential?
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Ed = V3000 x 0.8 = V2400 V = VMoving in the direction of the field is a decrease of potential, so
-2400 V = V.
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14. What is the change in electrical potential energy of the electron in problem 13?
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PEE = qV
PEE = (-1.6 x 10-19)(-2400)
PEE = 3.88 x 10-16 J
The electron is being moved in the direction it does not “want” to go. That is an increase in potential energy.
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15. A helium nucleus (composed of two protons and two neutrons) is moved 24 cm in the direction of an electric field with a strength of 9000 N/C. What is the change in electric potential?
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Ed = V9000 x 0.24 = V2160 V = VMoving in the direction of the field is a decrease of potential, so
-2160 V = V.
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16. What is the change in electrical potential energy of the helium nucleus in problem 15?
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PEE = qV
PEE = (1.6 x 10-19)(2)(-2160)(There are two protons in one helium nucleus.)
PEE = -6.912 x 10-16
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