Chapter 18 Pretest Capacitance and Potential

1. The amount of charge that can be placed on a capacitor does not depend on: A) the area of the plates, B) the conducting material of the plates, C) the distance separating the plates, D) the insulating material between the plates.

1. The amount of charge that can be placed on a capacitor does not depend on: A) the area of the plates, B) the conducting material of the plates, C) the distance separating the plates, D) the insulating material between the plates.

2. Dielectric strength: A) indicates the potential gradient a material will withstand without being punctured by a spark discharge, B) is a dimensionless quantity, C) is directly related to the amount of space needed between capacitor plates, D) uses a vacuum as the standard of comparison.

2. Dielectric strength: A) indicates the potential gradient a material will withstand without being punctured by a spark discharge, B) is a dimensionless quantity, C) is directly related to the amount of space needed between capacitor plates, D) uses a vacuum as the standard of comparison.

3. Two point charges of 4. 3 μC and 5. 6 μC are separated by 0. 50 m 3. Two point charges of 4.3 μC and 5.6 μC are separated by 0.50 m. What is the electrical potential energy of this two charge system?

PE = k q1 q2 / d PE = (9 x 109)(4.3 x 10-6)(5.6 x 10-6)/(0.5) PE = 0.43344 J The charges are the same so the energy is positive.

4. A uniform field is 600 N/C and is directed straight up from the ground. The potential at a distance of 20 m above ground is 4500 V. What is the potential 8 m above ground?

Ed = V, so the potential difference between 20 m and 8 m is 600 x 12 = 7200 V. Since the field is directed upward, the potential must be 7200 V higher at 8 m: 4500 + 7200 = 11700 V.

5. What is the potential at a distance of 0 5. What is the potential at a distance of 0.5 m from a point charge of 7.0 μC? μ μ

V = k q / d V = (9 x 109)(7.0 x 10-6) / 0.5 V = 126000 V

6. A 0. 50 μF capacitor is connected to a 12 V battery 6. A 0.50 μF capacitor is connected to a 12 V battery. What charge is stored on the capacitor? μ μ

q = VC q = 12 x 0.5 = 6 μC or q = 12(0.5 x 10-6)= 6 x 10-6 C

7. A 0. 25 μF capacitor is connected to a 9. 0 V battery 7. A 0.25 μF capacitor is connected to a 9.0 V battery. How much potential energy is stored on the capacitor?

PEE = ½ CV2 PEE = ½ 0. 25(9)2 PEE = 10. 125 μJ or, PEE = ½ (0 PEE = ½ CV2 PEE = ½ 0.25(9)2 PEE = 10.125 μJ or, PEE = ½ (0.25x 10-6)(9)2 PEE = 1.0125 x 10-5 J

1. Compare electric potential and electrical potential difference.

Electric potential and electrical potential difference are the same.

I meant to ask you to compare electric potential ENERGY and electrical potential difference. Electrical potential difference is electric potential energy divided by charge.

2. Two capacitors, 0.200 μF and 0.500 μF, are connected in parallel and charged to a potential difference of 80.0 V. What is the total charge acquired?

Total capacitance of capacitors in parallel is the sum of the capacitances. CT = 0.2 + 0.5 = 0.7 μF q = VC q = 80 x 0.7 = 56 μC or q = 80(0.7 x 10-6)= 5.6 x 10-5 C

3. The capacitors of Item 2 are discharged and connected in series 3. The capacitors of Item 2 are discharged and connected in series. A charge of +50.0 μC is transferred to the ungrounded terminal. Calculate the total potential difference across the two capacitors.

Total reciprocal of the capacitance of capacitors in series is the sum of the reciprocals of the capacitances. 1/CT = 1/0.2 + 1/0.5 1/CT = 5 + 2 1/CT = 7 1/CT = 1/7 μF 50 μC = V(1/7 μF) 350 V = V