Two 11-cm-diameter electrodes 0.42 cm apart form a parallel-plate capacitor. The

Question

Two 11-cm-diameter electrodes 0.42 cm apart form a parallel-plate capacitor. The electrodes are attached by metal wires to the terminals of a 15 V battery. After a long time, the capacitor is disconnected from the battery but is not discharged.

(A)What are the charge on each electrode, the electric field strength inside the capacitor, and the potential difference between the electrodes right after the battery is disconnected? E=?, Q=?, delta V=?

(B)What are the charge on each electrode, the electric field strength inside the capacitor, and the potential difference between the electrodes after insulating handles are used to pull the electrodes away from each other until they are 0.84 cm apart? Q=?, V=?, delta V=?

(C)What are the charge on each electrode, the electric field strength inside the capacitor, and the potential difference between the electrodes after the original electrodes are expanded until they are 22 cm in diameter? Q=?, V=?,delta V=?

Explanation / Answer

Area = pi * 0.112 = 0.038 m2

Seperation, d = 0.42 cm = 0.0042 m

V = 15 V

a) C = epsilon A/d = (8.85*10-12 * 0.038) / (0.0042) = 8.007 *10-11 F

Q = CV = 8.007 *10-11 * 15 = 1.201 nC

E = V/d = 15 / 0.0042 = 3571.4 N/C

Potential difference = V =15 V

B) d = 0.0084 m

C = epsilon A/d = (8.85*10-12 * 0.038) / (0.0084) = 4.003 *10-11 F

Q = CV = 4.003 *10-11 * 15 = 0.6 nC

E = V/d = 15 / 0.0084 = 1785.7 N/C

Potential difference = V =15 V

C)

a)

A =  pi * 0.222 = 0.152 m2

Note, since seperation is not mentioned i am taking the original seperation of 0.42 cm

C = epsilon A/d = (8.85*10-12 * 0.152) / (0.0042) = 32.028 *10-11 F

Q = CV = 32.028 *10-11 * 15 = 4.804 nC

E = V/d = 15 / 0.0042 = 3571.4 N/C

Potential difference = V =15 V

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