Two 14-cm-diameter electrodes 0.48 cm apart form a parallel-plate capacitor. The

Question

Two 14-cm-diameter electrodes 0.48 cm apart form a parallel-plate capacitor. The electrodes are attached by metal wires to the terminals of a 17 V battery. What are the charge on each electrode, the electric field strength inside the capacitor, and the potential difference between the electrodes.

Part A

What are the charge on each electrode, the electric field strength inside the capacitor, and the potential difference between the electrodes while the capacitor is attached to the battery?

Q=? C

Part B

E=? V/m

Part C

(delta)V=? V

Part D

After insulating handles are used to pull the electrodes away from each other until they are 1.1 cm apart? The electrodes remain connected to the battery during this process.

Q=? C

Part E

E=? V/m

Part F

(delta)V=? V

Part G

After the original electrodes (not the modified electrodes of part D) are expanded until they are 21 cm in diameter while remaining connected to the battery?

Q=? C

Part H

E=? V/m

Part I

(delta)V=? V

If you could show how to do it and the answer it would amazing! Thank you!

Explanation / Answer

Here ,

part A)

capacitance of capacitor , C = Area*epsilon/d

C = 8.854 *10^-12 * pi*(0.07)^2/.0048

C = 2.84 *10^-11 F

Q = C*V

Q = 2.84 *10^-11 * 17

Q = 4.826 *10^-10 C

the charge on the capacitor is 4.826 *10^-10 C

part B)

electric field , E = V/d

E = 17/.0048

E = 3542 N/C

the electric field is 3542 N/C

part C)

the potential difference across the electrodes is 17 V

part D)

fpr d = 1.1 cm

capacitance of capacitor , C = Area*epsilon/d

C = 8.854 *10^-12 * pi*(0.07)^2/.011

C = 1.239 *10^-11 F

Q = C*V

Q = 1.239 *10^-11 F * 17

Q = 2.106 *10^-10 C

the charge on the capacitor is 2.106 *10^-10 C

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