Question: Two 14.0cm -diameter electrodes 0.56cm apart form a parallel-plate cap

Question

Question: Two 14.0cm -diameter electrodes 0.56cm apart form a parallel-plate capacitor. The electrodes are attached by metal wires to the terminals of a 13V battery. After a long time, the capacitor is disconnected from the battery but is not discharged.

A: What is the charge on each electrode right after the battery is disconnected? (q1, q2)

B: What is the electric field strength inside the capacitor right after the battery is disconnected? (E= )

C: What is the potential difference between the electrodes right after the battery is disconnected? (deltaV= )

D: What is the charge on each electrode after insulating handles are used to pull the electrodes away from each other until they are 1.9cm apart? (q1, q2)

E: What is the electric field strength inside the capacitor after insulating handles are used to pull the electrodes away from each other until they are 1.9cm apart? (E= )

F: What is the potential difference between the electrodes after insulating handles are used to pull the electrodes away from each other until they are 1.9cm apart? (V= )

Explanation / Answer

A)

capacitance , C = A*epsilon/d

C = pi*(.07)^2 * 8.854 *10^-12 / .0056

C = 2.433 *10^-11 F

as Q = CV

Q = 2.433 *10^-11 * 13

Q = 3.164 *10^-10 C

the charge on electrods is q1 = 3.164 *10^-10 C

q2 = -3.164 *10^-10 C

B)

E = V/d

E = 13/.0056

E = 2321.4 N/C

C)
potential stays the same ,

V = 13 V

D)

as the battery is disconneted , the charge on electrodes stays the same ,

q1 = 3.164 *10^-10 C

q2 = -3.164 *10^-10 C

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