Two 11.0 cm -diameter electrodes 0.58 cm apart form a parallel-plate capacitor.

Question

Two 11.0 cm -diameter electrodes 0.58 cm apart form a parallel-plate capacitor. The electrodes are attached by metal wires to the terminals of a 15 Vbattery. After a long time, the capacitor is disconnected from the battery but is not discharged.

Part A

What is the charge on each electrode right after the battery is disconnected?

Enter your answers separated by a comma. Express number using three significant figures.

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Part B

What is the electric field strength inside the capacitor right after the battery is disconnected?

Express your answer using three significant figures.

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Part C

What is the potential difference between the electrodes right after the battery is disconnected?

Express your answer using three significant figures.

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Two 11.0 cm -diameter electrodes 0.58 cm apart form a parallel-plate capacitor. The electrodes are attached by metal wires to the terminals of a 15 Vbattery. After a long time, the capacitor is disconnected from the battery but is not discharged.

Part A

What is the charge on each electrode right after the battery is disconnected?

Enter your answers separated by a comma. Express number using three significant figures.

Q1,Q2 = C

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Part B

What is the electric field strength inside the capacitor right after the battery is disconnected?

Express your answer using three significant figures.

E = V/m

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Part C

What is the potential difference between the electrodes right after the battery is disconnected?

Express your answer using three significant figures.

VC = V

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Explanation / Answer

a)

C = A*epsilonnot/d = (pi*11*11*10^-4*8.85*10^-12/(0.58*10^-2))/4 = 1.45*10^-11 farad

Q = C*V =  5.8*10^-11*15 = 2.175*10^-10 C and - 2.175*10^-10C

b)

E = Q/(A*epsilonnot) = 2.175*10^-10*4/(pi*11*11*10^-4*8.85*10^-12) = 2586.07 V/m

c)

potential diff. = 15 V

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