Two 10-cm-diameter electrodes 0.60cm apart form a parallel-plate capacitor. The

Question

Two 10-cm-diameter electrodes 0.60cm apart form a parallel-plate capacitor. The electrodes are attached by metal wires to the terminals of a 17V battery. After a long time, the capacitor is disconnected from the battery but is not discharged.

Part 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?

Q=

Part B

Express your answer to two significant figures and include the appropriate units.

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Part C

Express your answer to two significant figures and include the appropriate units.

Part D

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 1.0cm apart?

Express your answer to two significant figures and include the appropriate units.

Part E

Express your answer to two significant figures and include the appropriate units.

Part F

Express your answer to two significant figures and include the appropriate units.

Part G

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 (not the modified electrodes of part b) are expanded until they are 20cm in diameter?

Express your answer to two significant figures and include the appropriate units.

Part H

Express your answer to two significant figures and include the appropriate units.

Part I

Express your answer to two significant figures and include the appropriate units.

Explanation / Answer

Area of electrodes=pie*R^2=pie*0.05^2=7.85*10^-3
d=6*10^-3 m

A: Q=CV
C=epsilon*A/d
epsilon = 8.85x10^-12 F/m
=> C=1.15*10^-11
=> Q=CV=1.955*10^-10

B:
E=Q/(A*epsilon)=2814.06
C:
V=17

D:
d=0.01m
C=6.9*10^-13
E:
E=Q/(A*epsilon)=9.9
F:
V=17 V
G:=> Area =4*7.85*10^-3
=> new capacitance=4*1.15*10^-11=4.6*10^-11
Q=CV=7.82*10^-10
H:
E=Q/(A*epsilon)=2.475
I:
V=17V

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