Two 10.0-cm-diameter electrodes 0.50cm apart form a parallel-plate capacitor. Th

Question

Two 10.0-cm-diameter electrodes 0.50cm apart form a parallel-plate capacitor. The electrodes are attached by metal wires to the terminals of a 14V battery.

Part A

What is the charge on each electrode while the capacitor is attached to the battery?

Enter your answers numerically separated by a comma.

Part B

What is the electric field strength inside the capacitor while the capacitor is attached to the battery?

Part D

What is the charge on each electrode after insulating handles are used to pull the electrodes away from each other until they are 1.3cm apart? The electrodes remain connected to the battery during this process.

Enter your answers numerically separated by a comma.

Part 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 1.3cm apart? The electrodes remain connected to the battery during this process.

Part 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 1.3cm apart? The electrodes remain connected to the battery during this process.

Explanation / Answer

I am assuming the plates of the capacitor are circular disks
? = 8.854 x 10^-12 Farad per meter
C = ?A/d , where ? = 8.854 x 10^-12, A = ?5^2 = 78.53 cm^2
and d = 0.50 cm
C = 13.9 pF where p is 10^-12
a) Q = CV =194.6 pCoulombs
b) E = V/d = 14/0.50cm = 2,800 Volts per meter
c) 14 Volts
I am assuming that the capacitor in the following questions is still connected to the battery
d) Now C decreases to
C = ?o*A/d = 8.854x10^-12*?*r^2/d = 5.349x10^-12F

the capacitance will decrease to5.35 pF
The charge Q = CV = (5.35 pF)(14) = 74.9 pCoulombs
e) E = V/d = 14/1.3 = 1076.92Volts per meter
f) 14 Volts
If the capacitor is removed from the battery before the plates are separated

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