Two 12-cm-diameter electrodes 0.54cm apart form a parallel-plate capacitor. The

Question

Two 12-cm-diameter electrodes 0.54cm apart form a parallel-plate capacitor. The electrodes are attached by metal wires to the terminals of a 14V battery. What are the charge on each electrode, the electric field strength inside the capacitor, and the potential difference between the electrodes:

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?

B. After insulating handles are used to pull the electrodes away from each other until they are 1.5cm apart? The electrodes remain connected to the battery during this process.

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

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Explanation / Answer

a) Q = CV = e0 A V/d = 8.85E-12*pi*0.06^2*14/0.54E-2 = 2.59E-10 C

E = v/d = 14/0.54E-2 = 2593 N/C

V = 14 V

b) now

Q = CV = e0 A V/d = 8.85E-12*pi*0.06^2*14/1.5E-2=9.34 E-11 C

E = 14/1.5E-2=933 N/C

V = 14 V

c. now Q=8.85E-12*pi*0.105^2*14/0.54E-2=7.95E-10 C

E doesnt change E = 2593 N/C

V = 14 V

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