A parallel-plate air capacitor is to store charge of magnitude 220 PC on each pl

Question

A parallel-plate air capacitor is to store charge of magnitude 220 PC on each plate when the potential difference between the plates is 35.0 V. If the area of each plate is 6.80 cm^2. what is the separation between the plates? Express your answer with the appropriate units. If the separation between the two plates is double the value calculated in part (a), what potential difference is required for the capacitor to store charge of magnitude 220 PC on each plate? Express your answer with the appropriate units.

Explanation / Answer

1) since C = q / V ie., eo A / d = q / V , where 'd' is the separation required

so that d = eo A V / q = 8.85x10^-12 x 6.8x10^-4 x 35/ 220 x10^-12 = 9.57 x10^-4 m

2) potential difference required V = q / C = q d / eo A = 220x10^-12 x 2*9.57x10^-4 / 8.85x10^-12 x 6.8x10^-4

V = 69.97 volt

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