A parallel-plate air capacitor has a capacitance of 830 pF. The charge on each p

Question

A parallel-plate air capacitor has a capacitance of 830 pF. The charge on each plate is 1.80 µC. (a) What is the potential difference betweenthe plates?
1 V

(b) If the charge is kept constant, what will be the potentialdifference between the plates if the separation is quadrupled?
2 V

(c) How much work is required to quadruple the separation?
3 J (a) What is the potential difference betweenthe plates?
1 V

(b) If the charge is kept constant, what will be the potentialdifference between the plates if the separation is quadrupled?
2 V

(c) How much work is required to quadruple the separation?
3 J

Explanation / Answer

a) Q=CV =>V = Q/C = 1.80C / 830pF = 2168.67V b) since C = A/d, when d->4d, C->C/4 V=Q/C/4 = 4* 2168.67V = 8674.68V

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