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in Physics by (69.4k points)

A parallel plate capacitor of separation 'd' is deformed as shown, with l1 , l2 >> d2 or d1 or d for the capacitance to remain unchanged.

(a)  l1 / d1 = l2 / d2

(b)  l1 / d2 = l2 / d1 .

(c)  (d1 l2 + d2 l1)/ d1 d2 = (l2 + l1)/d

(d)  l1 d2 + l2 d1 = (l2 + l1) d

1 Answer

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Best answer

Correct option (c)  (d1 l2 + d2 l1)/ d1 d2 = (l2 + l1)/d

Explanation:

Consider the capacitance of the deformed area alone,

1/C = 1/C1 + 1/C2

C is the capacitance for length l1 + l2 , C1 for length l1 and C2 for l2 .

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