On plate of a capacitor is connected to a spring as shown in fig. Area of both the plates is A. In steady state, separation between the plates is 0 8 d (spring was in its natural length and the distance between the plates was d when the capacitor was uncharged). The force constant of the spring is approximately

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4ε0AV2d3

2ε0AVd2

6ε0V2Ad3

ε0AV32d3

answer is A.

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Detailed Solution

The electrostatic attraction between the plates = spring force ∴ q22ε0A=kx or (CV)22ε0A=k[d−0⋅8d]or ε0A0⋅8d2V22ε0A=k(0⋅2d)or k=ε0AV20⋅256d3≈4ε0AV2d3.

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