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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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By Expert Faculty of Sri Chaitanya
a
4ε0AV2d3
b
2ε0AVd2
c
6ε0V2Ad3
d
ε0AV32d3

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detailed solution

Correct option is A

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.

ctaimg

Similar Questions

Two identical capacitors, have the same capacitance C. One of them is charged to potential V1 and the other to V2. The negative ends of the capacitors are connected together. When the positive ends are also connected, the decrease in energy of the combined system is

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