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Q.

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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a

$\frac{ε_{0} {AV}^{3}}{2 d^{3}}$

b

$\frac{6 ε_{0} V^{2}}{{Ad}^{3}}$

c

$\frac{4 ε_{0} {AV}^{2}}{d^{3}}$

d

$\frac{2 ε_{0} AV}{d^{2}}$

answer is A.

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Detailed Solution

The electrostatic attraction between the plates = spring force
$∴ \frac{q^{2}}{2 ε_{0} A} = kx$ or $\frac{(CV)^{2}}{2 ε_{0} A} = k [d - 0 \cdot 8 d]$
or $\frac{{(\frac{ε_{0} A}{0 \cdot 8 d})}^{2} V^{2}}{2 ε_{0} A} = k (0 \cdot 2 d)$
or $k = \frac{ε_{0} {AV}^{2}}{0 \cdot 256 d^{3}} \approx \frac{4 ε_{0} {AV}^{2}}{d^{3}}$ .

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