Plate X of a capacitor is fixed and the other plate Y is attached to an ideal spring of force constant $k$ whose other end is attached to a fixed support as shown in figure. Initially a battery of emf $V_{0}$ is connected to the plates of capacitor and the spring and capacitor plates are in steady equilibrium position with separation between the plates being $d$ . From this position, if plate Y is displaced slightly as shown and released, then choose the correct option(s).[Take plate area as A and mass of plate as m]

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In equilibrium the spring will have an extension of $\frac{{Aε}_{0} V_{0}^{2}}{{2d}^{2} k}$

Heat is produced in the circuit in subsequent motion

Plate attached with spring would oscillate in SHM with time period $2 π \sqrt{\frac{{md}^{3}}{2 {kd}^{3} - {Aε}_{0} V_{0}^{2}}}$

Plate attached with spring would oscillate in SHM with time period $2 π \sqrt{\frac{{md}^{3}}{{kd}^{3} - {Aε}_{0} V_{0}^{2}}}$

answer is A, B, D.

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Detailed Solution

ABD
Capacitance
$C = \frac{{Aε}_{0}}{d} and Q = \frac{{Aε}_{0} V_{0}}{d}, F = \frac{{Aε}_{0} V_{0}^{2}}{2 d^{2}}$
At equilibrium ${kΔx}_{0} = \frac{{Aε}_{0} V_{0}^{2}}{2 d^{2}}$
$\Rightarrow {Δx}_{0} = \frac{{Aε}_{0} V_{0}^{2}}{2 d^{2} k}$
Now if small shift x has been given then
$F_{(restoring)} = \frac{{Aε}_{0} V_{0}^{2}}{2 {(d + x)}^{2}} - ({Δx}_{0} - x) k \Rightarrow \frac{{Aε}_{0} V_{0}^{2}}{2 d^{2}} {(1 + \frac{x}{d})}^{- 2} - {kΔx}_{0} + kx ∴ F_{(restoring)} = kx + \frac{{Aε}_{0} V_{0}^{2}}{2 d^{2}} (- 2 . \frac{x}{d}) k_{eq} x = x (\frac{{kd}^{3} - {Aε}_{0} V_{0}^{2}}{d^{3}})$