A capacitance balance is shown in the figure. The balance has an ideal block pulley system attached on one side and a parallel plate capacitor that has a variable gap width on the other side. Assume upper plate of the capacitor has negligible mass. When the separation between the plates is $\frac{1}{\sqrt{5}}$ meter, the attractive force between the plates balances the pulley block system. Then find value of $V_{0}$ in volts (EMF of battery). (Take area of the plate = $\frac{20}{3 ε_{0}} m^{2}$ and $g = 10 m / s^{2}$ )

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Detailed Solution

$\begin{array}{l} 10 g - T = 10 a \\ T - 5 g = 5 a \\ \Rightarrow T = \frac{200}{3} N \\ ∴ 4 T = \frac{800}{3} N \end{array}$

Thus , Tension in upper rope $= \frac{800}{3} N$

Force on one plate due to other $= F = Q E = (C V_{0}) (\frac{σ}{2 ε_{0}}) = (C V_{0}) (\frac{C V_{0}}{2 A ε_{0}}) = \frac{{V_{0}}^{2} A ε_{0}}{2 d^{2}} [∵ C = \frac{ε_{0} A}{d} a n d σ = \frac{Q}{A} = \frac{C V_{0}}{A}]$
By balancing Torque about ‘O’
$\begin{array}{l} \frac{800}{3} (4) = \frac{V_{0}^{2} ε_{0} A}{2 d^{2}} (4) \\ \Rightarrow \frac{800}{3} = \frac{V_{0}^{2} ε_{0} 20}{2 (\frac{1}{5}) 3 ε_{0}} \\ \Rightarrow V_{0} = 4 V o l t s \end{array}$