A parallel plate capacitor is charged by a DC supply of 500V. Plate separation is 1mm. If capacitor is lowered in to water with water filling in the gap between the plates, then change in pressure at any point between the plates is ${kε}_{0} \times 10^{13}$ (N/m²). Find the value of K, given $ε_{r}$ (for water 0 = 81).

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Detailed Solution

Electrostatic pressure = Electrostatic force per unit area = Potential energy per unit volume

$\{Note |\frac{dU}{dx}| = F, \frac{dU}{Adx} = \frac{F}{A}\}$

Electrostatic pressure at A

When water fills up to point A, electrostatic pressure will be

Change in pressure = $p_{2} - p_{1} = \frac{1}{2} ε_{0} (ε - 1) \frac{V^{2}}{d^{2}}$

$\Rightarrow p_{2} - p_{1} = ε_{0} \times 10^{13}$

Hence, K = 1.

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