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In the given circuit, When switch S is opened, the plates are uncharged and are separated by distance d=8.00mm. Battery is of 100V and springs are identical with spring constant K. Capacitance of capacitor with separation d=8mm is 2 $μ$ F. When switch S is closed, the distance between plates is reduced to 4mm. Find the sum of spring constants of springs(in N/m).

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answer is 2500.

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Detailed Solution

When switch is closed, force on one spring is

$F = \frac{Q^{2}}{2 ε_{0} A} = \frac{4 C^{2} (ΔV)^{2}}{2 ε_{0} A} = \frac{2 C^{2} (ΔV)^{2}}{(\frac{ε_{0} A}{d}) \cdot d} = \frac{2 C (ΔV)^{2}}{d}$

On spring stretches by distance, $x = \frac{d}{4}$

so, $k = \frac{F}{x} = \frac{2 C (ΔV)^{2}}{d} \times \frac{4}{d} = \frac{8 C (ΔV)^{2}}{d^{2}}$ =2500N/m

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