Two charged capacitors have their outer plates fixed and inner plates connected by a spring of force constant $k$ . The magnitude of charge on each capacitor is $q$ and sign of charge is shown in figure. Find the extension in the spring at equilibrium.

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We know that spring force F = k x will balance electrostatic force of attraction. Therefore, q 2 2 A ε 0 = k x ⇒ x = q 2 2 A ε 0 k

Physics - Electrostatics Question with Solution | TestHub

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