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 on each capacitor is q and sign of charge is shown in figure. Find the extension in the spring at equilibrium.

Options

A. `(q^(2))/(2A epsilon_(0)k)`
B. `(q^(2))/(4 A epsilon_(0) k)`
C. `(q^(2))/(A epsilon_(k))`
D. zero

Answer

A

Explanation

We know that spring force F = kx will balance electrostatic force of attraction.
Therefore `(q^(2))/(2 A epsilon_(0)) = kx `
`implies x = (q^(2))/(2 A epsilon_(0) k)`