Question

4. Two charged capacitors have their outer plates fixed and inner plates connected by a spring of force constant ‘K. The charge on each capacitor is q. Find the extension in the spring at equilibrium​

Answer 1

Answer:

The extension in the spring is q^2/2ϵ0Ak

Explanation:

Force F with which the plates of a capacitor attract each other is given by

F=1/2qE

where q is the charge on the capacitor and E, the electric field between the plates. Substitute E by σ/ϵ0 in the above equation.

F=1/2qσ/ϵ0=1/2qσA/ϵ0A=q^2/2ϵ0A

If the spring is elongated by a length x then

F=kx

Or

q^2/2ϵ0A =kx

⇒x=q^2/2ϵ0Ak

Hence the extension in the spring is q^2/2ϵ0Ak