Question

Two charged capacitors, have their outer plates fixed and inner plates connected by a spring of force constant ${}^{\prime }{k}^{\prime }$. The charge on each capacitor is $q$. Find the extension in the spring at equilibrium.

• A. $\frac{q^2}{2A{ϵ}_{0}k}$
• B. $\frac{q^2}{4A{ϵ}_{0}k}$
• C. $\frac{q^2}{A{ϵ}_{0}k}$
• D. $Zero$

Answer 1

The correct option is A $\frac{q^2}{2A{ϵ}_{0}k}$

Force F with which the plates attract each other is given by,

$F=\frac{1}{2}qE$

$E=\frac{\sigma }{{ϵ}_0}$

$\Rightarrow F=\frac{1}{2}q\frac{\sigma }{{ϵ}_0}$

$=\frac{1}{2}q\frac{\sigma A}{{ϵ}_{0}A}$

$=\frac{q^2}{2{ϵ}_{0}A}$

When the spring is elongated by a length x then,

$F=kx$

or,

$\frac{q^2}{2{ϵ}_{0}A}=kx$

$\Rightarrow x=\frac{q^2}{2{ϵ}_{0}Ak}$