Two identical blocks are kept on a frictionless horizontal table connected by a spring of stiffness $$k$$ and of original length $${\ell _0}.$$ A total charge $$Q$$ is distributed on the block such that maximum elongation of spring at equilibrium is equal to $$x.$$ Value of $$Q$$ is

Solution :
For maximum elongation charges on the blocks must be equal to $$\frac{Q}{2}$$ on each block.
$$\therefore \frac{1}{{4\pi {\varepsilon _0}}}\frac{{\frac{Q}{2}\frac{Q}{2}}}{{{{\left( {{\ell _0} + x} \right)}^2}}} = kx,\,\,Q = 2\left( {{\ell _0} + x} \right)\sqrt {4\pi {\varepsilon _0}kx} .$$