A block of mass m containing a net negative charge  $-q$ is placed on a frictionless horizontal table and is connected to a wall through an unstretched spring of spring constant K. If horizontal electric field E parallel to the spring is switched on, then the maximum compression of the spring is

Force on the block $F=qE$  is towards left
Let maximum compression in the spring be x. Then work done by electric force during compression = the increase of potential energy in the spring.    
                                                                                                            $$\begin{gathered} Fx=\frac{1}{2}K{x}^2-0 \\ Eqx=\frac{1}{2}k{x}^2 \\ x=\left(\frac{2Eq}{K}\right) \end{gathered}$$