A small non-conducting ball carrying positive charge 0.01 C hangs in equilibrium from an ideal spring of spring constant 100 N/m as shown. Now, a vertically downward electric field of magnitude 2000 V/m is switched on. In its subsequent motion, the maximum kinetic energy (in Joules) of the ball is_______.

 

Initially, the extension in the spring is $\Delta x_i=\frac{mg}{k}$

After the electric field is switched on, it applies a downward force on the ball, and hence the equilibrium extension in the spring is now $\Delta x_f=\frac{mg+qE}{k}$

So, the amplitude of the oscillations of the ball is $\Delta x_f-\Delta x_i=\frac{qE}{k}$

And the angular frequency of the oscillations is $\omega =\sqrt{\frac{k}{m}}$

So, the maximum kinetic energy is $K{E}_{MAX}=\frac{1}{2}m{\omega }^2{A}^2=\frac{1}{2}k{A}^2=\frac{q^2{E}^2}{2k}=\frac{{\left(\left(0.01\right)\left(2000\right)\right)}^2}{2\left(100\right)}=2J$