$\mathrm{E}_{\mathrm{y}}=-\frac{\delta \mathrm{V}}{\delta \mathrm{y}}=-4 \mathrm{\,Vm}^{-1}$
$a_{x}=\frac{q E_{1}}{m}=\frac{1 \times 10^{-6} \times 3}{0.1}$
$=-3 \times 10^{-5} \mathrm{\,ms}^{-2}$
$a_{y}=\frac{q E_{y}}{m}=\frac{1 \times 10^{-6} \times 4}{0.1}$
$=-4 \times 10^{-5} \mathrm{\,ms}^{-2}$
Time taken to cross the $X$ - axis
Using $s=u t+\frac{1}{2} a t^{2}$
$3.2=\frac{1}{2} \times 4 \times 10^{-5} \times t^{2}$
$t=400 \mathrm{\,s}$

Reason : Potential due to charge of outer shell remains same at every point inside the sphere.

