A uniform electric field of strength E exists in a region. An electron (charge –q, mass m) enters a point A with velocity V $\widehat{j}$. It moves through the electric field & exits at point B with same speed as shown in the figure.Then: 

As velocity along y-axis remains unchanged, so there should not be any electric field along y axis.

As velocity along x axis is increasing, so force on the electron must be along +x direction, so electric field must be towards –x direction.
(A) Velocity along x axis at B:
From A $\to$  B
  ${\text{V}}_{\text{x}}{\text{= u}}_{\text{x}}{\text{+ a}}_{\text{x}}\text{t}$
Or   ${\text{V}}_{\text{x}}\text{= 0 + }\left(\frac{\text{eE}}{\text{m}}\right)\left(\frac{\text{d}}{\text{V}}\right)\Rightarrow V_x=\frac{\text{eEd}}{\text{mV}}$
Where,  $\text{E=}\frac{{\text{2amv}}^{\text{2}}}{{\text{ed}}^{\text{2}}}\Rightarrow \therefore {\text{V}}_{\text{x}}\text{=}\frac{\text{2aV}}{\text{d}}$
(D) Net velocity vector at B

$$\begin{gathered} {\overrightarrow{V}}_B=V_x\stackrel{\wedge }{i}+V_y\stackrel{\wedge }{j} \\ {\overrightarrow{V}}_B=\frac{2aV}{d}\stackrel{\wedge }{i}+V\stackrel{\wedge }{j} \end{gathered}$$
  
(B) Rate of work done at B = Power =  $\overrightarrow{F}\times {\overrightarrow{V}}_B$
$\text{=}\left(eE\hat{i}\right)\text{.}(\frac{\text{2aV}}{\text{d}}\hat{i}+V\hat{j})$ . 
$\text{=eE}\left(\frac{\text{2aV}}{\text{d}}\right)$ ; Where,  $\text{E = }\frac{{\text{2amV}}^{\text{2}}}{{\text{ed}}^{\text{2}}}$
$\Rightarrow \therefore \text{P=}\frac{{\text{4ma}}^{\text{2}}{\text{V}}^{\text{3}}}{{\text{d}}^{\text{3}}}$  
(C) Rate of work done at A:
  $P_A=\overrightarrow{F}\times \overrightarrow{V_A}$
 $\text{=}\left(e\overrightarrow{E}\hat{i}\right)\times \left(V\hat{j}\right)=0$