A particle of charge + q and mass m moving under the influence of a uniform electric field E i and a uniform magnetic field B k follows trajectory from P to Q as shown in fig. The velocities at P and Q are v i and 

- 2 v j respectively. Which of the following statements (s) is/are correct

(i) $\mathrm{W}=(\mathrm{K}.\mathrm{E}.{)}_{\mathrm{Q}}-(\mathrm{K}.\mathrm{E}.{)}_{\mathrm{P}}$
$\therefore \mathrm{qE}\times 2\mathrm{a}=\frac{1}{2}\mathrm{m}(2\mathrm{v}{)}^2-\frac{1}{2}{\mathrm{mv}}^2$
or $\mathrm{qE}\times 2\mathrm{a}=\frac{1}{2}\mathrm{m}\times 3{\mathrm{v}}^2$
$\mathrm{E}=\frac{3}{4}\left(\frac{{\mathrm{mv}}^2}{\mathrm{qa}}\right)$
(ii) The rate of work done by electric field at P
$\begin{array}{l}=\mathrm{Fv}=\mathrm{qEv}\\ =\mathrm{q}\left[\frac{3}{4}\left(\frac{{\mathrm{mv}}^2}{\mathrm{qa}}\right)\mathrm{v}\right]=\frac{3}{4}\left(\frac{{\mathrm{mv}}^3}{\mathrm{a}}\right)\end{array}$
(iii) At Q, the force due to electric field is q E along X-axis. Here velocity v is along negative y-axis. Hence rate of doing work F_e .v= 0            $\left(\because \mathrm{\theta }={90}^{\circ }\right)$
The force due to magnetic field
F_m = q (v x B)
Here F_m- is also perpendicular to velocity vector v,
$\therefore$ Rate of doing work in magnetic field
F_m. v = 0