\(\mathrm{W}=\mathrm{k}_{\mathrm{f}}-\mathrm{k}_{\mathrm{i}}\)

\(q E(2 a-0)=\frac{1}{2} m(2 V)^{2}-\frac{1}{2} m V^{2}\)

\(\mathrm{q} \mathrm{E} 2 \mathrm{a}=\frac{3}{2} \mathrm{mV}^{2}\)

\(\mathrm{E}=\frac{3}{4} \frac{\mathrm{mv}^{2}}{\mathrm{qa}}\)

Option \((\mathrm{B})\)

Rate of work done \(\mathrm{P}=\overrightarrow{\mathrm{F}} \cdot \overrightarrow{\mathrm{V}}=\mathrm{FV} \cos \theta=\mathrm{FV}\)

Power \(=q \mathrm{EV}\)

Power \(=\mathrm{q}\left(\frac{3}{4} \frac{\mathrm{mV}^{2}}{\mathrm{qa}}\right) \mathrm{V}\)

Power \(=9 \frac{3}{4} \frac{\mathrm{mV}^{3}}{\mathrm{qa}}\)

Power \(=\frac{3}{4} \frac{\mathrm{mV}^{3}}{\mathrm{a}}\)

Option \((\mathrm{C})\)

Angle between electric force and velocity is \(90^{\circ},\) hence rate of work done will be zero at \(\mathrm{Q} .\)

Option (D)

Initial angular momentum \(\mathrm{L}_{\mathrm{i}}=\mathrm{mVa}\)

Final angular momentum \(\mathrm{L}_{\mathrm{f}}=\mathrm{m}(2 \mathrm{V})\) (2a)

Change in angular momentum \(\mathrm{L}_{\mathrm{f}}-\mathrm{L}_{\mathrm{i}}=3 \mathrm{mVa}\)