Physics - Electrostatics Question with Solution | TestHub

As velocity in y direction remain same there should be No field in y-direction time taken to reach $(2\mathrm{a},\mathrm{d})$ from $(\mathrm{a},0)=\frac{{\mathrm{s}}_{\mathrm{y}}}{{\mathrm{v}}_{\mathrm{y}}}=\frac{\mathrm{d}}{\mathrm{v}}$

$$\begin{array}{rlrlrlr}& s_x=\frac{1}{2}{a}_x{t}^2\Rightarrow a=\frac{1}{2}\frac{eE}{m}\frac{d^2}{v^2}\Rightarrow \vec{E}=\frac{2am{v}^2}{e{d}^2}& \vec{E}=\frac{-2am{v}^2}{e{d}^2}(\hat{i})& \vec{v}=\overrightarrow{v_x}\hat{i}+{\vec{v}}_y\hat{j}=\left(\frac{2av}{d}\right)\hat{i}+v_y\hat{j}\left[{\vec{v}}_x=a_{x}t\right]& P=\vec{F}\cdot \vec{v}& P_B=\frac{2am{v}^2}{d^2}\times \frac{2av}{d}\Rightarrow \frac{4am{v}^3}{d^3}& P_A=0\left[\text{ As angle between }G\text{ and }\vec{v}\text{ is }90^{\circ }\right]\end{array}$$