A charged particle (mass $m$ and charge $q$) moves along the $x$-axis with velocity $V_0$. When it passes through the origin it enters a region having uniform electric field $\vec{E} = -E\hat{j}$ which extends up to $x = d$. Equation of path of the particle in the region $x > d$ is:
Answer: (B) $y = \dfrac{qEd}{mV_0^{2}}\left(\dfrac{d}{2}-x\right)$
Inside the field the acceleration is $\dfrac{qE}{m}$ along $-y$, so $y = -\dfrac{qE}{2m}\left(\dfrac{x}{V_0}\right)^{2}$.
At $x = d$: $y_d = -\dfrac{qEd^{2}}{2mV_0^{2}}$ and slope $\dfrac{dy}{dx} = -\dfrac{qEd}{mV_0^{2}}$.
Beyond $x = d$ the particle moves in a straight line with this slope:
$$y = -\frac{qEd^{2}}{2mV_0^{2}} - \frac{qEd}{mV_0^{2}}(x-d) = \frac{qEd}{mV_0^{2}}\left(\frac{d}{2}-x\right)$$
Solution by Sreeraj P, M.Sc Physics