1. Motion inside the electric field region

The particle enters the region at the origin with initial velocity along $+x$: $v_x = V_0, \qquad v_y(0)=0$

The electric field is $\vec E = -E\hat j$ so the force on charge $q$ is $\vec F = q\vec E = -qE\hat j$ Hence acceleration is only along $y$: $a_x=0, \qquad a_y = -\frac{qE}{m}$

Since there is no $x$-acceleration, the $x$-velocity remains constant: $x = V_0 t$

The field exists up to $x=d$, so the time spent inside the field is $t_1 = \frac{d}{V_0}$

2. Coordinates and velocity at $x=d$

At $t=t_1$, the $y$-displacement is $y_1 = \frac{1}{2}a_y t_1^2 = \frac{1}{2}\left(-\frac{qE}{m}\right)\left(\frac{d}{V_0}\right)^2$ $y_1 = -\frac{qEd^2}{2mV_0^2}$

The vertical velocity at exit is $v_{y1} = a_y t_1 = -\frac{qE}{m}\cdot \frac{d}{V_0} = -\frac{qEd}{mV_0}$

So at $x=d$, the particle has:

• position $\left(d,\,-\dfrac{qEd^2}{2mV_0^2}\right)$
• velocity components $v_x=V_0$, $v_y=-\dfrac{qEd}{mV_0}$

3. Motion in the region $x>d$

For $x>d$, there is no electric field, so the particle moves in a straight line with constant velocity.

Equation of straight line through $(d,y_1)$ with slope $\frac{dy}{dx} = \frac{v_y}{v_x} = \frac{-qEd/(mV_0)}{V_0} = -\frac{qEd}{mV_0^2}$

Thus, $y-y_1 = -\frac{qEd}{mV_0^2}(x-d)$

Substitute $y_1=-\dfrac{qEd^2}{2mV_0^2}$: $y + \frac{qEd^2}{2mV_0^2} = -\frac{qEd}{mV_0^2}(x-d)$

$y = -\frac{qEd}{mV_0^2}(x-d) - \frac{qEd^2}{2mV_0^2}$

Factor out $\dfrac{qEd}{mV_0^2}$: $y = \frac{qEd}{mV_0^2}\left[-(x-d)-\frac d2\right]$ $y = \frac{qEd}{mV_0^2}\left(\frac d2 - x\right)$

4. Compare with options

The obtained equation is $\boxed{y = \frac{qEd}{mV_0^2}\left(\frac d2 - x\right)}$

This matches Option B.

5. Comparison with stored correct answer

Stored correct answer: B

Derived answer: B

So they agree.