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Step-by-Step Solution
Step 1: Identify the Regions of Motion
The charged particle (mass $m$, charge $q$) moves initially along the x-axis with a constant velocity $V_0$. It enters a region $0 \le x \le d$ where there is a uniform electric field
$ \vec{E} = -\,E \,\hat{j} $. Beyond $x = d$, the electric field is zero again.
Step 2: Forces and Equations of Motion in the Region $0 \le x \le d$
In the region $0 \le x \le d$, the electric force on the particle is
$ F = q \vec{E} = q \left(-E \,\hat{j}\right) = -qE \,\hat{j} $.
Since the force acts along the $-y$ direction, the motion in the $x$-direction remains uniform, while the $y$-motion is accelerated.
a) $x$-direction
$ \displaystyle x(t) = V_0 \,t, $
because the velocity along $x$ remains $V_0$ and there is no acceleration in $x$-direction.
b) $y$-direction
The acceleration along $y$ is
$ a_y = \frac{F_y}{m} = \frac{-\,qE}{m}. $
Since the particle starts at $y = 0$ with zero initial velocity along $y$ at $t=0$:
$$
y(t) = u_y\,t + \frac{1}{2} a_y t^2 = 0 + \frac{1}{2}\left(\frac{-qE}{m}\right) t^2 = -\frac{qE}{2m}\,t^2.
$$
Step 3: Time Spent by the Particle in the Field Region
The particle travels from $x = 0$ to $x = d$ under the uniform electric field. Using $x(t) = V_0 t$, the time $t_d$ to reach $x = d$ is
$$
t_d = \frac{d}{V_0}.
$$
Step 4: Velocity Components at $x = d$
Horizontal velocity at $x = d$:
$$
v_x(d) = V_0 \quad (\text{unchanged}).
$$
Vertical velocity at $x = d$: From $v_y = a_y t$, we have
$$
v_y(d) = \left(\frac{-qE}{m}\right) t_d = \left(\frac{-qE}{m}\right)\left(\frac{d}{V_0}\right) = -\,\frac{q E d}{m\,V_0}.
$$
Also, the $y$-position at $x = d$ is
$$
y(d) = -\frac{qE}{2m}\left(t_d\right)^2 = - \frac{q E}{2m} \left(\frac{d}{V_0}\right)^2
= -\,\frac{q E d^2}{2\, m\, V_0^2}.
$$
Step 5: Motion Beyond $x = d$
For $x > d$, there is no electric field acting on the particle. Hence, the particle moves under uniform velocity components which it had at $x = d$:
a) $x$-direction:
$$
v_x = V_0, \quad \text{so} \quad x(t) - d = V_0 \,\Delta t,
$$
where $\Delta t = t - t_d$ is the time elapsed after crossing $x = d$.
b) $y$-direction:
$$
v_y = -\,\frac{q E d}{m\,V_0}, \quad \text{and} \quad
y - y(d) = v_y \,\Delta t.
$$
Step 6: Express $y$ in Terms of $x$ for $x > d$
To eliminate time, use the relation $x - d = V_0\,\Delta t$, which gives
$ \Delta t = \frac{x - d}{V_0}. $
Then
$$
y - y(d) = \left(-\,\frac{q E d}{m\,V_0}\right) \left(\frac{x - d}{V_0}\right).
$$
Recall that $y(d) = -\,\frac{q E d^2}{2\,m\,V_0^2}$. Substituting that,
\[
y = -\,\frac{q E d^2}{2\,m\,V_0^2} \;-\; \frac{q E d}{m\,V_0^2} (x - d).
\]
Simplify:
\[
y = -\,\frac{q E d^2}{2\,m\,V_0^2} \;-\; \frac{q E d}{m\,V_0^2} x + \frac{q E d^2}{m\,V_0^2}.
\]
Combine like terms:
\[
y = \frac{q E d^2}{m\,V_0^2} \left( - \frac{1}{2} \right) \;-\; \frac{q E d}{m\,V_0^2} x
+ \frac{q E d^2}{m\,V_0^2}.
\]
\[
y = -\,\frac{q E d^2}{2\,m\,V_0^2} + \frac{q E d^2}{m\,V_0^2} - \frac{q E d}{m\,V_0^2} x.
\]
\[
y = \frac{q E d^2}{2\,m\,V_0^2} - \frac{q E d}{m\,V_0^2} x.
\]
Factor out $\frac{q E d}{m\,V_0^2}$:
\[
y = \frac{q E d}{m\,V_0^2} \left( \frac{d}{2} - x \right).
\]
This matches the correct answer given:
$$
y = \frac{q E d}{m\,V_0^2} \Bigl(\,\tfrac{d}{2} - x\Bigr).
$$
Final Answer
$ y = \frac{q\,E\,d}{m\,V_0^2}\Bigl(\tfrac{d}{2} - x\Bigr). $
Reference Image from Provided Solution:
