P19R2015 – Turning Around in E

As the particle enters the region with the electric field, it experiences the electric force which slows it down. If $v_0$ is small enough, the electric force will slow the particle to a stop before causing it to turn around. If $v_0$ is too large, the particle will reach the end of the region with a nonzero velocity and then exit it and keep moving (at constant speed) to the right. The greatest value $v_{max}$ of $v_0$ is therefore achieved when the particle turns around exactly at $x=L$.

To determine $v_{max}$, we first draw the free-body diagram of the charge and derive its acceleration by applying Newton’s Second Law.

By Newton’s Second Law, the acceleration $a_x$ acting on the charged particle is equal to

\begin{aligned}
F_{net\ x}=ma_x\ \ \ \ \ &\Rightarrow \ \ \ \ \ -qE=ma_x \\
\\
&\Rightarrow \ \ \ \ \ a_x=-\frac{qE}{m}
\end{aligned}

Method 1: solve $v_x\left(t\right)=0$ and substitute into $x\left(t\right)$

The particle’s initial velocity is $v_{0x}=v_0$ and its initial position is $x_0=0$. Its kinematic equation are therefore

\left\{ \begin{array}{c}
\displaystyle{x\left(t\right)=\frac{1}{2}a_xt^2+v_{0x}t+x_0=-\frac{qE}{2m}t^2+v_0t} \\
\\
\displaystyle{v_x\left(t\right)=a_xt+v_{0x}=-\frac{qE}{m}t+v_0 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ }  \end{array}
\right.

The particle turns around at a time $t_1$ such that $v_x\left(t_1\right)=0$ and we solve for $t_1$ as follows

\begin{aligned}
v_x\left(t_1\right)=0\ \ \ \ &\Rightarrow \ \ \ \ -\frac{qE}{m}t_1+v_0=0 \\
\\
&\Rightarrow \ \ \ \ \ t_1=\frac{mv_0}{qE}
\end{aligned}

The initial speed $v_0$ therefore determines how long it takes the particle to turn around and therefore how far across the region it travels. Particularly, for small values of $v_0$, the particle will turn around closer to the left edge of the region, while for very large values of $v_0$ it will slow down while traveling in the region but will reach the right edge with a nonzero velocity and therefore leave the region never to return.

Thus, the largest speed $v_0$ for which the particle still turns around in the region is the speed for which the particle turns around just as it reaches the right edge of the region, before it exits. Therefore, the maximum initial speed $v_{max}$ for which the particle turns around in the region is such that the particle turns around at $x=L$. In other words, for $\displaystyle{t_1=\frac{mv_{max}}{qE}}$ we must have $x\left(t_1\right)=L$.

Writing out this condition below, we can then solve for $v_{max}$ as follows

\begin{aligned}
x\left(t_1\right)=L\ \ \ \ \ &\Rightarrow \ \ \ \ -\frac{qE}{2m}t^2_1+v_{max}t_1=L \\
\\
&\Rightarrow \ \ \ \ -\frac{qE}{2m}{\left(\frac{mv_{max}}{qE}\right)}^2+v_{max}\left(\frac{mv_{max}}{qE}\right)=L \\
\\
&\Rightarrow \ \ \ \ -\frac{mv^2_{max}}{2qE}+\frac{mv^2_{max}}{qE}=L \\
\\
&\Rightarrow \ \ \ \ -\frac{mv^2_{max}}{2qE}+\frac{2mv^2_{max}}{2qE}=L \\
\\
&\Rightarrow \ \ \ \ \frac{mv^2_{max}}{2qE}=L \\ 
\\
&\Rightarrow \ \ \ \ \ v_{max}=\sqrt{\frac{2qEL}{m}}
\end{aligned}

The maximum value of the speed $v_0$ for which the particle turns around before exiting the region is therefore equal to

\boxed{v_{max}=\sqrt{\frac{2qEL}{m}}}

Method 2: using $v^2_{fx}=v^2_{0x}+2a_x\mathrm{\Delta }x$

This method is much faster due to the fact that it directly relates speed to distance traveled without requiring to first solve for time $t_1$ and then solving $x\left(t_1\right)=L$ as we did before.

The argument that for $v_0=v_{max}$ the particle turns around at $x=L$ still holds and we therefore write the following kinematic equation that $v_{max}$ must satisfy.

\begin{aligned}
v^2_f=v^2_0+2a_x\mathrm{\Delta }x\ \ \ \ \ &\Rightarrow \ \ \ \ \ 0^2=v^2_{max}-\frac{2qEL}{m} \\
\\
&\Rightarrow \ \ \ \ \ v_{max}=\sqrt{\frac{2qEL}{m}}
\end{aligned}

The maximum value of the speed $v_0$ for which the particle turns around before exiting the region is therefore equal to

\boxed{v_{max}=\sqrt{\frac{2qEL}{m}}}

If $\displaystyle{v_0=\frac{v_{max}}{2}}$, the particle will turn around before the end of the region at a distance $L_1$ from the left end.

Method 1:

The particle turns around at a time $t_1$ such that $v_x\left(t_1\right)=0$ and we solve

\begin{aligned}
v_x\left(t_1\right)=0\ \ \ \ \ &\Rightarrow \ \ \ \ \ -\frac{qE}{m}t_1+v_0=0 \\
\\
&\Rightarrow \ \ \ \ \ \ t_1=\frac{mv_0}{qE} \\
\\
&\Rightarrow \ \ \ \ \ \ t_1=\frac{mv_{max}}{2qE} \\
\\
&\Rightarrow \ \ \ \ \ \ t_1=\frac{m}{2qE}\sqrt{\frac{2qEL}{m}}=\sqrt{\frac{mL}{2qE}}
\end{aligned}

The distance $L_1$ at which the particle turns around is then derived by computing $x\left(t_1\right)$ and is equal to

\begin{aligned}
L_1&=x\left(t_1\right) \\
\\
&=-\frac{qE}{2m}t^2_1+v_0t_1 \\
\\
&=-\frac{qE}{2m}\left(\frac{mL}{2qE}\right)+\frac{v_{max}}{2}\sqrt{\frac{mL}{2qE}} \\
\\
&=-\frac{L}{4}+\frac{1}{2}\sqrt{\frac{2qEL}{m}}\sqrt{\frac{mL}{2qE}} \\
\\
&=-\frac{L}{4}+\frac{L}{2} \\
\\
&=\frac{L}{4}
\end{aligned}

If $\displaystyle{v_0=\frac{v_{max}}{2}}$, the particle will turn around at a distance $L_1$ from the left end equal to

\boxed{L_1=\frac{L}{4}}

Method 2:

The particle will turn around at a distance $L_1$ that satisfies the following equation

\begin{aligned}
v^2_f=v^2_0+2a_x\mathrm{\Delta }x\ \ \ \ &\Rightarrow \ \ \ \ \ 0^2={\left(\frac{v_{max}}{2}\right)}^2-\frac{2qEL_1}{m} \\
\\
&\Rightarrow \ \ \ \ \ \frac{v^2_{max}}{4}=\frac{2qEL_1}{m} \\
\\
&\Rightarrow \ \ \ \ \ \frac{qEL}{2m}=\frac{2qEL_1}{m} \\
\\
&\Rightarrow \ \ \ \ \ L_1=\frac{L}{4}
\end{aligned}

The particle will turn around at a distance $L_1$ from the left end equal to

\boxed{L_1=\frac{L}{4}}