P19R2015 – Projectile Motion Through E

We form the ratio $mg/F_E$ and compute it below

\boxed{\frac{mg}{F_E}=\frac{mg}{qE}=\frac{1.67\times {10}^{-27}\cdot 9.8}{1.6\times {10}^{-19}\cdot {10}^6}\approx {10}^{-13}}

Thus, we conclude that $mg={10}^{-13}\ F_E$ and that the weight force can safely be neglected in front of the electric force. The motion of the particle through the electric field will therefore be the same is it only experienced the electric force.

To describe the motion of the charged particle between the plates, we start by deriving its kinematic equations and apply Newton’s Second Law to determine the acceleration is experiences.

Neglecting the weight force in front of the electric force, we draw the free-body diagram of the particle below

and applying Newton’s Second Law in the $x$ and in the $y$ direction, we derive that the components of the particle’s acceleration are

\begin{aligned}
\left\{ \begin{array}{c}
F_{net\ x}=ma_x \\
F_{net\ y}=ma_y \end{array}
\right.\ \ \ \ \ \ \ &\Rightarrow \ \ \ \ \ \ \left\{ \begin{array}{c}
0=ma_x \ \ \ \ \\
\\
-qE=ma_x \end{array}
\right. \\
\\
&\Rightarrow \ \ \ \ \ \ \ \left\{ \begin{array}{c}
a_x=0 \ \ \ \ \ \ \\
\\
\displaystyle{a_y=-\frac{qE}{m}} \end{array}
\right.
\end{aligned}

The horizontal motion of the particle is therefore uniform (constant speed) while its vertical motion is uniformly accelerated (constant acceleration). Its path through the electric field will be parabolic.

By property of projectile motion, when the particle reaches its highest point its vertical velocity is zero while its horizontal velocity – which remains constant throughout the motion – is equal to $v_0{\mathrm{cos} \left(\theta \right)\ }$.

Thus, the particle’s velocity vector is horizontal when it reaches its highest point, with magnitude

\boxed{v_x=v_0{\mathrm{cos} \left(\theta \right)\ }}

The components of acceleration of the particle are known, and the components of its initial velocity are $v_{0x}=v_0{\mathrm{cos} \left(\theta \right)\ }$ and $v_{0y}=v_0{\mathrm{sin} \left(\theta \right)\ }$ in the $x$ and $y$ directions respectively. Keeping in mind that the initial position of the particle is $x_0=y_0=0$, we derive the following kinematic equations

\left\{ \begin{array}{c}
x\left(t\right)=v_0{\mathrm{cos} \left(\theta \right)\ }t \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \  \\ 
\\
\displaystyle{y\left(t\right)=-\frac{qE}{2m}t^2+v_0{\mathrm{sin} \left(\theta \right)\ }t} \ \ \ \ \end{array}
\right.\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \left\{ \begin{array}{c}
v_x\left(t\right)=v_0{\mathrm{cos} \left(\theta \right)\ }\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \  \\ 
\\
\displaystyle{v_y\left(t\right)=-\frac{qE}{m}t+v_0{\mathrm{sin} \left(\theta \right)\ }} \ \  \end{array}
\right.

The particle reaches its highest point at time $t_1$ for which $v_y\left(t_1\right)=0$ and we solve for $t_1$ as follows

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

Thus, the particle reaches its highest point at time $t_1$ equal to

\boxed{t_1=\frac{mv_0{\mathrm{sin} \left(\theta \right)\ }}{qE}}

To determine the length $D$, we first find the time at which the particle exits the region. This occurs at time $t_2$ such that $y\left(t_2\right)=0$ because the entrance point and the exit point have the same vertical position. Thus, we solve

\begin{aligned}
y\left(t_2\right)=0\ \ \ \ \ &\Rightarrow \ \ \ \ \ \ -\frac{qE}{2m}t^2_2+v_0{\mathrm{sin} \left(\theta \right)\ }t_2=0 \\
\\
&\Rightarrow \ \ \ \ \ \ t_2\left(-\frac{qE}{2m}t_2+v_0{\mathrm{sin} \left(\theta \right)\ }\right)=0
\end{aligned}

Solving for $t_2$, we get two solutions

t_2=0\ \ \ \ \ \ \ \ and\ \ \ \ \ \ \ \ t_2=\frac{2mv_0{\mathrm{sin} \left(\theta \right)\ }}{qE}

Ruling out $t_2=0$ (since that is the instant when the particle enters the region), we conclude that the particle exits the region at time

t_2=\frac{2mv_0{\mathrm{sin} \left(\theta \right)\ }}{qE}

In that amount of time, the particle has traveled a distance $D$ horizontally and we conclude that

\boxed{D=x\left(t_2\right)=v_0{\mathrm{cos} \left(\theta \right)\ }\left(\frac{2mv_0{\mathrm{sin} \left(\theta \right)\ }}{qE}\right)=\frac{2mv^2_0{\mathrm{sin} \left(\theta \right)\ }{\mathrm{cos} \left(\theta \right)\ }}{qE}}

Note: the time found is double the amount of time it takes the particle to reach its maximum height because it exits at the same height that it entered.

The particle reaches its maximum height at time

t_1=\frac{mv_0{\mathrm{sin} \left(\theta \right)\ }}{qE}

Its vertical height at that instant is equal to

\boxed{y\left(t_1\right)=-\frac{qE}{2m}t^2_1+v_0{\mathrm{sin} \left(\theta \right)\ }t_1=-\frac{qE}{2m}{\left(\frac{mv_0{\mathrm{sin} \left(\theta \right)\ }}{qE}\right)}^2+v_0{\mathrm{sin} \left(\theta \right)\ }\left(\frac{mv_0{\mathrm{sin} \left(\theta \right)\ }}{qE}\right)=\frac{mv^2_0{{\mathrm{sin}}^2 \left(\theta \right)\ }}{2qE}}