P19B2016 – Charged Plate & Point Charge

The ball experiences the electric force which is given by ${\overrightarrow{F}}_E=Q_2\overrightarrow{E}$. Since the charge on the ball is negative, we know that the electric force will point in the direction opposite the electric field. In addition, since $Q_1$ is negative, we know that the electric field created by the large plate is perpendicular to the plate, directed toward the plate and uniform with a magnitude $E=\sigma /2{\varepsilon }_0$. This is summarized in the figure below.

The electric force acting on the ball is therefore directed away from the plate (makes sense, the ball is repelled by the plate) and is given by

\boxed{{\overrightarrow{F}}_E=\left|Q_2\right|E\ \overrightarrow{x}=\frac{\left|Q_2\right|\sigma }{2{\varepsilon }_0}\ \overrightarrow{x}=\frac{\left|Q_2Q_1\right|}{2A{\varepsilon }_0}\ \overrightarrow{x}=\frac{\left(-1\right)\cdot \left(-4\right)}{2\cdot 8\pi \cdot {\varepsilon }_0}\ \overrightarrow{x}=\frac{1}{4\pi {\varepsilon }_0}\ \overrightarrow{x}}

Answer: b

Newton’s Third Law tells us that the force of the ball on the plate is the same as the force of the plate on the ball in magnitude.

The charge $Q_1$ of the plate is positive and therefore the electric field ${\overrightarrow{E}}_1$ created by the plate is perpendicular to the plate, directed to the left and uniform.

The charge $Q_2$ of the ball is negative and therefore it creates a radially inward electric field ${\overrightarrow{E}}_2$. On the $x$ axis to the left of the plate, the electric field created by the ball is horizontal and points right.

The total electric field at a point a distance $x$ to the left of the plate (on the $x$ axis) is given by the principle of superposition and is equal to

{\overrightarrow{E}}_{tot}={\overrightarrow{E}}_1+{\overrightarrow{E}}_2=\left(E_2-E_1\right)\hat{x}=\left(\frac{\left|Q_2\right|}{4\pi {\varepsilon }_0{\left(x-x{ball}\right)}^2}-\frac{\sigma }{2{\varepsilon }_0}\right)\hat{x}

To find the position $x$ where the electric field cancels, we solve

\begin{aligned}
E_{tot}=0\ \ \ \ \ &\Rightarrow \ \ \ \ \frac{\left|Q_2\right|}{4\pi {\varepsilon }_0{\left(x-x_{ball}\right)}^2}-\frac{Q_1}{2{\varepsilon }_0A}=0 \\ 
\\ &\Rightarrow \ \ \ \ \ x-x_{ball}=\pm \sqrt{\frac{\left|Q_2\right|A}{2\pi Q_1}} \\
\\
&\Rightarrow \ \ \ \ \ x=x_{ball}\pm \sqrt{\frac{\left|Q_2\right|A}{2\pi Q_1}}
\end{aligned}

This yields two solutions

\begin{aligned}
x=&x_{ball}-\sqrt{\frac{\left|Q_2\right|A}{2\pi Q_1}}=\frac{1}{2}-\sqrt{\frac{3\cdot 8\pi }{2\pi \cdot 3}}=\frac{1}{2}-\sqrt{4}=-1.5\ m \\
\\
x=&x_{ball}+\sqrt{\frac{\left|Q_2\right|A}{2\pi Q_1}}=\frac{1}{2}+\sqrt{\frac{3\cdot 8\pi }{2\pi \cdot 3}}=\frac{1}{2}+\sqrt{4}=2.5\ m
\end{aligned}

The first value of $x$ corresponds to a location that $1.5\ m$ to the left of the plate (because $x$ is negative) and we conclude that the electric field cancels at the location

\boxed{x=-1.5\ m}

From our answer to question 3, we know that the only two locations where the electric field cancels are $x=-1.5\ m$ and $x=2.5\ m$. Since the ball is only $0.5\ m$ away from the plate (to the right) we conclude that there is no position in between the plate and the ball where the electric field cancels.

Yes, based on our answer to question 3, we know that the electric field cancels at $x=2.5\ m$ to the right of the plate, i.e. $2\ m$ to the right of the ball.

Answer: c

Whatever the sign of the charge $Q_1$, the electric field it creates at $Q_3$ will be perpendicular to the plate.

Whatever the sign of the charge $Q_2$, the electric field it creates at $Q_3$ will be at $45{}^\circ $ with respect to the horizontal.

Since the net force acting on the electron is horizontal and directed to the right, we conclude that $Q_2$ must be zero otherwise it would have a $y$ component. In addition, since the net force is to the right and the electron has a negative charge, we conclude that the plate must be negatively charged to repel the electron and therefore $Q_1$ must be negative.

Answer: b

As explained previously, the charge $Q_2$ must be zero if the net force is to be purely horizontal (and therefore not have any vertical component).

Answer: c

$Q_3$ is negative since it is an electron.