P19S2014 – Infinite Sheet & Point Charge

The infinite sheet of charge creates a uniform electric field ${\overrightarrow{E}}_{sheet}$ that is horizontal and points away from the sheet $\left(\sigma >0\right)$. The electric force acting on $q_1>0$ due to the sheet alone is therefore horizontal and points in the $+x$ direction since ${\overrightarrow{F}}_E=q_1{\overrightarrow{E}}_{sheet}$.

The charge $q_2$ can either attract or repel $q_1$ by exerting an electric force $\overrightarrow{F_2}=q_1\overrightarrow{E_2}$. However, for the net force acting on $q_1$ to be only in the $y$-direction, the $x$-component of $\overrightarrow{F_2}$ must cancel out with $\overrightarrow{F_E}$ as shown. This means that the charge $q_2$ must attract the charge $q_1$ which requires that $q_2$ be negative

The charge $q_2$ is negative.

The magnitude of ${\overrightarrow{F}}_E$ is given by

F_E=\frac{q_1\sigma }{2{\varepsilon }_0}

The $x$-component of ${\overrightarrow{F}}_2$ must equal the magnitude of ${\overrightarrow{F}}_E$ so that the net force on $q_1$ has no $x$-component, and we write

F_2{\mathrm{cos} \left(\theta \right)\ }=F_E\ \ \ \ \Rightarrow \ \ \ \ \ \frac{|q_1q_2|}{4\pi {\varepsilon }_0r^2}{\mathrm{cos} \left(\theta \right)\ }=\frac{q_1\sigma }{2{\varepsilon }_0}

The distance between the two point charges is equal to $d=5\ m$ (3-4-5 triangle) and $\mathrm{cos}\mathrm{}(\theta )$ can be derived from geometry as shown.

Thus, the charge $q_2$ is given by

\begin{aligned}
\frac{\left|q_1q_2\right|}{4\pi {\varepsilon }_0r^2}{\mathrm{cos} \left(\theta \right)\ }=\frac{q_1\sigma }{2{\varepsilon }_0}\ \ \ \ &\Rightarrow \ \ \ \ \ \frac{\left|q_2\right|}{4\pi {\varepsilon }_0\cdot 25}\cdot \frac{4}{5}=\frac{2}{2\cdot 25\pi {\varepsilon }_0} \\
\\
&\Rightarrow \ \ \ \ \ \ q_2=-5\ C
\end{aligned}

The charge $q_2$ is therefore equal to

\boxed{q_2=-5 \ C}