P20B2018 – Charged Wire and Point Charge

Answer: e

We draw the different possible scenarios below to determine which one would yield a net force along the $+y$ direction.

For the net force to be exclusively in the $+y$ direction, the $x$-components of the forces must cancel and the $y$-components of forces must be directed in the $+y$ direction. This occurs if $\lambda < 0$ and $Q < 0$ as seen above.

Recalling that the electric field created by an infinite wire with linear charge density $\left|\lambda \right|$ is given by $\displaystyle{E_{wire}=\frac{\left|\lambda \right|}{2\pi {\varepsilon }_0r}}$, we draw the free-body diagram of the electron below.

Since the horizontal component of the net force acting on the electron must be zero, we write

\begin{aligned}
F_{net\ x}=0\ \ \ \ &\Rightarrow \ \ \ \ \ F_Q{\mathrm{cos} \left(\theta \right)\ }-F_{wire}=0 \\
\\
&\Rightarrow \ \ \ \ \ \frac{k\left|Q\right|e}{25d^2}\cdot {\mathrm{cos} \left(\theta \right)\ }=\frac{e\left|\lambda \right|}{2\pi {\varepsilon }_0d} \\
\\
&\Rightarrow \ \ \ \ \ \frac{4\left|Q\right|}{125\ d^2}=\frac{\left|\lambda \right|}{2\pi {\varepsilon }_0d} \\
\\
&\Rightarrow \ \ \ \ \ \left|Q\right|=\frac{125\left|\lambda \right|d}{2}
\end{aligned}

Thus, we conclude that the magnitude of the charge $\left|Q\right|$ is equal to

\boxed{\left|Q\right|=\frac{125\left|\lambda \right|d}{2}}