-- ELECTRIC FIELDS --
-- GAUSS'S LAW --
-- ELECTRIC POTENTIAL --
-- CAPACITORS --
-- DC CIRCUITS --
-- MIDTERM 1 - STUDY GUIDE --
-- MAGNETISM --
-- INDUCTION --
-- ELECTROMAGNETIC WAVES --
-- OPTICS --
-- MIDTERM 2 - STUDY GUIDE --
-- INTERFERENCE & DIFFRACTION --
-- QUANTUM MECHANICS --
-- RELATIVITY --
-- STUDY GUIDE - FINAL --

P20B2018 – Charged Wire and Point Charge

Charged Wire & Point Charge

An infinitely long vertical wire with given linear charge density $\left|\lambda \right|$ is placed at $x=-d$. A point charge with charge $\left|Q\right|$ is placed at $\left(4d,0\right)$. It is found that when an electron is placed at $\left(0,3d\right)$ the net force it feels is only in the $+y$ direction. Neglect gravity.

A. Multiple Choice

Which is true above the signs of the charges?

  • a. $\lambda $ is positive and $Q$ is positive.
  • b. $\lambda $ is positive and $Q$ is negative.
  • c. $\lambda $ is zero and $Q$ is negative.
  • d. $\lambda $ is negative and $Q$ is positive.
  • e. $\lambda $ is negative and $Q$ is negative.
View answer

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.

B. Find the magnitude of the point charge $\left|Q\right|$ in terms of the other variables and known constants.

View answer

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}}