Question

A +2 nC charge is located at (0,8.20) cm and a -4nC charge is located (3.23, 0) cm.Where would a -9 nC charge need to be located in order that the electric field at the origin be zero? Find the distance r from the origin of the third charge.

Answer 1

The electric field of a point charge is given by:

$E=\frac{kq}{r^2}$

For the charge of +2 nC

$E_{+2}=\frac{k(2\times10^{-9}C)}{(8.20^2)}$

For the charge of -4nC:

$E_{-4}=\frac{k(-4\times10^{-9}C)}{(3.23^2)}$

Since the field at the origin must be zero, then:

$\begin{gathered} E_{+2}=E_{-9} \\ so: \\ \frac{k(2\times10^{-9}C)}{(0.082^2)}=\frac{k(-9\times10^{-9})}{y^2} \\ solve_{\text{ }}for_{\text{ }}y: \\ y=\sqrt{\frac{(0.082^2)(-9\times10^{-9})}{(2\times10^{-9})}} \\ y\approx0.1739 \end{gathered}$

And:

$\begin{gathered} E_{-4}=E_{-9} \\ \frac{k(-4\times10^{-9}C)}{(0.0323^2)}=\frac{k(-9\times10^{-9})}{x^2} \\ solve_{\text{ }}for_{\text{ }}x: \\ x=\sqrt{\frac{(0.0323^2)(-9\times10^{-9})}{(-4\times10^{-9})}} \\ x=-0.04845 \end{gathered}$

Therefore, the coordinates of the -9 nC charge are:

(-0.04845,0.1739)

And the distance r is:

$\begin{gathered} r=\sqrt{(-0.04845)^2+(0.1739)^2} \\ r=0.1805cm \end{gathered}$