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=(kq)/(r^2)`

For the charge of +2 nC

`E_(+2)=(k(2*10^(-9)C))/((8.20^2))`

For the charge of -4nC:

`E_(-4)=(k(-4*10^(-9)C))/((3.23^2))`

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

`\begin{gathered} E_(+2)=E_(-9) \\ so: \\ (k(2*10^(-9)C))/((0.082^2))=(k(-9*10^(-9)))/(y^2) \\ solve_{\text{ }}for_{\text{ }}y: \\ y=\sqrt{((0.082^2)(-9*10^(-9)))/((2*10^(-9)))} \\ y\approx0.1739 \end{gathered}`

And:

`\begin{gathered} E_(-4)=E_(-9) \\ (k(-4*10^(-9)C))/((0.0323^2))=(k(-9*10^(-9)))/(x^2) \\ solve_{\text{ }}for_{\text{ }}x: \\ x=\sqrt{((0.0323^2)(-9*10^(-9)))/((-4*10^(-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=√((-0.04845)^2+(0.1739)^2) \\ r=0.1805cm \end{gathered}`