Question

Two charges are located on the x-axis. A 22 µC charge is located at the origin and a 47 µC charge is located at x = 4.0 m. What is the electric field at x = 3.00 m?

Answer 1

Electric field =$\frac{kq}{r^{2}}$

$E_{1}=\frac{8.99\times10^{9}\times22\times10^{-6}}{3^{2}} = 22\times10^{3} N/C$

$E_{2}=\frac{8.99\times10^{9}\times47\times10^{-6}}{1^{2}} = 423\times10^{3} N/C$

The field at $x=3$ is,

$E_{1}-E_{2}= (423\times10^{3}) - (22\times10^{3})=4.0\times10^{5}N/C$ towards $-x$ direction.