Question

A 20 pC charge is placed at the origin and a -30 pC is places 3m to the right of it. What is the magnitude and direction of the electric field 7m meters from the origin?

Answer 1

Answer

Electric field due to any point charge is given

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

Electric field due 20 pC charge

$E_1=\frac{9\times10^9\times20\times10^{-12}}{7\times7}(-i)$

$=3.7\times10^{-3}(-i)$ V/m

$E_2=\frac{9\times10^9\times30\times10^{-12}}{4\times4}(i)$

= $=16.9\times10^{-3}(i)$ V/m

So resultant

The magnitude and direction of the electric field 7m meters from the origin

Is

E=$E_1+E_2$

$=13.2\times10^{-3}(i)$ v/m

Direction in positive x direction.