Question

Point charge q 1 = +2.00µC is located @ x = 0, y = 10cm and q 2 = -

2.00µC is located @ x = -5.0cm, y = -5.0cm. Find the net electric field

at the origin.

Answer 1

Answer

Electric field due to Point charge q 2 = +2.00µC is

$E_1=\frac{9\times10^9\times2\times10^{-6}}{(10^2)}$ (-j)

=180(-j) V/m on origin

Electric field due to Point charge q 2 = +2.00µC is

$E_2=\frac{9\times10^9\times2\times10^{-6}}{(5^2+5^2)}$ (-i-j)

=360(-i-j) V/m on origin

So net electric field on origin is

$E=E_1+E_2$

=360(-i)+540(-j)