Question

What is the electric field strength at a point in space 4 m due east of a particle with a charge of +8 μC? What

is the direction of the field here?

Answer 1

Answer

Electric field at a point in space can be given as

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

So

$E=\frac{(9\times10^9) (8\times10^{-6}) }{4^2}\\E=4500N/C$

Direction will be towards east.