Question

calculate the electric Field intensity at a point 15 cm from a charge of 10× 10^ -3C

Answer 1

By the definition of the electric field intensity, we have:

$$E=\dfrac{kQ}{r^2}=\dfrac{9\cdot10^9\ \dfrac{Nm^2}{C^2}\cdot10\cdot10^{-3}\ C}{(0.15\ m)^2}=4.0\cdot10^9\ \dfrac{N}{C}.$$