Question

Calculate the magnitude and direction of the electric field 0.54 m from a +8.75 nC point charge.

Answer 1

We can find the magnitude of the electric field as follows:

$$E=\dfrac{kQ}{r^2},$$ $$E=\dfrac{9\cdot10^9\ \dfrac{Nm^2}{C^2}\cdot8.75\cdot10^{-9}\ C}{(0.54\ m)^2}=270.1\ \dfrac{N}{C}.$$

Since the electric charge is positive, the electric field directed away from the electric charge.