Question

1. What is the magnitude of the electric field E at a field point 2.0 m from a point charge q=4.0 nC?

Answer 1

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

Where: E is electric field

r is distance=2M

q is charge=4nC

$k = constant = 9×10^{9}$

$E= \frac{9×10^9×4×10^{-9}}{2^2}=9N/C$

E=9N/C