Question

A particle has charge -3.00 nC.

a) Find the magnitude and direction of the electric field due to this particle at a point 0.250 m directly above it.

b) at what distance from this particle does its electric field have a magnitude of 12.0 N/C?

Answer 1

(a)

$$E=\dfrac{k|q|}{r^2}=\dfrac{9\cdot10^9\ \dfrac{Nm^2}{C^2}\cdot|-3\cdot10^{-9}\ C|}{(0.25\ m)^2}=432\ \dfrac{N}{C}.$$

The electric field directed toward the negative charge (in upward direction).

(b)

$$r=\sqrt{\dfrac{k|q|}{E}}=\sqrt{\dfrac{9\cdot10^9\ \dfrac{Nm^2}{C^2}\cdot|-3\cdot10^{-9}\ C|}{12\ \dfrac{N}{C}}}=1.5\ m.$$