Question

A particle has charge -3.00nC.

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

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)

$$E=k\frac{q}{r^2}$$

$$E=9*10^9*\frac{3.00*10^{-9}}{0.250^2}=432\:\rm N/C$$

Directed radially toward the charge.

(b)

$$r_2=r_1\sqrt{\frac{E_2}{E_1}}=0.250\sqrt{\frac{12.0}{432}}=0.0417\:\rm m$$