Question

A particle has charge -3 nC. A) Find the magnitude and direction of the electric field due to this particle at a point 0.25 m directly above it. B) At what distance from this particle does its electric field have a magnitude of 12 N/C?

Answer 1

$a) E=kq/r^2, k=9\times 10^9$

$E=-\frac{(9\times 10^9 \times3\times 10^-9)}{0.25^2}=-$ $432N/C$

b) $12=\frac{9\times 10^9 \times 3\times 10^-9}{r^2}$

$r^2=27/12=2.25$

$r=1.5m$