Question

1.Find the potential at a distance of 0.07 m from a -6 nC charge.

2.What would be the PE of a -5 nC charge placed at this point?

3.What is the electric field strength at a point 40 cm from the a charge of 5x10^-9 C?

Answer 1

(a) potential

$V=\frac{kq}{r}$

$$V=-\frac{9\times10^9\times6\times10^{-9}}{0.07}=-771.42volt$$

(b) PE

$U=-qV$

$$U=((-5\times10^{-9})\times(-771.42))=3.85\times10^{-6}J$$

(C)

Electric field

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

$E=\frac{9\times10^9\times5\times10^{-9}}{0.4^2}=281.25N/C$