Answer to Question #309234 in Physics for Justin

Question

A positive charge, q = +20 nC, is on the y-axis at y = +4.00 cm.

(a) Determine the magnitude and direction of the electric field at the origin.

(b) What will be the magnitude and direction of the electric field at the origin if the charge is -20 nC?

Answer 1

(a)

$$E=k\frac{|q|}{r^2}=9*10^9*\frac{20*10^{-9}}{0.04^2}=1.1*10^5\:\rm N/C$$

The field is directed toward negative y-axis.

(b)

$$E=k\frac{|q|}{r^2}=9*10^9*\frac{20*10^{-9}}{0.04^2}=1.1*10^5\:\rm N/C$$

The field is directed toward positive y-axis.