Answer to Question #308247 in Physics for Hailey

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 of 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.