Question

A charge has a value of 5.5x10^-6C. Find the value of the electric field 0.02m away from it.

Answer 1

Apply the formula:

E= kQ/d^2

Where:

E = electric field

k = constant = 9 x 10^9 Nm^2/C^2

Q= charge = 5.5x10^-6C

d= distance = 0.02m

Replacing:

$E=\frac{(9\times10^{^9})(5.5\times10^{^{-6}})}{(0.02)^{^2}}$

E = 123,750,000 N/C