Question

A 2.40 μC charge is subject to a 3.00 mN force due to an Electric Field. What is the magnitude of the Electric Field at the location of the charge?

Answer 1

Electric field at the location of the charge is 1250 N/C

Explanation:

Electric field is the ratio of force and charge.

Force, F = 3.00 mN = 3 x 10⁻³ N

Charge, q = 2.40 μC = 2.40 x 10⁻⁶ C

We have

                 $E=\frac{F}{q}\\\\E=\frac{3\times 10^{-3}}{2.40\times 10^{-6}}\\\\E=1250N/C$

Electric field at the location of the charge is 1250 N/C