Question

When placed near another charge, a 20 microcoulomb charge experiences an attractive force of 0.080 N. What is the electric field strength at that location?

Answer 1

Answer:

E = 4000 N/C

Explanation:

Given the following data;

Force = 0.080 N.

Charge, q = 20 microcoulomb = 20 * 10^-6 = 2 * 10^-5 Coulombs

To find the electric field strength;

Mathematically, the electric field strength is given by the formula;

Electric field strength = force/charge

Substituting into the formula, we have;

E = 0.080/0.00002

E = 4000 N/C