1. A +4 nC charge is placed at a distance r from another charge. If the charge experiences a force of 6000 N, what is the electric field intensity E ?

1500 N/C

Explanation:

From the question,

Applying,

E =  F/q Equation 1

Where E = Electric Field intensity, F = Force experienced by the charge, q = Charge

Given: F = 6000 N, q = +4nC

Substitute these value into equation 1

E = 600/4

E = 6000/4

E = 1500 N/C

Hence, the electric field intensity is 1500 N/C.