Question

A point charge gives rise to an electric field with magnitude 1 N/C at a distance of 3 m. If the distance is increased to 6 m, then what will be the new magnitude of the electric field?

Answer 1

The magnitude of the final electric field is 0.5 N/C.

Step-by-step explanation:

Given;

initial electric field, E₁ = 1 N/C

initial distance moved by the charge, r₁ = 3 m

final distance moved by the charge, r₂ = 6 m

let the magnitude of the final electric field = E₂

The electric potential created by the charge is given as;

V = E₁r₁ = E₂r₂

`E_2 = (E_1r_1)/(r_2) \\\\E_2 = (1 N/C\ * \ 3m)/(6m) \\\\E_2 = 0.5 \ N/C`

Therefore, the magnitude of the final electric field is 0.5 N/C.