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

Answer:

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

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

$E_2 = \frac{E_1r_1}{r_2} \\\\E_2 = \frac{1 N/C\ \times \ 3m}{6m} \\\\E_2 = 0.5 \ N/C$

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