Question

What is the magnitude of a point charge that would create an electric field of 1.18 N/C at points 0.822 m away?

Answer 1

Answer:

q = 8.85 x 10⁻¹¹ C

Explanation:

given,

Electric field, E = 1.18 N/C

distance, r = 0.822 m

Charge magnitude = ?

using formula of electric field.

$E = \dfrac{kq}{r^2}$

k is the coulomb constant

$q= \dfrac{Er^2}{k}$

$q= \dfrac{1.18\times 0.822^2}{9\times 10^9}$

  q = 8.85 x 10⁻¹¹ C

The magnitude of charge is equal to q = 8.85 x 10⁻¹¹ C