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.
[tex]E = \dfrac{kq}{r^2}[/tex]
k is the coulomb constant
[tex]q= \dfrac{Er^2}{k}[/tex]
[tex]q= \dfrac{1.18\times 0.822^2}{9\times 10^9}[/tex]
q = 8.85 x 10⁻¹¹ C
The magnitude of charge is equal to q = 8.85 x 10⁻¹¹ C
