Answer:
The magnitude of the electric field experienced by the charge is 4 x 10¹⁰ N/C.
Explanation:
Given;
magnitude of the charge, q = 4 mC = 4 x 10⁻³ C
distance of the charge, r = 3 cm = 0.03 m
The magnitude of the electric field is calculated as follows;
[tex]E = \frac{F}{q} = \frac{Kq}{r^2} \\\\[/tex]
where;
K is Coulomb's constant, = 9 x 10⁹ Nm²/C²
[tex]E = \frac{Kq}{r^2}\\\\E = \frac{(9\times 10^9)(4 \times 10^{-3})}{0.03^2} \\\\E = 4 \times 10^{10} \ N/C[/tex]
Therefore, the magnitude of the electric field experienced by the charge is 4 x 10¹⁰ N/C.
