Answer:
F = 3340.8 N
Explanation:
given,
q₁ = + 29 μ C
q₂ = +3.2 mC
distance,r = 50 cm = 0.5 m
Force between two charge = ?
[tex]F = \dfrac{kq_1q_2}{r^2}[/tex]
k = 9 x 10⁹ N.m²/ C²
[tex]F = \dfrac{9\times 10^9\times 29\times 10^{-6}\times 3.2 \times 10^{-3}}{0.5^2}[/tex]
F = 3340.8 N
The magnitude of force between the two charges is equal to F = 3340.8 N
