Answer:
0.355 m
Explanation:
Data provided in the question:
Charges, q₁ = 53.0 µC = 53.0 × 10⁻⁶ C and q₂ = 44.0 µC = 44.0 × 10⁻⁶ C
Repulsive force exerted by the charges, F = 166 N
The value of Coulomb constant, k = 8.98755 × 10⁹ N·m²/C²
Now,
The Force exerted between the charges is given as = [tex]\frac{kq_1q_2}{r^2}[/tex]
Thus,
166 N = [tex]\frac{8.98755\times10^9\times53\times10^{-6}44\times10^{-6}}{r^2}[/tex]
or
r² = [tex]\frac{8.98755\times10^9\times53\times10^{-6}44\times10^{-6}}{166}[/tex]
or
r = 0.355 m
