Question

Two electrostatic point charges of +53.0 µC and +44.0 µC exert a repulsive force on each other of 166 N. What is the distance between the two charges? The value of the Coulomb constant is 8.98755 × 109 N · m2 /C 2 . Answer in units of m

Answer 1

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 = $\frac{kq_1q_2}{r^2}$

Thus,

166 N = $\frac{8.98755\times10^9\times53\times10^{-6}44\times10^{-6}}{r^2}$

or

r² = $\frac{8.98755\times10^9\times53\times10^{-6}44\times10^{-6}}{166}$

or

r = 0.355 m