Question

Calculate the force between two charges each of $20$ micro coulombs in a medium of relative permittivity $4$. The distance between the two charges is $40cm.$

Answer 1

$F=\frac{1}{4\pi \epsilon k}\times \frac{\left(20\times {10}^{-6}\right)\left(20\times {10}^{-6}\right)}{{\left(0.4\right)}^2}$

$=\frac{9\times {10}^9}{4}\times \frac{400\times {10}^{-12}\times 10\times 10}{0.4\times 0.4}$

$=\frac{9\times 4\times {10}^3\times {10}^{-12}\times {10}^9}{4\times 16}$

$F=0.5625N$