The force of repulsion between two point positive charges 5μ μ C and 8μ μ C separated at a distance of 0.02 map art is. [14πεo 1 4 π ε o = 9 x 109Nm2C-2]

The force of repulsion between two point positive charges 5$\mu$C and 8$\mu$C separated at a distance of 0.02 map art is. [$\frac{1}{4\pi {\epsilon }_o}$ = 9 x 109Nm2C-2]

Answer Details

F =$\frac{1}{4\pi {\epsilon }_o}$ x $\frac{q_1{q}_2}{r^2}$

F =$\frac{5\times 10\times 8\times 10}{0.002\times 0.002}$ x 9 x 10

F = 900N