Question

When two charge of value 2c and 8c are placed at a distance of 20m. Then where should a third charge of 1c is placed in between them so that it is in equilibrium?

Answer 1

Hi your answer

$\\ q_1 = q_2 = q = 1.0 C distance between = 2 km = 1\times 10^3m \\so, force = \frac{kq_1q_2}{r^2} F = \frac{(9\times 10^9) \times 1 \times 1}{(2\times10^3)^2} = \frac{9\times10^9}{2^2\times 10^6} = 2,25 \times 10^3 N \\ The weight of body = mg = 40\times 10N = 400N \\ So, \frac{wt of body}{force between ch arg es} = (\frac{2.25\times10^3}{4\times 10^2})^{-1} = (5.6)^{-1} = \frac{1}{5.6}$