Question

A -3C charge is placed at a distance of 10cm from another charge of -5C. If a test charge is placed somewhere in between the line connecting the first two charges, what is the ratio between their distances to the test charge. Assume that the ratio must be greater than or equal to 1

Answer 1

Gives

$q_1=-3C\\q_2=-5C\\l=10m$

$\frac{kqQ_1}{x^2}=\frac{kqQ_2}{(l-x)^2}$

$\frac{kq\times3}{x^2}=-\frac{kq\times5}{(10-x)^2}$

$3x^2-5(10-x)^2\\3x^2=-5(100+x^2-20x)$

$8x^2-100x+500=0$

x=5.6m

Third point charge put at x=5.6m due to -3C charge

Ratio

$\frac{l}{r}=\frac{10}{5.6}=1.78>1$