Question

2 charges of 4pc and 2pc by separated by a distance of 2m where should a charge of 1pc should be placed between the 2 so the three charges are in equilibrium​

Answer 1

Explanation:

Two point charges +4q and +2pc

Kept apart =2m

Two cases will be needed.

Given charges, +4pc, +2pc.

let, the third charge of 1pc be placed at a distance x from 4pc and hence its distance from 2pc and (2−x)cm for the system to remain in equilibrium F1=F2

where

$F _{1} = k\frac{(4pc×1pc)}{ {x}^{2} } \\</p><p>F2= \frac{K(4pc×2pc)}{ {(2−x)}^{2} }$

now evaluate yourself

thanks for asking