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Q.

Two fixed point charges $+ 4 e$ and $+ e$ unit are separated by a distance ‘a’ The third point charge should be placed at t distance $\frac{R a}{3}$ from $+ 4 e$ charge for it to be in equilibrium. Then find ‘R’ ----

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Detailed Solution

$F_{1} = F_{2}$

$f r o m l a r g e r c h a r g e,$

$x' = \frac{d}{\sqrt{\frac{q_{s m a l l}}{q_{l a r g e}}} + 1}$

$x = \frac{a}{\sqrt{\frac{e}{4 e}} + 1} = \frac{2 a}{3}$

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