Question

The distance between two charge $+4Q$ and $+Q$ is $r$. What should be the nature, magnitude and position of a charge to be kept in between the two charges on the line joining them to keep the charges in equilibrium?

Answer 1

Let the new charged be placed at distance R from q

force on p due to q =$Fq-p$ =$8.99\times 109\times \frac{q\times p}{R^2}$

force on p due to 4q=$F_{4}q-p$=$8.99\times 109\times \frac{4q\times p}{r-R^2}$

for equilibrium

$F_q-p$ =$F_{4}q-p$

$8.99\times 109\times \frac{q\times p}{R^2}$ =$8.99\times 109\times \frac{4q\times p}{r-R^2}$

$R=\frac{r}{3}$

Position of the new charge