A charge q is placed at the centre of line joining two equal charges each equal to Q. Show that system of three charges will be in equillibriumif q=-Q/4.

Dear Student,

Let the charges Q and Q be placed at a distance r from each other

To be in equilibrium force between charges Q and Q should be equal to charge Q and q

Thus force between Q and Q will be

^{$F_1=\frac{1}{4\pi {\epsilon }_0}\frac{Q^2}{r^2}$}

Force between q and Q will be

$F_2=\frac{1}{4\pi {\epsilon }_0}\frac{Qq}{{\left({\displaystyle \frac{r}{2}}\right)}^2}$

By condition of equilibrium,

F₁ + F₂ = 0

$$\begin{gathered} \frac{1}{4\pi {\epsilon }_0}\frac{Q^2}{r^2}+\frac{1}{4\pi {\epsilon }_0}\frac{Qq}{{\left({\displaystyle \frac{r}{2}}\right)}^2}=0 \\ \Rightarrow \frac{Q^2}{r^2}=-\frac{4Qq}{r^2} \\ \Rightarrow q=\frac{-Q}{4} \end{gathered}$$

Thus for equilibrium of the system q = -Q/4.

Regards,