Question

Two free point charges +4Q and +Q are placed at a distance r. A third charge q is so placed such that all the three are in equilibrium.

• A. q is placed at a distance $\frac{1}{3}r$ from 4Q
• B. q is placed at a distance $\frac{1}{2}r$ from Q
• C. $q=\frac{4Q}{9}$
• D. $q=-\frac{4Q}{9}$

Answer 1

The correct option is D $q=-\frac{4Q}{9}$


For equilibrium $\frac{4Qq}{(r-x{)}^2}=\frac{qQ}{x^2}$
$\Rightarrow x=\frac{r}{3}$
For $x=\frac{r}{3}$
For all the charges to be in equilibrium
$\frac{4Q^2}{r^2}+\frac{4Qq}{(r-x{)}^2}+\frac{Qq}{x^2}=0$ ..........(1)
Substituting $x=\frac{r}{3}$ in (1), we get $q=\frac{-4Q}{q}$