Question

To keep system of 3 charges in equilibrium as shown in figure, $q_0$ must be

• A. $q_0=\frac{-4}{9}q$
• B. $q=\frac{-q}{3}$
• C. $q_0=\frac{-2}{5}q$
• D. $q_0=\frac{-q}{2}$

Answer 1

The correct option is A $q_0=\frac{-4}{9}q$


For equilibrium of system force on all charges should be zero.
$\frac{kq{q}_0}{x_1^2}=\frac{k{q}_{0}4q}{x_2^2}\Rightarrow x_2^2=4x_1^2\Rightarrow x_2=2x_1$
$\frac{kq{q}_0}{x_1^2}=\frac{kq\times 4q}{(x_1+x_2{)}^2}\Rightarrow \frac{q_0}{x_1^2}=\frac{4q}{(3x_1{)}^2}\Rightarrow q_0=\frac{4}{9}q$
Charge $q_0$ must be opposite in nature as that of the charges kept at the ends.