Question

Two charges. $$-Q$$ and $$-Q$$ are a distance $$l$$ apart. These two charges are free to move but do not because there is a third (fixed) charge nearby. What must be the magnitude of the third charge and its placement in order for the first two to be in equilibrium?

Answer 1

Correct option is B. $$0.4$$ Q
The electric force on third $$Q$$ due to charge first $$Q_0$$ is,
$$F_{31}=k\dfrac{Q_0Q}{x^2}$$..........(1)
The electric force on third $$Q$$ due to charge second $$3Q_0$$ is,
$$F_{32}=k\dfrac{3Q_0Q}{(1-x)^2}$$..........(1)
The charges are in equilibrium so, the net electric force on the third charge is zero, that is
$$F_{31}-F_{32}=0$$
Substitute the equilibrium (1) and (2( in the above equation.
$$k\dfrac{Q_0Q}{x^2}-k\dfrac{3Q_0Q}{(1-x)^2}=0$$.
$$\dfrac{1}{x^2}-\dfrac{3}{(1-x)^2}=0$$
$$\dfrac{1}{x^2}=\dfrac{3}{(1-x)^2}$$
Rearrange the above equation for $$x$$.
$$x=\dfrac{l}{\sqrt{3}+1}=0.366l$$
The three charges are in equilibrium and the net force on the first charge due to the third charge and the second charge is
$$k\dfrac{Q_0Q}{x^2}=k\dfrac{3Q^2_0}{l^2}$$

$$Q=3Q_0\dfrac{x^2}{l^2}$$
$$=Q_0\dfrac{3}{(\sqrt{3}+1)^2}=0.402Q_0$$