$(a)\;1,2\qquad(b)\;1,2,3\qquad(c)\;1,2,4\qquad(d)\;1,2,3,4$

Answer : (c) 1 ,2 ,4

Explanation :

In the neighbourhood of $\;Q_{1} \;$ potential is +ve . Hence $\;Q_{1}\;$ is +ve .

In the neighborhood of $\;Q_{2}\;$ potential is -ve. Hence $\;Q_{2}\;$ is -ve.

Now at point B potential is zero . Let B is at distance x from $\;Q_{2}\;.$

Then ,

$\large\frac{kQ_{1}}{r+x}+\large\frac{Q_{2}}{x}=0$

$\large\frac{Q_{1}}{Q_{2}}=-\large\frac{r+x}{x}$

$\large\frac{|Q_{1}|}{|Q_{2}|}=\large\frac{r}{x} +1 > 1$

$|Q_{1}| > |Q_{2}|$

Equilibrium points are those points at which $\;\large\frac {\partial V}{\partial x}\;$ is zero and not those at which V is zero . At A & B $\;\large\frac{\partial V}{\partial x}\;$ is not zero .

Therefore these are not equilibrium points .

C is a point of unstable equilibrium because $\;\large\frac{\partial V}{\partial x}\; $ is zero at C and C is a point of maxima of V .

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