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The curve represents the distribution of potential along the straight line joining the two charges $\;Q_{1}\;$ and $\;Q_{2}\;$ (separated by a distance r ) then which of the following are correct 1. $\;|Q_{1}| > |Q_{2}|\;$ 2. $\;Q_{1}\;is\;+ve\;$ 3. $\;A\;$ and $\;B\;$ are equibrium points 4. $\;C\;$ is point of unstable equilibrium

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

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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 .
answered Feb 13, 2014 by yamini.v
 

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