A) At a distance of r/3 from 2q
B) At a distance of 2/73 from 2q
C) At a distance of r/16 from 2q
D) None of the above
Correct Answer: A
Solution :
Let the third charge - q is placed at a distance of x from the charge 2g. Then, the potential energy of the system is \[\left( \frac{1}{2},\frac{3}{2} \right)\] Here,\[\left( \frac{1}{2},\frac{1}{2} \right)\] U is minimum, where \[\left( \frac{1}{2},\pm \sqrt{2} \right)\]is maximum Let \[{{y}^{2}}-kx+8=0,\] For y to be maximum, \[\frac{1}{8}\] \[\frac{1}{4}\] \[{{(y-2)}^{2}}=(x-1),\]\[x=1+xy\frac{dy}{dx}+\frac{{{(xy)}^{2}}}{2!}+{{\left( \frac{xy}{dx} \right)}^{2}}+\frac{{{(xy)}^{3}}}{3!}{{\left( \frac{dy}{dx} \right)}^{3}}\] But at \[y={{\log }_{e}}(x)+C\] positive i.e. at \[y={{({{\log }_{e}}x)}^{2}}+C\] is minimum or U is maximum. So, there cannot be any point between them where the potential energy is minimum.
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