Charges $2 q$ and $8 q$ are placed at the end points $A$ and $B$ respectively of a $9 c m$ long straight line. A third charge $q$ is placed at point $C$ of $A B$ such that the potential energy of the system is minimum. Find he distance of $C$ from $A$ in cm .

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Detailed Solution

Charges 2q and 8q are placed at the end points A and B repsectivel

$U = \frac{1}{4 π ε_{0}} \times \frac{1}{10^{- 2}} [\frac{(8 q) (2 q)}{9} + \frac{(2 q) (q)}{x} + \frac{(q) (8 q)}{(9 - x)}] = \frac{10^{2} q^{2}}{4 π ε_{0}} [\frac{16}{9} + \frac{2}{x} + \frac{8}{9 - x}]$

For potential energy to be minimum,

$\frac{d U}{d x} = 0 \Rightarrow - \frac{2}{x^{2}} + \frac{8}{(9 - x)^{2}} = 0 or \frac{9 - x}{x} = 2 or x = 3 c m$

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