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Two charges each of +q coulomb are place...

Two charges each of `+q` coulomb are placed along line. A third charge -q is placed between them. At what position will the system be in equilibrium ?

Text Solution

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For charge -q to be in equilbrium, for an-q due to +Q at point A should be equal and opposite to that due to +Q at point B.
i.e., `(1)/(4pi epsi_(0)) (Qq)/(x^(2)) =(1)/(4 pi epsi_(0))(Q.q)/((r-x)^(2))`
`x^(2)=(r-x)^(2)`
`x=r-x`
`x=(r)/(2)`
`:.` for equilirium, -q must be kept at the middle of the line joining the A and B.
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Knowledge Check

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