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Question
Two charges q and – 3q are placed fixed on x-axis separated by distance ‘d’. Where should a third charge 2q be placed such that it will not experience any force?
Solution
At P: on 2q, Force due to q is to the left and that due to –3q is to the right.
∴ `(2q^2)/(4 piε_0 x^2) = (6q^2)/(4 piε_0 (d + x)^2`
∴ `(d + x)^2 = 3x^2`
∴ `2x^2 - 2dx - d^2` = 0
`x = d/2 + (sqrt(3)d)/2`
(–ve sign would be between q and – 3q and hence is unacceptable.)
`x = d/2 + (sqrt(3)d)/2 = d/2 (1 + sqrt(3))` to the left of q.
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