Two point charges Q each are placed at a distance d apart. A third point charge q is placed at a distance x from midpoint on the perpendicular bisector. The value of x at which charge q will experience the maximum coulomb’s force is

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x=d

$x = \frac{d}{2 \sqrt{2}}$

$x = \frac{d}{\sqrt{2}}$

$x = \frac{d}{2}$

answer is D.

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Detailed Solution

We have from the given figure $F_{n e t} = 2 F \cos θ$

$F_{n e t} = 2 \frac{K Q q}{x^{2} + \frac{d^{2}}{4}} . \frac{x}{{(x^{2} + \frac{d^{2}}{4})}^{1 / 2}} F_{n e t} = 2 \frac{K Q q x}{{(x^{2} + \frac{d^{2}}{4})}^{3 / 2}} F o r m a x i m u m F_{n e t} = \frac{d F_{n e t}}{d x} = 0 \Rightarrow x \times - \frac{3}{2} {(x^{2} + \frac{d^{2}}{4})}^{- 5 / 2} .2 x + {(x^{2} + \frac{d^{2}}{4})}^{- 3 / 2} = 0 \Rightarrow {(x^{2} + \frac{d^{2}}{4})}^{- 5 / 2} {(- 3 x^{2} + x^{2} + \frac{d^{2}}{4})}^{- 3 / 2} = 0 \Rightarrow 2 x^{2} = \frac{d^{2}}{4} \Rightarrow x^{2} = \frac{d^{2}}{8} \Rightarrow x = \frac{d}{2 \sqrt{2}}$

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