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  • Two point charges  each are placed at a distance d apart. A third point charge q is placed at a distance x from mid-point on t

Two point charges \mathrm{Q} each are placed at a distance d apart. A third point charge q is placed at a distance x from mid-point on the perpendicular bisector. The value of x at which charge q will experience the maximum Coulomb's force is :

Option: 1

\mathrm{x=d}


Option: 2

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


Option: 3

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


Option: 4

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


Answers (1)

best_answer

\mathrm{\text { Fnet } =2 F \cos \theta} \\

\mathrm{=\frac{2 \times K \times Q q}{\left[\left(\frac{d}{2}\right)^{2}+x^{2}\right]} \times \frac{x}{\sqrt{\left(\frac{d}{2}\right)^{2}+x^{2}}} }\\

\mathrm{=\frac{2 K Q q x}{\left(\frac{d^{2}}{4}+x^{2}\right)^{3 / 2}}}

For Fnet to be maximum

\mathrm{\frac{d (\text { Fnet })}{d x}=0}

On solving we get

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

Hence the correct option is 4

Posted by

Irshad Anwar

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