Question

Two particles A and B, each with a charge Q, are placed a distance d apart. Where should a particle of charge q be placed on the perpendicular bisector of AB, so that it experiences maximum force? What is the magnitude of this maximum force?

Answer 1

Explanation:

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Answer 2

A particle of charge q be placed on the perpendicular bisector of AB, so that it experiences maximum force at $\frac{16 k \theta q x}{\left(4 x^{2}+d^{2}\right)^{\frac{3}{2}}}$.

The magnitude of this maximum force is given by $d=\frac{d}{2 \sqrt{2}}$.

Explanation:

Given data in the question ;

Force on the particle of charge ' q ' at ' c ' is only the portion of 2 forces x

Let the q charge be set on AB's perpendicular bisector at a distance x.

So,  

$\mathrm{f}_{\mathrm{onc}}=\mathrm{F}_{\mathrm{CB}} \sin \theta+\mathrm{F}_{\mathrm{Ac}} \sin \theta$    

But.

$2 \mathrm{F}_{\mathrm{CB}} \sin \theta=2 \frac{K Q q}{x^{2}+\left(\frac{d}{2}\right)^{2}} \times \frac{X}{\sqrt{x^{2}+\frac{d^{2}}{4}}}$

               $=\frac{2 k \theta q x}{\left(x^{2}+\frac{d^{2}}{4}\right)^{\frac{3}{2}}}$        

              $=\frac{16 k \theta q x}{\left(4 x^{2}+d^{2}\right)^{\frac{3}{2}}}$

For a maximum force :

$\frac{d F}{d x}=0$

$\frac{d}{d x}\left(\frac{16 k \theta q x}{\left(4 x^{2}+d^{2}\right)^{\frac{3}{2}}}\right)=0$

$k\left[\frac{4 x^{2}+d^{2}-x\left(\frac{3}{2}\left(\left(4 x^{2}+d^{2}\right)^{\frac{1}{2}}\right) 8 x\right)}{\left(4 x^{2}+d^{2}\right)^{3}}\right]=0$

$\frac{k\left(4 x^{2}+d^{2}\right)^{\frac{1}{2}}\left[\left(4 x^{2}+d^{2}\right)^{3}-12 x^{2}\right]}{\left(4 x^{2}+d^{2}\right)^{3}}=0$

$\left(4 x^{2}+d^{2}\right)^{3}=12 \mathrm{x}^{2}$

$16 x^{4}+d^{4}+8 x^{2} d^{2}=12 x^{2}$

$d^{4}+8 x^{2} d^{2}=0$

$d^{2}=0$

$d^{2}+8 x^{2}=0$

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

Thus, the magnitude of maximum force is $d=\frac{d}{2 \sqrt{2}}$.