Question

Pls pls pls solve the problem... ​

Answer 1

Answer:

Proved

Explanation:

The distance between the third change be $\sqrt{d^2+(\frac{R}{2})$

The two Forces actes on the charge 'q', are $F1$ and $F2$

$F1=F2=\frac{1}{4\pi e0 } *\frac{qq1}{d^2+(\frac{R}{2})^2 }$

also from geometry we get that,

cosΘ=$\frac {d} {\sqrt{d^2+(\frac{R}{2})^2 } }$

Net force $F=F1$ cosΘ+$F2$ cosΘ=$\frac{2qq1d}{4\pi e0 } *[d^2+(\frac{R}{2} )]^-3/2$

For force to be maximum, $\frac{dF}{d(d)} =0$

On solving we get,

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

Hence proved!

Hope it help you!

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Answer 2

The distance between the third change be \sqrt{d^2+(\frac{R}{2})

The two Forces actes on the charge 'q', are F1F1 and F2F2

F1=F2=\frac{1}{4\pi e0 } *\frac{qq1}{d^2+(\frac{R}{2})^2 }F1=F2=

4πe0

1

∗

d

2

+(

2

R

)

2

qq1

also from geometry we get that,

cosΘ=\frac {d} {\sqrt{d^2+(\frac{R}{2})^2 } }

d

2

+(

2

R

)

2

d

Net force F=F1F=F1 cosΘ+F2F2 cosΘ=\frac{2qq1d}{4\pi e0 } *[d^2+(\frac{R}{2} )]^-3/2

4πe0

2qq1d

∗[d

2

+(

2

R

)]

−

3/2

For force to be maximum, \frac{dF}{d(d)} =0

d(d)

dF

=0

On solving we get,

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

2

2

R