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  • Help me answer: For a uniformly charged ring radius R, the electric field on its axis has the largest magnitude at a distance h from its centre. Then value of h is:

For a uniformly charged ring radius R, the electric field on its axis has the largest magnitude at a distance h from its centre. Then value of h is:

  • Option 1)

      \frac{R}{\sqrt{5}}

      

     

  • Option 2)

    \frac{R}{\sqrt{2}}

  • Option 3)

    R

  • Option 4)

    R\sqrt{2}

Answers (3)

best_answer

 

E and V at a point P that lies on the axis of ring -

\dpi{100} E_{x}=\frac{kQx}{\left ( x^{2}+R^{2} \right )^{\frac{3}{2}}}  ,   V=\frac{kQ}{\left ( x^{2}+R^{2} \right )^{\frac{1}{2}}}

The electric field due to ring on its axis

\dpi{100} E_{x}=\frac{kQx}{\left ( x^{2}+R^{2} \right )^{\frac{3}{2}}}

here x = h

\dpi{100} E_{h}=\frac{kQh}{\left ( h^{2}+R^{2} \right )^{\frac{3}{2}}}

For finding maximum find \frac{dE}{dh} and equate to zero

\Rightarrow kQ (h^{2} + R^{2})^{\frac{3}{2}} = kQh. \frac{3}{2} (h^{2} + R^{2})^{\frac{1}{2}} . 2h

h^{2} + R^{2} = 3 h^{2}

2h^{2} = R^{2}

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

 

 

 


Option 1)

  \frac{R}{\sqrt{5}}

  

 

Option 2)

\frac{R}{\sqrt{2}}

Option 3)

R

Option 4)

R\sqrt{2}

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