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  • Help me understand! A circular ring is charged uniformly. It has radius R , charge +Q . Electric field will be max. w.r.t X (distance).

A circular ring is charged uniformly. It has radius R , charge +Q . Electric field will be max. w.r.t X (distance).

  • Option 1)

    X = 0

  • Option 2)

    X = R

  • Option 3)

    X = \pm \frac{R}{\sqrt2}

  • Option 4)

    X = infinity

 

Answers (1)

best_answer

As we learnt ,

 

Max E and V at a point on the axis of a uniformly charged ring -

If   x= \pm \frac{R}{\sqrt{2}}

E_{max}=\frac{Q}{6\sqrt{3}\pi \varepsilon _{0}R^{2}}  ,   V_{max}=\frac{Q}{2\sqrt{6}\pi \varepsilon _{0}R}

- wherein

 

 For E to be max

\frac{dE}{dx} = 0 \; \Rightarrow \frac{d}{dx}\left [ \frac{KQ}{\left (R^{2}+X^{2} \right )^{\frac{3}{2}}} \right ] = 0

E_{max} at X = \pm \frac{R}{\sqrt{2}}

 


Option 1)

X = 0

Option 2)

X = R

Option 3)

X = \pm \frac{R}{\sqrt2}

Option 4)

X = infinity

Posted by

Avinash

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