JEE MAIN - Physics (2019 - 9th January Morning Slot - No. 18)
For a uniformly charged ring of radius R, the electric field on its axis has the largest magnitude at a distance h from its center. Then value of h is :
$${R \over {\sqrt 5 }}$$
$${R \over {\sqrt 2 }}$$
R
R$$\sqrt 2 $$
Explanation
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Electric field on the axis of the ring,
$$E = {{KQh} \over {{{\left( {{R^2} + {h^2}} \right)}^{{3 \over 2}}}}}$$
For maximum electric field,
$${{dE} \over {dh}} = 0$$
$$ \Rightarrow $$ $$h = {R \over {\sqrt 2 }}$$
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