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Considered a uniformly charged ring of radius R find the point on the axis where the electric field is maximum

Sanketbhole , 6 Years ago

Grade 12

1 Answers

Eshan

Last Activity: 6 Years ago

Dear student,

Electric field due to a charged ring at a distance $x$ from it is given as

$E=\dfrac{kQx}{(x^2+R^2)^{3/2}}$
The derivative of E wrt x is zero at the point where the electric field is maximum.

Hence $\dfrac{dE}{dx}=kQ(\dfrac{(x^2+R^2)^{3/2}.(1)-x\dfrac{3}{2}(x^2+R^2)^{1/2}.(2x)}{x^2+R^2})=0$
$\implies x=\dfrac{R}{\sqrt{2}}$

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