For a uniformly charged ring of radius R,the | vedclass

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Question

(B) The electric field $E$ at a distance $x$ from the center of a uniformly charged ring of radius $R$ is given by $E = \frac{k Q x}{(x^2 + R^2)^{3/2}

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$R/\sqrt{2}$

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(B) The electric field $E$ at a distance $x$ from the center of a uniformly charged ring of radius $R$ is given by $E = \frac{k Q x}{(x^2 + R^2)^{3/2}}$.To find the distance $h$ where the electric field is maximum,we set the derivative with respect to $x$ to zero: $\frac{dE}{dx} = 0$.Using the quotient rule: $\frac{d}{dx} \left[ \frac{x}{(x^2 + R^2)^{3/2}} \right] = \frac{(x^2 + R^2)^{3/2} - x \cdot \frac{3}{2}(x^2 + R^2)^{1/2} \cdot 2x}{(x^2 + R^2)^3} = 0$.This simplifies to $(x^2 + R^2) - 3x^2 = 0$,which gives $R^2 - 2x^2 = 0$.Therefore,$x^2 = R^2/2$,which implies $x = R/\sqrt{2}$.Thus,the maximum magnitude occurs at $h = R/\sqrt{2}$.

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

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