The maximum value of the electric field on th | vedclass

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Question

(B) The electric field on the axis of a charged ring at a distance $x$ from the center is given by $E = \frac{1}{4\pi \epsilon_0} \frac{Qx}{(x^2 + R^2

Vedclass · Accepted Answer

$\frac{1}{4\pi \epsilon_0} \frac{2Q}{3\sqrt{3}R^2}$

Vedclass · Answer

(B) The electric field on the axis of a charged ring at a distance $x$ from the center is given by $E = \frac{1}{4\pi \epsilon_0} \frac{Qx}{(x^2 + R^2)^{3/2}}$.To find the maximum electric field,we differentiate $E$ with respect to $x$ and set it to zero: $\frac{dE}{dx} = 0$.This condition is satisfied when $x = \frac{R}{\sqrt{2}}$.Substituting $x = \frac{R}{\sqrt{2}}$ into the expression for $E$:$E_{\max} = \frac{1}{4\pi \epsilon_0} \frac{Q(R/\sqrt{2})}{((R/\sqrt{2})^2 + R^2)^{3/2}}$$E_{\max} = \frac{1}{4\pi \epsilon_0} \frac{QR/\sqrt{2}}{(R^2/2 + R^2)^{3/2}} = \frac{1}{4\pi \epsilon_0} \frac{QR/\sqrt{2}}{(3R^2/2)^{3/2}}$$E_{\max} = \frac{1}{4\pi \epsilon_0} \frac{QR/\sqrt{2}}{(3\sqrt{3}R^3 / 2\sqrt{2})} = \frac{1}{4\pi \epsilon_0} \frac{2Q}{3\sqrt{3}R^2}$.

The maximum value of the electric field on the axis of a charged ring having charge $Q$ and radius $R$ is:

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