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Grade:12th pass

1 Answers

Eshan
askIITians Faculty 2095 Points
2 years ago
Dear student,

Electric field at a distance x from the center of charged ring along its axis is given as-

E=\dfrac{kQx}{(R^2+x^2)^{3/2}}
Hence electric field is maximum when\dfrac{dE}{dx}=0
\implies (R^2+x^2)^{3/2}-\dfrac{3}{2}x(R^2+x^2)^{1/2}.2x=0
\implies x=\dfrac{R}{\sqrt{2}}

Hence potential at that point=\implies \dfrac{kQ}{\sqrt{R^2+\dfrac{R^2}{(\sqrt{2})^2}}}=\dfrac{\sqrt{2}kQ}{\sqrt{3}R}=\dfrac{Q}{2\sqrt{6}\pi\epsilon_0R}

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