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A thin wire ring of radius "r" carries a charge q. Find the magnitude of the electric field strength on the axis of the ring as a function of distance L from the centre. Find the same for L gtgt r. Find maximum field strength and the corresponding distance L. |
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Answer» Solution :Due to a ring ELECTRIC FIELD strength at a distance "L" from its centre on it can be given as `E = (qL)/(4 PI epsilon_0(L^2 + r^2)^(3//2)) to (1)` For `L gtgt r` we have `E = 1/(4 pi epsilon_0) q/(L^2)` Thus the ring behaves like a point charge. For `F_("Max") , (dE)/(dL) = 0` , From equation (1) we get `(dE)/(dL) = q/(4 pi epsilon_0) [((r^2 + L^2)^(3//2) - 3/2(r^2 + L^2)^(1//2) . 2L^2)/((r^2 + L^2)^3)] = 0` `(r^2 + L^2)^(3//2) = 3/2 (r^2 + L^2)^(1//2) xx 2L^2` On SOLVING we get `L = r/(sqrt2) to (2)` Substituting the value of "L" in equation (1) we get `E = 1/(4 pi epsilon_0) xx (q(r//sqrt(2)))/((r^2 + r^2//2)^(3//2)) = q/(6sqrt(3) pi epsilon_0 r^2)` |
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