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A thin wire 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 gt gt 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)/(4pi epsi_(0)(L^(2) + r^(2))^(3//2)) rarr (1)` For `L gt gt t` we have `E= (1)/(4pi in_(0))(Q)/(L^(2))` Thus the ring behaves like a POINT charge. For `E_("Max") , (dE)/(dL) = 0`. From equation (1) we get `(dE)/(dL) = (q)/(4pi epsi_(0)) [((r^(2) + L^(2))^(3//2)- (3)/(2) (r^(2) + L^(2))^(1//2)2L)/((r^(2) + L^(2))^(3))]=` `(r^(2) + L^(2))^(3//2) = (3)/(2) (r^(2) + L^(2))^(1//2) xx 2L` On solving we get `L = (r )/(sqrt2) rarr (2)` Substituting the value of "L.. in equation (1) we get `E= (1)/(4pi in_(0)) xx (q (r//sqrt2))/((r^(2) + r^(2)//2)^(3//2)) = (q)/(6 sqrt3 pi in_(0)r^(2))` |
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