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A point charge q is located at the centre of a thin ring of radius R withuniformly distributed charge "-q". Find the magnitude of the electric field strength vector at the point lying on the axis of the ring at a distance x from its centre if x gt gt R. |
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Answer» Solution :Electric field at P due ring `E_(1)=(qx)/(4PI in_(0) (x^(2)+R^(2))^(3//2))` (towards centre) Electric field at P due to +q `E_(2)=(q)/(4pi in_(0)x^(2))` (away from centre) Thus net field at P is `E_("Net") =E_(1)+E_(2)` `=(q)/(4pi in_(0)) [(1)/(x^(2)) -(x)/((x^(2)+R^(2))^(3//2))]` For `x GT gt R` `E_("net")=(q[(x^(2)+R^(2))^(3//2)-x^(3)])/(4pi in_(0)x^(2)(x^(2)+R^(2))^(3//2))` `=(3qR^(2))/(4pi in_(0) 2x^(4))=(3qR^(2))/(8pi in_(0)x^(4))` (Using Binomial approximation). |
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