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Derive an expression for field intensity due to a uniformly charged ring at a point on the axis. |
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Answer» Solution :A ring of radius R is charged uniformly with linear density `lambda=(dq)/(dl)`. Consider a small element of length dl. The FIELD intensity at the axial point P is `dE=(1)/(4pi epsi_(0)) (dq)/(r^(2))`. This can be resolved into `dE sin theta`- the perpendicular component and `dE cos theta-` the parallel component. The perpendicular component gets cancelled by SYMMETRY. The TOTAL field is `E=int dE cos theta=int (dq)/(4pi epsi_(0) r^(.2)) cos theta=int (lambda.dl)/(4pi epsi_(0) r^(.2))cos theta` The quantities r., `cos theta and lambda` are constants, as we look hte coil from P. Hence total field is `E=(lambda cos theta)/(4pi epsi_(0) r^(.2)) int dl=(lambda. cos theta)/(4 pi epsi_(0) r^(.2)) xx 2pi R` But `2pi R lambda="total charge , q"` Hence `E=q/(4pi epsi_(0) r^(.2)) cos theta=(q)/(4pi epsi_(0)) 1/r^(.2) xx r/(r.)=(qr)/(4pi epsi_(0)r^(.3))=(qr)/(4pi epsi_(0) (R^(2)+r^(2))^(2/3))` When `R lt lt r, E=(q)/(4pi epsi_(0) r^(2))` 
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