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Find the electric field on the axis of a charged disc. |
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Answer» Solution :Consider a disc of uniform surface charge density `'sigma'`. Let US calculate the electric field due to a charged ring of radius r. from the centre and having a width, dr. Here we are using the expression for electric field INTENSITY for a charged ring of radius r at a point on the axis at a distance x from the centre. `E=(kxq)/((x^(2)+r^(2))^(3//2))` directed along the axis outwards from the centre. Thus electric field due to elementary ring. `dE=(kxdq)/((x^(2)+r^(2))^(3//2))`, directed along the line OP. Now, the area of the ring, `dS=2pir.dr` Charge on the elementary ring `implies dq=sigma2pirdr`, Thus, electric field due to entire disc `|E|_(p)=intdE=kxunderset(0)overset(R)INT(sigma2pirdr)/((x^(2)+r^(2))^(3//2))=(1)/(4piin_(0))x.sigmapiunderset(0)overset(R)int(2r)/((x^(2)+r^(2))^(3//2))dr` `E=(sigma)/(2in_(0))[1-(x)/((x^(2)+R^(2))^(1//2))]=(sigma)/(2epsilon_(0))(1-costheta)`, where `theta` = semi vertical angle subtend by the disc at P. Also, as `Rrarroo,E=(sigma)/(2in_(@)`, which is the electric field in front of an infinite plane SHEET of charge. 
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