A charge is distributed uniformaly over is ring of radius 'a' .Obtain an expression for the electric intensity E at a point on the axis of the ring. Hence show that for points at large distances from the ring, it behaves like a point charge.

Explanation

Consider a uniform circular ring of radius .a. carrying a charge Q distributed uniformaly over its surface Let P be a point situated at a distance r from the centre of ring along its axis.
Consider an element of length Dl around point A of the ring carrying a charge Dq `=(Q)/(2 pi A ) Delta l `
The electrical field at point P due to this element is given as
` " "| oversetto (Delta E) | =(1)/( 4 pi in _0) .(Delta q)/((AP)^(2)) =(1)/( 4 pi in _0) .(Q)/( 2 pi a ( a^(2) +r^(2)) `
The electric field ` oversetto (Delta E )`, directed along the direction AP, subtends an angle ` theta ` with the axis of ring and can be resolved into two components namely (i) `Delta E cos theta ` along the axis of ring , and (ii) ` Delta E sin theta ` normal to the axis of ring.
It is clear from symmetry that the normal components `Delta E sin theta ` due to mutually opposite charges elements at A and B nullify each other. Hence , net electric field due to whole ring will be
` E= sum Delta E cos theta =sum ((1)/(4 pi in _0) .(QDelta l )/(2 pi a (a^(2) +r^(2)) ). (r)/( (a^(2) +r^(2))^(1//2)) ) `
` =(1)/(4 pi in_0) .(Qr)/( 2pi a (r^(2) +a^(2))^(3//2)) .sum Delta l =(1)/( 4 pi in _0) . (Qr)/( 4 pi a(r^(2)+a^(2) )^(3//2)). 2pi a =(1)/( 4 pi in _0) .(Qr)/( (r^(2) +a^(2)) ^(3//2))`
The field `oversetto E ` is directed along the axis OP of the charged ring.
If ` gtgt a,` then the above relation may be expressed as
` oversetto E= (1)/( 4 pi in _0) .(Q)/(r^(2)) hat r `
This shows that for far points at long distances from the ring. it behaves lika a point charges.
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