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Question

A charge is distributed uniformly over a 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 point's at large distances from the ring, it behaves like a point charge.

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Solution

Consider a point P on the axis of uniformly charged ring at a distance x from its centre O. Point P is at distance r=a2+x2 from each element dl of ring. If q is total charge on ring, then, charge per metre length, λ=q2πa.
The ring may be supposed to be formed of a large number of ring elements.
Consider an element of length dl situated at A.
The charge on element, dq=λdl
The electric field at P due to this element
dE1=14πϵ0dqr2=14πϵ0λdlr2, along PC
The electric field strength due to opposite symmetrical element of length dl at B is dE2=14πϵ0dqr2=14πϵ0λdlr2, along PD
If we resolve dE1 and dE2 along the axis and perpendicular to axis, we note that the components perpendicular to axis are oppositely directed and so get cancelled, while those along the axis are added up. Hence, due to symmetry of the ring, the electric field strength is directed along the axis.
The electric field strength due to charge element of length dl, situated at A, along the axis will be
dE=dE1cosθ=14πϵ0λdlr2cosθ
But, cosθ=xr
dE=14πϵ0λdlxr3=14πϵ0λxr3dl
The resultant electric field along the axis will be obtained by adding fields due to all elements of the ring, i.e.,
E=14πϵ0λxr3dl=14πϵ0dl
But, dl= whole length of ring =2πa and r=(a2+x2)1/2
E=14πϵ0λx(a2+x2)3/22πa
As, λ=q2πa, we have E=14πϵ0(q2πa)x(a2+x2)3/22πaE=14πϵ0qx(a2+x2)3/2
or, E=14πϵ0qx(a2+x2)3/2, along the axis
At large distances i.e., x>>a,E=14πϵ0qx2
i.e., the electric field due to a point charge at a distance x.
For points on the axis at distances much larger than the radius of ring, the ring behaves like a point charge.
1673678_472038_ans_2e18f688d5d143c38fbf4f0508b59f6c.png

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