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Question

A thin wire ring of radius r carries a charge q. Find the magnitude of the electric field strength on the axis of the ring as a function of distance l from its centre. Investigate the obtained function at l>>r. Find the maximum strength magnitude and the corresponding distance l. Draw the approximate plot of the function E(l).

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Solution

From the symmetry of the condition, it is clear that, the field along the normal will be zero.
i.e. En=0 and E=E1
Now dE1=dq4πϵ0(R2+l2)cosθ
But dq=q2πRdx and cosθ=l(R2+l2)1/2
Hence
E=dEl=2πR0ql2πRdx4πϵ0(R2+l2)3/2
or E=14πϵ0ql(l2+R2)3/2
and for l>>R, the ring behaves like a point charge, reducing the field to the value,
E=14πϵ0ql2
For Emax, we should have dEdl=0
So, (l2+R2)3/232l(l2+R2)1/22l=0 or l2+R23l2=0
Thus l=R2 and Emax=q63πϵ0R2.
1784296_1508312_ans_91cf874f48a14de78e5551f9e1f4217a.jpg

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