Q.353. Maximum electric intensity along the axis of uniformly charged ring

Question 353.
Show that the maximum magnitude, Emax, of the electric field along the axis of a uniformly charged ring occurs at x = a/√2 and has value Q / [6√3 πεo a^2].

Answer 353.
Electric intensity at a distance x from the center of the ring on its axis is given by
E = [Q/(4πεo)] x / (x^2 + a^2)^(3/2)
=> E is maximum where the value of
f (x) = x / (x^2 + a^2)^(3/2) is maximum
For f (x) to be maximum, f '(x) = 0 and f"(x) < 0
ln [f (x)] = lnx - (3/2) ln(x^2 + a^2)
=> f '(x) / f (x) = 1/x - 3x / (x^2 + a^2)
f '(x) = 0
=> 1/x - 3x / (x^2 + a^2) = 0
=> (x^2 + a^2) - 3x^2 = 0
=> x^2 = a^2/2
=> x = ± a/√2
=> E is maximum for x = ± a/√2 and minimum at x = 0
Magnitude of Emax
= [Q/(4πεo)] * (a/√2) / (a^2/2 + a^2)^(3/2)
= [Q/(4πεo)] * (a/√2) / [a^3 * (3/2)^(3/2)]
= [Q/(4πεo)] * (1/a^2) * (1/√2) * (2/3) * √(2/3)
= Q / [6√3 πεo a^2].

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