Electric Field

Question

A uniformly charged thin ring of radius $R$ carries total charge $Q$. At what distance $x$ from the centre along the axis does the electric field magnitude reach its maximum value?

RankGuru · Accepted Answer

The electric field at a point on the axis of a uniformly charged ring at distance $x$ from the centre is given by $E(x) = \frac{kQx}{(x^2 + R^2)^{3/2}}$. This arises because all charge elements contribute equal axial components while their transverse components cancel by symmetry. To find the maximum, we differentiate $E(x)$ with respect to $x$ and set it to zero. Computing $\frac{dE}{dx} = kQ \frac{(x^2+R^2)^{3/2} - x \cdot \frac{3}{2}(x^2+R^2)^{1/2} \cdot 2x}{(x^2+R^2)^3} = 0$. The numerator condition gives $(x^2+R^2) - 3x^2 = 0$, so $R^2 = 2x^2$, yielding $x = R/\sqrt{2}$. Option $x = R$ is incorrect because differentiating at $x = R$ does not satisfy the zero-derivative condition. Option $x = R\sqrt{2}$ is incorrect because it results from inverting the relation $R^2 = 2x^2$. Option $x = 2R$ is incorrect because it has no derivation basis in this differentiation. This is a classic JEE Advanced calculus-based electrostatics problem testing both field derivation and optimization. Plausibility check: $x = R/\sqrt{2} \approx 0.707R$ is physically reasonable as the maximum should be between the centre (field is zero) and \infty (field is zero), closer to the ring where charge density is felt most strongly.

Electric Field

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