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MCQMediumJEE 2025Electric Field & Field Lines

JEE Physics 2025 Question with Solution

Consider a circular loop that is uniformly charged and has a radius 2\sqrt{2}. Find the position along the positive zz-axis of the cartesian coordinate system where the electric field is maximum if the ring was assumed to be placed in the xyxy-plane at the origin:

  • A

    a2\frac{a}{\sqrt{2}}

  • B

    a2\frac{a}{2}

  • C

    aa

  • D

    00

Answer

Correct answer:A

Step-by-step solution

Standard Method

Given: A uniformly charged ring of radius R=2R = \sqrt{2} lies in the xyxy-plane with center at the origin.

Find: The position on the positive zz-axis where the axial electric field is maximum.

The magnitude of the electric field on the axis of a uniformly charged ring is

E(z)=14πε0 Q z(R2+z2)3/2E(z)=\frac{1}{4\pi\varepsilon_0}\,\frac{Q\,z}{\left(R^2+z^2\right)^{3/2}}

For finding the location of maximum field, the constant factor Q4πε0\frac{Q}{4\pi\varepsilon_0} does not affect the result. So maximize

f(z)=z(R2+z2)3/2f(z)=\frac{z}{\left(R^2+z^2\right)^{3/2}}

Derivative Condition

Using logarithmic differentiation,

ddz[ln⁡f(z)]=ddz(ln⁡z−32ln⁡(R2+z2))=0\frac{d}{dz}\left[\ln f(z)\right]=\frac{d}{dz}\left(\ln z-\frac{3}{2}\ln\left(R^2+z^2\right)\right)=0

Hence,

1z−32⋅2zR2+z2=0\frac{1}{z}-\frac{3}{2}\cdot\frac{2z}{R^2+z^2}=0

which gives

1z−3zR2+z2=0\frac{1}{z}-\frac{3z}{R^2+z^2}=0

So,

(R2+z2)−3z2=0(R^2+z^2)-3z^2=0 R2−2z2=0R^2-2z^2=0 z2=R22z^2=\frac{R^2}{2}

Thus,

z=R2z=\frac{R}{\sqrt{2}}

Since the point lies on the positive zz-axis, take the positive value.

Now substitute R=2R=\sqrt{2}:

z=22=1z=\frac{\sqrt{2}}{\sqrt{2}}=1

the solution concludes the maximum occurs at z=1z=1. Since the options are written in terms of aa, this corresponds to the general result z=a2z=\frac{a}{\sqrt{2}} for a ring of radius aa. Therefore, the correct option is A. The raw listed answer disagrees with the worked solution, and the worked solution is taken as authoritative.

Common mistakes

  • Using the ring radius directly as the position of maximum field is incorrect. The axial field of a charged ring is maximum at z=R2z=\frac{R}{\sqrt{2}}, not at z=Rz=R. Always differentiate the axial field expression before choosing the position.

  • Substituting R=2R=\sqrt{2} too early can hide the general relation. First derive z=R2z=\frac{R}{\sqrt{2}}, then substitute the numerical radius. This avoids confusion between the general option form and the numerical result.

  • Ignoring the positive-axis condition is a conceptual error. From z2=R22z^2=\frac{R^2}{2} one gets two roots, but the question asks for the point along the positive zz-axis, so only the positive root is valid.

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