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MCQEasyJEE 2023Electric Potential & Potential Energy

JEE Physics 2023 Question with Solution

The electric potential at the centre of two concentric half rings of radii R1R_1 and R2R_2, having the same linear charge density λ\lambda is:

  • A

    2λϵ0\frac{2\lambda}{\epsilon_0}

  • B

    λ2ϵ0\frac{\lambda}{2\epsilon_0}

  • C

    λ4ϵ0\frac{\lambda}{4\epsilon_0}

  • D

    λϵ0\frac{\lambda}{\epsilon_0}

Answer

Correct answer:B

Step-by-step solution

Standard Method

Given: Two concentric half rings have radii R1R_1 and R2R_2 and the same linear charge density λ\lambda.

Find: The electric potential at the centre.

The potential at the center due to a half ring is given by:

V=λ⋅πR4πϵ0R=λ4ϵ0V = \frac{\lambda \cdot \pi R}{4 \pi \epsilon_0 R} = \frac{\lambda}{4 \epsilon_0}

For two concentric half rings:

Vtotal=λ4ϵ0+λ4ϵ0=λ2ϵ0V_{\text{total}} = \frac{\lambda}{4 \epsilon_0} + \frac{\lambda}{4 \epsilon_0} = \frac{\lambda}{2 \epsilon_0}

Therefore, the correct option is B.

Using contribution of each half ring

Given: Two half rings of radii R1R_1 and R2R_2 carry the same linear charge density λ\lambda.

Find: Net potential at the common centre.

Potential at centre

V=14πϵ0qRV = \frac{1}{4\pi\epsilon_0}\frac{q}{R}

For a half ring, charge is

q=λ⋅πRq = \lambda \cdot \pi R

So for one half ring,

V=14πϵ0λπRR=λ4ϵ0V = \frac{1}{4\pi\epsilon_0}\frac{\lambda \pi R}{R} = \frac{\lambda}{4\epsilon_0}

This result is independent of radius, so both half rings contribute equally:

V1=λ4ϵ0,V2=λ4ϵ0V_1 = \frac{\lambda}{4\epsilon_0}, \qquad V_2 = \frac{\lambda}{4\epsilon_0}

Hence,

Vtotal=V1+V2=λ2ϵ0V_{\text{total}} = V_1 + V_2 = \frac{\lambda}{2\epsilon_0}

Thus, the electric potential at the centre is λ2ϵ0\frac{\lambda}{2\epsilon_0}.

Common mistakes

  • Using electric field instead of electric potential. At the centre, field contributions depend on direction, but potential is a scalar and adds directly. Compute potential of each half ring separately and then add.

  • Assuming the answer depends on R1R_1 or R2R_2. For a half ring, q=λπRq = \lambda \pi R and dividing by RR in the potential formula cancels the radius. So each half ring contributes the same potential.

  • Taking the arc length of a half ring incorrectly. The length of a half ring is πR\pi R, not 2πR2\pi R. Using the full circumference gives the wrong charge and hence the wrong potential.

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