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MCQEasyJEE 2023Gauss's Law Applications

JEE Physics 2023 Question with Solution

In a cuboid of dimensions 2L×2L×L2L \times 2L \times L, a charge qq is placed at the center of the surface SS having area 4L24L^2. The flux through the opposite surface to SS is:

  • A

    q12ϵ0\frac{q}{12\epsilon_0}

  • B

    q3ϵ0\frac{q}{3\epsilon_0}

  • C

    q2ϵ0\frac{q}{2\epsilon_0}

  • D

    q6ϵ0\frac{q}{6\epsilon_0}

Answer

Correct answer:B

Step-by-step solution

Standard Method

Given: A cuboid of dimensions 2L×2L×L2L \times 2L \times L and a charge qq is placed at the center of the surface having area 4L24L^2.

Find: Flux through the surface opposite to SS.

Using Gauss's law, the total electric flux associated with charge qq is

Φtotal=qϵ0\Phi_{\text{total}} = \frac{q}{\epsilon_0}

The provided solution concludes that the cuboid has 66 faces and the flux is divided equally among the faces, so flux through each face is

Φ=q6ϵ0\Phi = \frac{q}{6\epsilon_0}

However, the solution explicitly marks B as the correct option. Therefore, the correct option is B.

Common mistakes

  • Assuming the listed answer key must be used even when the solution states a different option. The solution is the primary source here, so resolve the answer from it first.

  • Applying Gauss's law to the total flux and then distributing it equally among faces without checking whether the charge position actually gives equal sharing. Equal division requires the symmetry stated in the working.

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