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NVAMediumJEE 2023Gauss's Law Applications

JEE Physics 2023 Question with Solution

As shown in figure, a cuboid lies in a region with electric field E=2x2i^−4yj^+6k^\mathbf{E} = 2x^2 \hat{i} - 4y \hat{j} + 6 \hat{k} N/C. The magnitude of charge within the cuboid is nϵ0n \epsilon_0 C. The value of nn is _____ (if dimensions of cuboid are 1×2×3 m31 \times 2 \times 3 \, \text{m}^3)

A cuboid in three-dimensional axes with origin marked at (0,0,0), and vertices indicated by (1,0,0), (0,2,0), and (0,0,3).

Answer

Correct answer:12

Step-by-step solution

Standard Method

Given: Electric field is

E=2x2i^−4yj^+6k^\mathbf{E} = 2x^2 \hat{i} - 4y \hat{j} + 6 \hat{k}

The cuboid extends from x=0x=0 to x=1x=1, y=0y=0 to y=2y=2, and z=0z=0 to z=3z=3.

Find: The value of nn if the enclosed charge magnitude is nϵ0n\epsilon_0.

A cuboid drawn along x, y, z axes with origin at (0,0,0) and edge endpoints labeled (1,0,0), (0,2,0), and (0,0,3).

Using Gauss's law in differential form, the net flux is obtained from the divergence of the field over the cuboid. From the extracted working,

ϕnet=−8×3+2×6=−12\phi_{\text{net}} = -8 \times 3 + 2 \times 6 = -12

Now apply Gauss's law:

ϕnet=qϵ0\phi_{\text{net}} = \frac{q}{\epsilon_0}

So,

−12=qϵ0-12 = \frac{q}{\epsilon_0}

Hence,

∣q∣=12ϵ0|q| = 12\epsilon_0

Therefore, the value of nn is 1212.

Using divergence explicitly

Given:

E=2x2i^−4yj^+6k^\mathbf{E} = 2x^2 \hat{i} - 4y \hat{j} + 6 \hat{k}

Find: Enclosed charge magnitude in the form nϵ0n\epsilon_0.

Compute divergence:

∇⋅E=∂∂x(2x2)+∂∂y(−4y)+∂∂z(6)\nabla \cdot \mathbf{E} = \frac{\partial}{\partial x}(2x^2) + \frac{\partial}{\partial y}(-4y) + \frac{\partial}{\partial z}(6) =4x−4+0= 4x - 4 + 0

Integrate over the cuboid volume:

ϕnet=∭(4x−4) dV\phi_{\text{net}} = \iiint (4x-4) \, dV

with 0≤x≤10 \le x \le 1, 0≤y≤20 \le y \le 2, 0≤z≤30 \le z \le 3.

ϕnet=∫01∫02∫03(4x−4) dz dy dx\phi_{\text{net}} = \int_0^1 \int_0^2 \int_0^3 (4x-4) \, dz \, dy \, dx =(2)(3)∫01(4x−4) dx= (2)(3) \int_0^1 (4x-4) \, dx =6[2x2−4x]01=6(2−4)=−12= 6 \left[2x^2 - 4x\right]_0^1 = 6(2-4) = -12

By Gauss's law,

q=ϵ0ϕnet=−12ϵ0q = \epsilon_0 \phi_{\text{net}} = -12\epsilon_0

Therefore, the magnitude is

∣q∣=12ϵ0|q| = 12\epsilon_0

So the required value is 1212.

Common mistakes

  • Using the electric field directly instead of its divergence for enclosed charge. Gauss's law for enclosed charge depends on net flux, not on the field at one point. Compute ∇⋅E\nabla \cdot \mathbf{E} or total outward flux through faces.

  • Forgetting that the question asks for the magnitude of charge. The flux comes out negative, so q=−12ϵ0q=-12\epsilon_0, but the required value is ∣q∣=12ϵ0|q|=12\epsilon_0.

  • Ignoring the cuboid dimensions while integrating. The limits x:0→1x:0\to1, y:0→2y:0\to2, z:0→3z:0\to3 determine the total flux. Missing these bounds gives an incorrect value of enclosed charge.

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