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MCQEasyJEE 2024Gauss's Law Applications

JEE Physics 2024 Question with Solution

An electric field E=2x i^ N/CE = 2x\, \hat{i} \, \text{N/C} exists in space. A cube of side 2 m2 \, \text{m} is placed in the space. The electric flux through the cube is:

  • A

    8 N m2/C8 \, \text{N m}^2/\text{C}

  • B

    12 N m2/C12 \, \text{N m}^2/\text{C}

  • C

    16 N m2/C16 \, \text{N m}^2/\text{C}

  • D

    20 N m2/C20 \, \text{N m}^2/\text{C}

Answer

Correct answer:C

Step-by-step solution

Standard Method

Given: Electric field is E⃗=2x i^\vec{E} = 2x\, \hat{i} and the cube has side a=2 ma = 2 \, \text{m}.

Find: The total electric flux through the cube.

Using Gauss's law for flux through a closed surface,

Φ=∮E⃗⋅dA⃗\Phi = \oint \vec{E} \cdot d\vec{A}

Since the field is only along the xx-direction, faces perpendicular to the yy and zz axes contribute zero flux.

Consider the cube from x=0x = 0 to x=2x = 2. The two relevant faces are at x=0x = 0 and x=2x = 2. Area of each face is

A=2×2=4 m2A = 2 \times 2 = 4 \, \text{m}^2

For the face at x=0x = 0,

E⃗=2(0) i^=0\vec{E} = 2(0)\, \hat{i} = 0

So the flux through this face is

Φ0=E⋅A=0×4=0 N m2/C\Phi_0 = E \cdot A = 0 \times 4 = 0 \, \text{N m}^2/\text{C}

For the face at x=2x = 2,

E⃗=2(2) i^=4 N/C\vec{E} = 2(2)\, \hat{i} = 4 \, \text{N/C}

Hence the flux through this face is

Φ2=E⋅A=4×4=16 N m2/C\Phi_2 = E \cdot A = 4 \times 4 = 16 \, \text{N m}^2/\text{C}

Therefore, total flux through the cube is

Φtotal=Φ0+Φ2=0+16=16 N m2/C\Phi_{\text{total}} = \Phi_0 + \Phi_2 = 0 + 16 = 16 \, \text{N m}^2/\text{C}

The correct option is C.

Face-to-face flux difference

Given: E⃗=2x i^\vec{E} = 2x\, \hat{i} and cube side 2 m2 \, \text{m}.

Find: Net flux through the cube.

Only the left and right faces contribute because the field has no yy or zz component. So directly use

Φ=ErightA−EleftA\Phi = E_{\text{right}} A - E_{\text{left}} A

with

A=4 m2,Eright=2(2)=4,Eleft=2(0)=0A = 4 \, \text{m}^2, \qquad E_{\text{right}} = 2(2) = 4, \qquad E_{\text{left}} = 2(0) = 0

Substituting,

Φ=4⋅4−0⋅4=16 N m2/C\Phi = 4 \cdot 4 - 0 \cdot 4 = 16 \, \text{N m}^2/\text{C}

Therefore, the correct option is C.

Common mistakes

  • Students often add contributions from all six faces. This is wrong because E⃗=2x i^\vec{E} = 2x\, \hat{i} has no component normal to the faces perpendicular to the yy and zz axes. Only the two faces normal to the xx-axis contribute.

  • A common mistake is taking the field to be constant everywhere as E=4 N/CE = 4 \, \text{N/C}. This is wrong because the field depends on position xx. Evaluate EE separately at x=0x = 0 and x=2x = 2.

  • Some students ignore the outward area direction and write both face contributions with the same sign. For a closed surface, flux depends on E⃗⋅dA⃗\vec{E} \cdot d\vec{A}. Use outward normals and then add the signed contributions correctly.

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