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Electrostatics question

2024 · 4 Apr · Shift 2 · Q75
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Electrostatics question

2024 · 4 Apr · Shift 2 · Q75

JEE MainPhysicsElectrostaticsMCQ+4 / −1
A charge qqq is placed at the center of one of the surface of a cube. The flux linked with the cube is:
  1. A
    q2ϵ0\frac{q}{2 \epsilon_0}2ϵ0​q​
  2. B
    Zero
  3. C
    q4ϵ0\frac{q}{4 \epsilon_0}4ϵ0​q​
  4. D
    q8ϵ0\frac{q}{8 \epsilon_0}8ϵ0​q​
View written solutionFree

Correct answer: A

  1. Use Gauss's law

    Gauss's law states: Φ=qenclosedε0\Phi = \frac{q_{\text{enclosed}}}{\varepsilon_0}Φ=ε0​qenclosed​​ where Φ\PhiΦ is the total electric flux through a closed surface.

  2. Understand the geometry

    The charge qqq is placed at the center of one face of the cube, i.e. on the boundary of the cube.

    To handle such cases, imagine placing an identical cube adjacent to the given cube, sharing that face. Then the charge lies exactly at the common face and becomes the center of the resulting rectangular box made of 2 identical cubes.

  3. Apply symmetry

    The combined closed surface (the rectangular box of 2 cubes) encloses the charge qqq at its center.

    Hence, total flux through the combined surface is: Φtotal=qε0\Phi_{\text{total}} = \frac{q}{\varepsilon_0}Φtotal​=ε0​q​

    By symmetry, the flux is equally divided between the two identical cubes.

    Therefore, flux through one cube is: Φcube=12⋅qε0=q2ε0\Phi_{\text{cube}} = \frac{1}{2}\cdot \frac{q}{\varepsilon_0} = \frac{q}{2\varepsilon_0}Φcube​=21​⋅ε0​q​=2ε0​q​

  4. Check options

    • A: q2ε0\frac{q}{2\varepsilon_0}2ε0​q​ ✅
    • B: 000 ❌
    • C: q4ε0\frac{q}{4\varepsilon_0}4ε0​q​ ❌
    • D: q8ε0\frac{q}{8\varepsilon_0}8ε0​q​ ❌
  5. Final answer

    q2ε0\boxed{\frac{q}{2\varepsilon_0}}2ε0​q​​

So the correct option is A.

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