Question

There are two point charges, one at the vertex and the other at the face of a cube as shown. Find the electric flux through the cube.

• A. \( \dfrac{3q}{\varepsilon_0} \)
• B. \( \dfrac{q}{\varepsilon_0} \)
• C. \( \dfrac{3q}{4\varepsilon_0} \)
• D. \( \dfrac{5q}{\varepsilon_0} \)

Hint:

A charge at a vertex contributes \( \tfrac{1}{8} \), on an edge \( \tfrac{1}{4} \), and on a face \( \tfrac{1}{2} \) of its total flux to a cube.

Answer 1

The Correct Option is C

Alternative Method (Imaginary Gaussian Surface & Symmetry Method):

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According to Gauss’s law, the total electric flux through any closed surface depends only on the net charge enclosed by that surface:

Φ = q_enclosed / ε₀

When charges are placed on boundaries, we imagine extending the geometry symmetrically to form a complete closed Gaussian surface.

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Step 1: Charge placed at the vertex

A charge 2q is located at a vertex of the cube. To apply Gauss’s law properly, imagine 7 more identical cubes attached around the vertex, forming a larger cube with the charge at its center.

Total flux through the larger cube:

Φ = 2q / ε₀

Since this larger cube consists of 8 identical small cubes, the flux through one cube is:

Φ₁ = (2q / ε₀) ÷ 8 = q / (4ε₀)

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Step 2: Charge placed at the center of a face

A charge q is placed at the center of one face of the cube. Now imagine attaching one identical cube to this face, so that the charge lies at the center of the combined rectangular solid.

Total flux through the combined closed surface:

Φ = q / ε₀

Since the solid consists of 2 identical cubes, the flux through the given cube is:

Φ₂ = (q / ε₀) ÷ 2 = q / (2ε₀)

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Step 3: Total electric flux through the cube

By the principle of superposition, total flux is the sum of individual contributions:

Φ = Φ₁ + Φ₂

Φ = q/(4ε₀) + q/(2ε₀)

Φ = 3q / (4ε₀)

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Final Answer:

Total electric flux through the cube = 3q / (4ε₀)