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

Concept:
According to Gauss’s law, the total electric flux through a closed surface is: \[ \Phi = \frac{q_{\text{enclosed}}}{\varepsilon_0} \] If a charge lies on a boundary (vertex, edge, or face), only a fraction of its flux contributes to the given surface, depending on symmetry.
Step 1: Contribution of the charge at the vertex. A charge \(2q\) is placed at a vertex of the cube. A vertex is shared by 8 identical cubes. Hence, effective charge enclosed by one cube: \[ q_1 = \frac{2q}{8} = \frac{q}{4} \] Corresponding flux: \[ \Phi_1 = \frac{q}{4\varepsilon_0} \]
Step 2: Contribution of the charge at the face. A charge \(q\) is placed at the center of a face. A face is shared by 2 identical cubes. Hence, effective charge enclosed: \[ q_2 = \frac{q}{2} \] Corresponding flux: \[ \Phi_2 = \frac{q}{2\varepsilon_0} \]
Step 3: Total electric flux through the cube. \[ \Phi = \Phi_1 + \Phi_2 \] \[ = \frac{q}{4\varepsilon_0} + \frac{q}{2\varepsilon_0} \] \[ = \frac{3q}{4\varepsilon_0} \]