Question

There are two point charges, one at the vertex and the other at the face centre as shown on the cube. 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:

For Gauss law problems, remember the sharing rules: vertex $\rightarrow \tfrac{1}{8}$, edge $\rightarrow \tfrac{1}{4}$, face $\rightarrow \tfrac{1}{2}$ of the charge contributes to one cube.

Answer 1

The Correct Option is C

Concept: Gauss’s law states that 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 that charge contributes to the flux through the given surface.
Step 1: Contribution of charge at the vertex A point charge $2q$ is placed at a vertex of the cube. A vertex is shared by $8$ identical cubes. Hence, the effective enclosed charge due to this charge is: \[ q_{\text{vertex}} = \frac{2q}{8} = \frac{q}{4} \]
Step 2: Contribution of charge at the face centre A point charge $q$ is placed at the centre of a face of the cube. A face is shared by $2$ cubes. Hence, the effective enclosed charge due to this charge is: \[ q_{\text{face}} = \frac{q}{2} \]
Step 3: Total enclosed charge \[ q_{\text{enclosed}} = \frac{q}{4} + \frac{q}{2} = \frac{q}{4} + \frac{2q}{4} = \frac{3q}{4} \] Step 4: Electric flux through the cube Using Gauss’s law: \[ \Phi = \frac{q_{\text{enclosed}}}{\varepsilon_0} = \frac{3q}{4\varepsilon_0} \]
Step 5: Hence, the electric flux through the cube is: \[ \boxed{\dfrac{3q}{4\varepsilon_0}} \]