Q234 / 240 · KCET 2026 · Physics
What will be the total electric flux through the faces of the cube as given in the figure with side of length '$a$' if a charge Q is placed at B midpoint of an edge of the cube (see figure)?
- A$\dfrac{Q}{8\varepsilon_0}$
- B$\dfrac{Q}{3\varepsilon_0}$
- Correct answerC$\dfrac{Q}{4\varepsilon_0}$
- D$\dfrac{Q}{2\varepsilon_0}$
Solution
According to Gauss's Law, the total electric flux through a closed surface is given by $\Phi = \dfrac{q_{\text{enclosed}}}{\varepsilon_0}$.
The charge $Q$ is placed at the midpoint of an edge of the cube. This charge is not completely enclosed by the single cube. To completely enclose the charge symmetrically, we can imagine placing three additional identical cubes such that all four cubes share this common edge.
The total charge enclosed by this larger Gaussian surface (made of 4 cubes) is $Q$. Therefore, the total electric flux through the combination of these 4 cubes is $\dfrac{Q}{\varepsilon_0}$.
By symmetry, the flux is shared equally among the 4 cubes. Thus, the total electric flux passing through the faces of the given single cube is:
$\Phi_{\text{cube}} = \dfrac{1}{4} \times \dfrac{Q}{\varepsilon_0} = \dfrac{Q}{4\varepsilon_0}$
Answer: $\dfrac{Q}{4\varepsilon_0}$
The charge $Q$ is placed at the midpoint of an edge of the cube. This charge is not completely enclosed by the single cube. To completely enclose the charge symmetrically, we can imagine placing three additional identical cubes such that all four cubes share this common edge.
The total charge enclosed by this larger Gaussian surface (made of 4 cubes) is $Q$. Therefore, the total electric flux through the combination of these 4 cubes is $\dfrac{Q}{\varepsilon_0}$.
By symmetry, the flux is shared equally among the 4 cubes. Thus, the total electric flux passing through the faces of the given single cube is:
$\Phi_{\text{cube}} = \dfrac{1}{4} \times \dfrac{Q}{\varepsilon_0} = \dfrac{Q}{4\varepsilon_0}$
Answer: $\dfrac{Q}{4\varepsilon_0}$
Hence, the correct answer is option: 'C'.