Question

What is the flux through a cube of side a if a point charge of q is at one of its corner?

• A. $ \frac{2q}{\varepsilon _0} $
• B. $ \frac{q}{8\varepsilon _0} $
• C. $ \frac{q}{\varepsilon _0} $
• D. $ \frac{q}{2\varepsilon _0}6a^2 $

Answer 1

The Correct Option is B

To determine the electric flux through a cube when a point charge is placed at one of its corners, we can use Gauss's Law. Gauss's Law states that the total electric flux (\Phi) through a closed surface is equal to the charge enclosed (q_{\text{enclosed}}) divided by the permittivity of the free space (\varepsilon_0):

\Phi = \frac{q_{\text{enclosed}}}{\varepsilon_0}

In this scenario, the point charge q is located at one corner of a cube. Because a cube has eight corners and the charge is at one corner, it effectively shares its influence with eight such hypothetical cubes. Thus, only \frac{1}{8} of the charge q is enclosed by one cube.

Therefore, the flux through the cube is given by:

\Phi = \frac{q}{8\varepsilon_0}

Let's clarify why the other options are incorrect:

• \frac{2q}{\varepsilon_0} is incorrect because it does not consider the distribution of the charge among the eight cubes.
• \frac{q}{\varepsilon_0} assumes the whole charge is inside one cube without sharing.
• \frac{q}{2\varepsilon_0}6a^2 suggests a surface integral based on an assumption that is not valid without considering other orientation factors of charging inside the cube.

Hence, the correct answer is \frac{q}{8\varepsilon_0}.