Question

Electric flux through cube of side 'a' enclosing charge 'q' is:

• A. $q/a^2$
• B. $q/\epsilon_0$
• C. Zero
• D. $2q/(a\epsilon_0)$

Hint:

Gauss's Law: Total flux = enclosed charge over epsilon-naught.

Answer 1

The Correct Option is B

Step 1: The law that applies.
Gauss's law tells us the total electric flux through any closed surface depends only on the charge sitting inside it: \[ \Phi = \frac{q_{enc}}{\varepsilon_0}. \]
Step 2: Why the shape does not matter.
The cube is just one example of a closed surface. Gauss's law does not care whether the surface is a cube, a sphere, or any odd shape; only the enclosed charge counts.
Step 3: Apply it to the cube.
The cube of side $a$ encloses a charge $q$, so \[ \Phi = \frac{q}{\varepsilon_0}. \]
Step 4: Notice what drops out.
The side length $a$ does not appear in the answer at all. Options that include $a$ are extra distractors.
Step 5: Conclusion.
The flux through the cube is $\frac{q}{\varepsilon_0}$. \[ \boxed{\dfrac{q}{\varepsilon_0}} \]