Question

A point charge + q is placed at the centre of a cube of side $l$. The electric flux emerging from the cube is

• A. $ \frac{ 6 q l^2 }{ \varepsilon_0} $
• B. $ \frac{ q }{ 6 l^2 \varepsilon_0} $
• C. zero
• D. $ \frac{ q }{ \varepsilon_0} $

Answer 1

The Correct Option is D

To solve this problem, we need to find the electric flux emerging from a cube when a point charge \( +q \) is placed at its center.

The concept used here is Gauss's Law, which relates the electric flux through a closed surface to the charge enclosed by that surface. Gauss's Law is mathematically expressed as:

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

where:

• \Phi is the electric flux through the closed surface,
• Q_{\text{enclosed}} is the total charge enclosed within the surface,
• \varepsilon_0 is the permittivity of free space.

In this case, the charge enclosed, Q_{\text{enclosed}}, is simply \( +q \) because the charge \( q \) is at the center of the cube. The cube is a symmetrical closed surface surrounding the charge.

Substituting the enclosed charge into Gauss's Law:

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

This result shows that the electric flux through the entire surface of the cube is \frac{q}{\varepsilon_0}.

Let's now evaluate the given options:

• \frac{6ql^2}{\varepsilon_0} - This suggests dependence on the side length l, which is incorrect because the flux depends only on the enclosed charge.
• \frac{q}{6l^2\varepsilon_0} - This is also incorrect because it introduces unwarranted factors related to the side length.
• Zero - This would be true if the net charge enclosed was zero, which is not the case here.
• \frac{q}{\varepsilon_0} - This is the correct answer, consistent with Gauss's Law for the enclosed charge.

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