Q.A point charge of $2.0\,\mu\text{C}$ is at the centre of a cubic Gaussian surface $9.0\,\text{cm}$ on edge. What is the net electric flux through the surface?

The problem asks for the net electric flux through a closed surface due to an enclosed point charge. This is a direct application of Gauss's Law.

The given point charge is $Q_{enc} = 2.0\,\mu\text{C} = 2.0 \times 10^{-6}\,\text{C}$. The dimensions of the cubic Gaussian surface ($9.0\,\text{cm}$ on edge) are irrelevant for calculating the total flux, as Gauss's Law depends only on the magnitude of the enclosed charge, not the shape or size of the surface, provided the charge is enclosed. …

Electric flux is a measure of the "flow" of the electric field through a given surface. Conceptually, it represents the number of electric field lines passing through that surface. For a closed surface, a positive flux indicates field lines leaving the surface (due to an enclosed positive charge), while a negative flux indicates field lines entering the surface (due to an enclosed negative charge).

Calculating electric flux directly using the definition $\Phi_E = \int \vec{E} \cdot d\vec{A}$ can be complex, especially for arbitrary surfaces or charge distributions. This is where Gauss's Law becomes incredibly powerful.

Gauss's Law provides a fundamental relationship between the electric flux through any closed surface and the net electric charge enclosed within that surface. It simplifies flux calculations significantly because it tells us that the total flux depends only on the enclosed charge, not on the specific shape, size, or location of the charges within the surface, nor on any charges outside the surface. This means that for a point charge, whether the Gaussian surface is a sphere, a cube, or an irregular blob, as long as it encloses the charge, the total flux will be the same.

Let's apply this principle to the given problem.

1. Identify the relevant physical law.

   The problem asks for the net electric flux through a closed surface due to an enclosed charge. This immediately points to Gauss's Law, which is the most direct way to calculate total electric flux through a closed surface.

   Gauss's Law states that the net electric flux ($\Phi_E$) through any closed surface is equal to the net electric charge ($q_{enc}$) enclosed within that surface divided by the permittivity of free space ($\epsilon_0$):

   $\Phi_E = \frac{q_{enc}}{\epsilon_0}$

2. Identify the given quantities.

   We are given:

   • The point charge $q = 2.0\,\mu\text{C}$.
   • The side length of the cubic Gaussian surface $L = 9.0\,\text{cm}$.

   We also need the value of the permittivity of free space:

   • $\epsilon_0 = 8.854 \times 10^{-12}\,\text{C}^2/(\text{N}\cdot\text{m}^2)$. …

Method: Gauss's Law (Direct Application)

Gauss's Law states that the net electric flux through any closed surface equals the charge enclosed divided by the permittivity of free space:

Steps

1. Identify the enclosed charge

   The point charge $Q = 2.0\ \mu\text{C} = 2.0 \times 10^{-6}\ \text{C}$ is at the centre of the cube.

   Therefore, $Q_{\text{enc}} = 2.0 \times 10^{-6}\ \text{C}$.

2. Recall the value of $\varepsilon_0$

   $\varepsilon_0 = 8.854 \times 10^{-12}\ \text{C}^2/\text{N·m}^2$ (use $8.85 \times 10^{-12}$ in calculations).

3. Apply Gauss's Law

4. Calculate …

Here are the most common mistakes students make when solving this exact Gauss’s law problem, along with how to avoid each.

1. Mistake: Confusing the shape of the Gaussian surface

The error:

Students think the cube’s shape matters — they try to compute flux through each face separately using $E \cdot A$, or worry about the edge length $9.0\,\text{cm}$.

Why it’s wrong:

Gauss’s law says the net flux through any closed surface depends only on the enclosed charge, not on the shape or size of the surface.

How to avoid:

• Immediately recall:

• The edge length is a distractor — ignore it for net flux.

2. Mistake: Forgetting to convert microcoulombs to coulombs

The error:

Plugging in $2.0$ instead of $2.0 \times 10^{-6}$.

Why it’s wrong:

The SI unit of charge is the coulomb (C). $\mu\text{C}$ means $10^{-6}\,\text{C}$.

How to avoid:

• Always write:

• Double-check unit prefixes before substituting into $\Phi = q/\varepsilon_0$.

3. Mistake: Using the wrong value of $\varepsilon_0$

The error:

Using $\varepsilon_0 = 8.85 \times 10^{-12}$ but forgetting units, or using $9 \times 10^9$ (which is $1/(4\pi\varepsilon_0)$) incorrectly.

Why it’s wrong:

Gauss’s law uses $\varepsilon_0$, not Coulomb’s constant $k$.

How to avoid:

• Memorise:

• If you accidentally use $k$, you’ll get flux in wrong units.

4. Mistake: Thinking flux depends on the position of the charge inside

The error:

Worrying that because the charge is “at the centre”, the flux is different than if it were off-centre.

Why it’s wrong:

As long as the charge is inside the closed surface, the net flux is the same — Gauss’s law is independent of location inside.

How to avoid:

• Remember:

only cares about whether the charge is inside, not where.

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