Q.A point charge causes an electric flux of $-1.0 \times 10^{3}\,\text{N m}^2/\text{C}$ to pass through a spherical Gaussian surface of $10.0\,\text{cm}$ radius centred on the charge.

Concept: Gauss's Law states that the total electric flux through a closed surface depends only on the enclosed charge, not on the size or shape of the surface.

Part (a): The flux through a Gaussian surface is given by

$\Phi_E = \frac{q_{\text{enc}}}{\epsilon_0}.$ …

Why Gauss's Law is the key

Gauss's Law tells us that the total electric flux $\Phi_E$ through any closed surface depends only on the net charge $q_{\text{enc}}$ enclosed by that surface:

where $\varepsilon_0 = 8.85 \times 10^{-12}\,\text{C}^2/(\text{N m}^2)$ is the permittivity of free space.

The beautiful insight here is that flux is a measure of how many field lines pierce the surface. A point charge emits a fixed number of field lines in all directions. No matter how large a sphere you draw around it, the same field lines pass through—they just spread out over a larger area, making the field weaker but keeping the total flux constant.

Part (a): Doubling the radius

1. The flux is independent of radius. Since the point charge remains at the center and the Gaussian surface still encloses the same charge, Gauss's Law guarantees the flux is unchanged.

2. Original flux: $\Phi_E = -1.0 \times 10^3\,\text{N m}^2/\text{C}$.

3. New flux when $r = 20.0\,\text{cm}$: Still $\Phi_E = -1.0 \times 10^3\,\text{N m}^2/\text{C}$.

The radius information ($10.0\,\text{cm}$ or $20.0\,\text{cm}$) is irrelevant to the flux calculation—it's a red herring. The flux depends only on the charge inside.

Part (b): Finding the point charge …

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 $\varepsilon_0$

3. Apply Gauss's Law

4. Calculate …

Common Mistakes with Gauss’s Law (and How to Avoid Them)

This problem is a classic test of Gauss’s Law:

The flux through a closed surface depends only on the charge enclosed — not on the size or shape of the surface.

✗ Mistake 1: Thinking flux changes when the radius changes

Why students do it:

They see the formula for electric field $E = \frac{kq}{r^2}$ and assume flux also depends on $r$.

Why it’s wrong:

Flux is $\Phi = \oint \vec{E} \cdot d\vec{A}$.

• If $r$ doubles, $E$ becomes $\frac{1}{4}$ as strong.
• But the surface area $A = 4\pi r^2$ becomes 4 times larger.
• The product $E \times A$ stays constant.

How to avoid:

Memorise the core idea: Gauss’s Law says flux depends only on the charge inside. The radius is irrelevant as long as the charge is still enclosed.

Correct answer for (a):

— unchanged.

✗ Mistake 2: Forgetting the sign of the charge

Why students do it:

They plug the magnitude of flux into $q = \varepsilon_0 \Phi$ and ignore the negative sign.

Why it’s wrong:

• Negative flux means the electric field points inward (toward the charge).
• That implies the charge is negative.

How to avoid:

Always carry the sign of $\Phi$ into the calculation. The sign of $q$ is the same as the sign of $\Phi$.

Correct calculation for (b):

✗ Mistake 3: Using the radius in the flux calculation

Why students do it:

They try to compute $E$ from $r$ and $q$, then integrate — a longer, unnecessary path.

Why it’s wrong:

• The problem gives you $\Phi$ directly.
• Using $r$ is extra work and introduces opportunities for error (unit conversion, squaring, etc.).

How to avoid:

When flux is given, use Gauss’s Law directly: $q = \varepsilon_0 \Phi$. The radius is a distractor.

✗ Mistake 4: Unit errors with $\varepsilon_0$

Why students do it: …