Q.Careful measurement of the electric field at the surface of a black box indicates that the net outward flux through the surface of the box is $8.0 \times 10^{3}\,\text{N m}^2/\text{C}$.

The key idea is Gauss's Law.

(a) The net charge inside the box:

1. Gauss's Law states that the net electric flux ($\Phi_E$) through any closed surface is directly proportional to the net charge ($Q_{enc}$) enclosed within that surface:

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

2. Rearranging the formula to solve for the net charge: $Q_{enc} = \Phi_E \epsilon_0$.
3. Substitute the given net outward flux $\Phi_E = 8.0 \times 10^{3}\,\text{N m}^2/\text{C}$ and the permittivity of free space $\epsilon_0 = 8.854 \times 10^{-12}\,\text{C}^2/(\text{N m}^2)$. …

At its heart, this problem is a direct application of Gauss's Law, a fundamental principle in electromagnetism. Gauss's Law provides a powerful way to relate the electric field on a closed surface to the total charge enclosed within that surface.

First, let's understand what "electric flux" means. Imagine electric field lines passing through a surface. Electric flux is a measure of the number of electric field lines piercing through that surface. A larger flux means more field lines are passing through. For a closed surface (like our "black box"), an outward flux means more field lines are leaving the surface than entering, implying a net positive charge inside. An inward flux means more field lines are entering, implying a net negative charge inside. If the net flux is zero, it means an equal number of field lines enter and leave, or no field lines at all, which points to a zero net charge inside.

Gauss's Law formalizes this idea:

This law is incredibly useful because it doesn't depend on the shape of the surface or the distribution of charges outside the surface; it only cares about the net charge inside.

Part (a): What is the net charge inside the box?

1. Identify the given information:

   We are given the net outward electric flux through the surface of the box:

   $\Phi_E = 8.0 \times 10^{3}\,\text{N m}^2/\text{C}$.

2. Recall Gauss's Law:

   The relationship between net flux and enclosed charge is given by:

3. Rearrange the formula to solve for the net enclosed charge ($q_{enc}$): Multiplying both sides by $\epsilon_0$, we get:

4. Substitute the known values and calculate: We use the given flux and the value of $\epsilon_0 = 8.854 \times 10^{-12}\,\text{C}^2/(\text{N m}^2)$.

5. Consider significant figures: The given flux has two significant figures ($8.0 \times 10^{3}$). Therefore, our answer should also be rounded to two significant figures.

Since the flux is outward and positive, the net charge inside must be positive.

Part (b): If the net outward flux through the surface of the box were zero, could you conclude that there were no charges inside the box? Why or Why not?

1. Apply Gauss's Law for zero net flux: If the net outward flux through the surface of the box were zero, then $\Phi_E = 0$. Using Gauss's Law:

This implies that:

2. Interpret "net charge is zero": …

Solution Method: Gauss's Law

Method Name

Gauss's Law — relates the net electric flux through a closed surface to the net charge enclosed.

Key Formula

Gauss's Law states:

Where:

• $\Phi_E$ = net electric flux (in $\text{N m}^2/\text{C}$)
• $Q_{\text{enc}}$ = net charge inside the closed surface (in C)
• $\varepsilon_0 = 8.854 \times 10^{-12} \, \text{C}^2/\text{N m}^2$ (permittivity of free space)

Steps for Part (a)

Step 1: Identify the given flux

Step 2: Apply Gauss's Law

Step 3: Substitute values

Step 4: Calculate

Answer (a):

Net charge inside the box = $7.1 \times 10^{-8} \, \text{C}$ (rounded to two significant figures)

Steps for Part (b)

Step 1: Recall Gauss's Law again

Step 2: If $\Phi_E = 0$, then …

Here are the common mistakes students make on this Gauss Law problem and how to avoid each.

Mistake 1: Forgetting the correct formula for Gauss Law

The error:

Students write $\phi = \frac{q}{\epsilon_0}$ but then plug in numbers without checking units or use the wrong value of $\epsilon_0$.

How to avoid:

Always write the exact form of Gauss Law:

Where:

• $\phi$ = net electric flux (in $\text{N m}^2/\text{C}$)
• $q_{\text{enc}}$ = net charge enclosed (in C)
• $\epsilon_0 = 8.854 \times 10^{-12} \, \text{C}^2/\text{N m}^2$

For part (a):

Rearrange:

Substitute:

Key: Always write the formula first, then substitute — never skip steps.

Mistake 2: Confusing “net flux” with “flux due to a single charge”

The error:

Students think that if flux is outward, the charge must be positive — which is correct — but they forget that net flux is the sum over the entire closed surface.

How to avoid:

Remember: Gauss Law says net outward flux = total enclosed charge divided by $\epsilon_0$. It does not tell you where inside the box the charge is located, only the net amount.

Mistake 3: Answering part (b) incorrectly — “Yes, no charges inside”

The error:

Students say: “If net flux is zero, then there are no charges inside the box.”

Why this is wrong:

A zero net flux means the net enclosed charge is zero. But you could have:

• No charges at all, or
• Equal amounts of positive and negative charge inside (net zero).

How to avoid:

State clearly:

If $\phi = 0$, then $q_{\text{enc}} = 0$. This does not imply there are no charges — only that the algebraic sum of charges is zero.

Correct answer for (b):

No, you cannot conclude there are no charges. There could be equal positive and negative charges inside, making the net charge zero.

Mistake 4: Ignoring the “closed surface” condition

The error: …