If the electric flux entering and leaving an enclosed surface respectively is phi_1 and phi_2, the e

Q: Why do we subtract the entering flux from the leaving flux?

By convention, the area vector for a closed surface points outward. Flux leaving the surface is in the same direction as the area vector (positive), while flux entering is in the opposite direction (negative). The net flux is their algebraic sum: $(+\phi_2) + (-\phi_1) = \phi_2 - \phi_1$.

Q: What if the entering flux is greater than the leaving flux?

If $\phi_1 > \phi_2$, the net flux $\phi_2 - \phi_1$ will be negative. According to Gauss's Law, this simply means the net enclosed charge $q$ is negative!

Q: Does the shape of the enclosed surface matter?

Not at all! Gauss's Law is universally true for any closed surface, regardless of its shape or size, as long as it completely encloses the charge.

Q: What exactly is $\varepsilon_0$?

$\varepsilon_0$ (epsilon naught) is the vacuum permittivity. It is a physical constant that represents the capability of a vacuum to permit electric field lines. Its value is approximately $8.85 \times 10^{-12} \text{ F/m}$.

Q: What if there are charges outside the surface?

Charges outside the surface do create electric field lines that pass through the surface, but every line that enters must also leave. Thus, their contribution to the *net* flux is exactly zero. Only the enclosed charge determines the net flux.

Question

If the electric flux entering and leaving an enclosed surface respectively is $\phi_1$ and $\phi_2$, the electric charge inside the surface will be

Options:

(A) $(\phi_2 - \phi_1)\varepsilon_0$

(B) $\frac{(\phi_1 + \phi_2)}{\varepsilon_0}$

(C) $\frac{(\phi_2 - \phi_1)}{\varepsilon_0}$

(D) $(\phi_1 + \phi_2)\varepsilon_0$

Sigma Percentile · Accepted Answer

### Visualizing the Flux

Imagine you are observing a mysterious, invisible 3D boundary in space—a closed surface. According to the problem, there is a flurry of electrical activity happening around it. Electric field lines are piercing into this surface, and some are emerging out of it.

In physics, we quantify this "flow" of electric field lines using a concept called **Electric Flux** ($\phi$). The problem states that the flux entering the surface is $\phi_1$, and the flux leaving the surface is $\phi_2$.

### The Sign Convention of Flux

Before we jump into any equations, we need to establish a strict rule. By standard convention in electromagnetism, the area vector of a closed surface always points **outward**, normal to the surface.

Because flux is the dot product of the electric field and the area vector ($\phi = \vec{E} \cdot d\vec{A}$):
- Field lines **leaving** the surface point in the same direction as the area vector, making the outward flux **positive**.
- Field lines **entering** the surface point opposite to the area vector, making the inward flux **negative**.

Therefore, the **net outward flux** ($\phi_{	ext{net}}$) through our closed surface is the total leaving flux minus the total entering flux:
$$ \phi_{	ext{net}} = \phi_2 - \phi_1 $$

### The Master Equation: Gauss's Law

Now, why do we care about the net flux? Because of one of the most elegant and powerful laws in physics: **Gauss's Law**.

Gauss's Law tells us that the net electric flux through any closed surface is directly proportional to the total electric charge enclosed within that surface. Mathematically, it is expressed as:
$$ \phi_{	ext{net}} = \frac{q_{	ext{enclosed}}}{\varepsilon_0} $$
where $\varepsilon_0$ is the permittivity of free space, a fundamental constant of nature.

### Final Calculation

We already determined our net flux, so let's substitute it into Gauss's Law:
$$ \phi_2 - \phi_1 = \frac{q}{\varepsilon_0} $$

To find the enclosed charge $q$, we simply rearrange the equation by multiplying both sides by $\varepsilon_0$:
$$ q = (\phi_2 - \phi_1)\varepsilon_0 $$

And there we have it! The total charge hiding inside the surface is simply the difference between the leaving and entering flux, scaled by the permittivity of free space.

Animated Solution for Physics - Electrostatics: If the electric flux entering and leaving an enclosed surface respectively is $ϕ_{1}$ and $ϕ_{2}$ , the electric charge inside the surface will be

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Visualized Solution

$V i s u a l i z in g t h e F l ux$

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The Sigma Insight: Electric Field Lines, Flux and Gauss's Law

Visualizing the Flux

The Sign Convention of Flux

The Master Equation

Final Calculation

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