Gausss law

Electric Flux and Gauss’s Law

Gauss’s law connects the total electric flux through a closed surface to the charge trapped inside that surface.

Key Equations

Gauss’s law:

$$\oint \vec{E}\cdot d\vec{A}=\frac{{Q}_{\text{enc}}}{{\epsilon }_{0}}$$

Variables:

• $\oint \vec{E}\cdot d\vec{A}$ = total electric flux through the closed surface (N·m²/C)
• $\vec{E}$ = electric field at each point on the surface (N/C)
• $d\vec{A}$ = infinitesimal area element, directed outward from the surface (m²)
• ${Q}_{\text{enc}}$ = total charge enclosed by the surface (C)
• ${\epsilon }_{0}$ = permittivity of free space, $8.85\times {10}^{-12}$ C²/(N·m²)

SI unit: N·m²/C (for the flux side); C (for the charge side)

ProportionalityThe total flux is directly proportional to the enclosed charge. Double the enclosed charge, and the flux doubles.

Calculus form: The circle on the integral sign ($\oint $) means the integration runs over a closed surface — every patch of area on that surface is included.

Reference sheet status: On the reference sheet.

The symbol $\oint $ denotes a surface integral over a closed surface. “Closed” means the surface has no holes — it completely surrounds a volume, like the skin of a balloon. The quantity $d\vec{A}$ is an infinitesimal patch of that surface, and its direction is the outward normal [pointing perpendicular to the surface, away from the enclosed volume]. So $\vec{E}\cdot d\vec{A}$ picks out the component of the electric field that pierces through each patch.

Gauss’s law says: add up the field-through-surface contributions from every tiny patch, and the total equals ${Q}_{\text{enc}}/{\epsilon }_{0}$. Charges outside the surface contribute zero net flux, because their field lines enter one side and exit the other. Only the charge inside determines the total flux.

This is Maxwell’s first equation — one of four equations that fully describe electromagnetism.

MisconceptionStudents often think a larger Gaussian surface produces more flux. The total electric flux through a Gaussian surface is independent of the size of the surface if the enclosed charge remains constant. A bigger surface has weaker $E$ at each patch but more area, and these effects cancel exactly.

Exam TipIf a question changes the surface radius but not the enclosed charge, the flux stays the same.

Examiner InsightFRQs frequently ask: “What is the flux through the surface?” Many students set up a full integral when the answer is simply ${Q}_{\text{enc}}/{\epsilon }_{0}$. Gauss’s law gives the total flux directly from the enclosed charge.

Exam TipTo find total flux, identify ${Q}_{\text{enc}}$ first — the integral may not be necessary.

BoundaryThe AP exam only expects quantitative application of Gauss’s law to point charges and distributions with spherical, cylindrical, or planar symmetry.

Know these cases for the AP exam.

Gaussian Surfaces and Symmetry

Choosing the right Gaussian surface is the key skill that turns a difficult integral into a simple product.

A Gaussian surface is a three-dimensional, closed surface — an imaginary boundary you draw around the charge distribution. It is not a physical object. You choose its shape to exploit the symmetry of the charge distribution so that the dot product $\vec{E}\cdot d\vec{A}$ simplifies.

Gaussian surfaces are typically constructed such that the field is either perpendicular or parallel to different regions of the surface:

• When $\vec{E}$ is perpendicular to a surface patch ($\vec{E}∥d\vec{A}$), the dot product becomes $E dA$.
• When $\vec{E}$ is parallel to a surface patch ($\vec{E}⟂d\vec{A}$), the dot product is zero.

These two cases are what make the integral tractable.

In every case, the strategy is the same. Pick a surface where $E$ is constant over the sections that contribute flux. Then $\oint \vec{E}\cdot d\vec{A}$ collapses to $E$ times the area of those sections.

Finding Enclosed Charge from Charge Density

When charge is spread out rather than concentrated at a point, the enclosed charge ${Q}_{\text{enc}}$ must be found by integrating the charge density over the region inside the Gaussian surface.

Key Equations

Charge from linear density:

$$Q=\int \lambda dℓ$$

Charge from surface density:

$$Q=\int \sigma dA$$

Charge from volume density:

$$Q=\int \rho dV$$

Variables:

• $\lambda $ = linear charge density (C/m)
• $\sigma $ = surface charge density (C/m²)
• $\rho $ = volume charge density (C/m³)
• $dℓ$ = infinitesimal length element (m)
• $dA$ = infinitesimal area element (m²)
• $dV$ = infinitesimal volume element (m³)

Reference sheet status: On the reference sheet (general forms).

If the charge density is uniform (constant), the integral simplifies to a product: $Q=\lambda L$, $Q=\sigma A$, or $Q=\rho V$. If the density varies with position, you must set up and evaluate the integral. Express the density entirely in terms of the integration variable, determine limits from the geometry, then integrate.

For a sphere of radius $R$ with volume charge density $\rho (r)$ that depends on the radial distance $r$, the volume element in spherical coordinates is $dV=4\pi {r}^{2} dr$. So:

$${Q}_{\text{enc}}=\int_{0}^{R}\rho (r) 4\pi {r}^{2} dr$$

For a cylinder of radius $R$ and length $L$ with $\rho (r)$, the volume element is $dV=2\pi rL dr$. So:

$${Q}_{\text{enc}}=\int_{0}^{R}\rho (r) 2\pi rL dr$$

MisconceptionStudents often use the total charge of the entire object when only a portion lies inside the Gaussian surface. If your Gaussian surface of radius $r$ sits inside a charged sphere of radius $R$, integrate only from $0$ to $r$, not $0$ to $R$.

Exam TipAlways match the upper limit of integration to the radius of the Gaussian surface, not the physical object.

Worked Example: Enclosed Charge from Non-Uniform Density

Scenario: A solid insulating sphere of radius $R=0.10$ m carries a volume charge density $\rho (r)=\beta r$, where $\beta =5.0\times {10}^{-5}$ C/m⁴. Determine the total charge enclosed by a Gaussian sphere of radius $r=0.060$ m.

Step-by-step solution:

Equation used

$${Q}_{\text{enc}}=\int_{0}^{r}\rho ({r}^{′}) 4\pi {r}^{′}{}^{2} d{r}^{′}$$

Given

$$\rho ({r}^{′})=\beta {r}^{′}$$

$$\beta =5.0\times {10}^{-5}{\text{ C/m}}^{4}$$

$$r=0.060\text{ m}$$

Working — substitute $\rho ({r}^{′})=\beta {r}^{′}$

$${Q}_{\text{enc}}=\int_{0}^{r}\beta {r}^{′}\cdot 4\pi {r}^{′}{}^{2} d{r}^{′}=4\pi \beta \int_{0}^{r}{r}^{′}{}^{3} d{r}^{′}$$

Evaluate the integral using the power rule:

$$\int_{0}^{r}{r}^{′}{}^{3} d{r}^{′}=\frac{{r}^{4}}{4}$$

$${Q}_{\text{enc}}=4\pi \beta \cdot \frac{{r}^{4}}{4}=\pi \beta {r}^{4}$$

Substitute values:

$${Q}_{\text{enc}}=\pi (5.0\times {10}^{-5})(0.060{)}^{4}$$

$$(0.060{)}^{4}=1.296\times {10}^{-5}$$

$${Q}_{\text{enc}}=\pi (5.0\times {10}^{-5})(1.296\times {10}^{-5})$$

$${Q}_{\text{enc}}=2.04\times {10}^{-9}\text{ C}$$

Answer

$${Q}_{\text{enc}}\approx 2.04\text{ nC}$$

Interpretation: Only the charge inside the Gaussian sphere at $r=0.060$ m contributes. The remaining charge between $r$ and $R$ has no effect on the flux through that surface.

Applying Gauss’s Law: Spherical Symmetry

The most common AP application of Gauss’s law uses spherical symmetry — a charge distribution where the electric field depends only on the radial distance $r$ from the center and points radially outward (or inward).

Derivation: Electric Field Outside a Uniformly Charged Sphere

Starting point: Gauss’s law.

$$\oint \vec{E}\cdot d\vec{A}=\frac{{Q}_{\text{enc}}}{{\epsilon }_{0}}$$

Step 1 — Choose the Gaussian surface

Draw a concentric sphere of radius $r>R$, where $R$ is the sphere’s radius. The charge distribution is spherically symmetric, so a spherical Gaussian surface matches that symmetry.

Step 2 — Apply the symmetry argument

By spherical symmetry, $\vec{E}$ is radial and constant in magnitude at every point on the Gaussian surface. At every patch, $\vec{E}$ is parallel to $d\vec{A}$ (both point radially outward). So $\vec{E}\cdot d\vec{A}=E dA$.

Step 3 — Simplify the integral

Because $E$ is constant over the surface:

$$\oint E dA=E\oint dA=E(4\pi {r}^{2})$$

Step 4 — Calculate the enclosed charge

All of the sphere’s charge $Q$ lies inside the Gaussian surface:

$${Q}_{\text{enc}}=Q$$

Step 5 — Solve for $E$

$$E(4\pi {r}^{2})=\frac{Q}{{\epsilon }_{0}}$$

Result

$$E=\frac{Q}{4\pi {\epsilon }_{0}{r}^{2}}=\frac{kQ}{{r}^{2}}$$

Outside a uniformly charged sphere, the electric field is identical to that of a point charge $Q$ at the center. The field is inversely proportional to ${r}^{2}$: doubling the distance reduces the field to one quarter.

For a Gaussian surface inside a uniformly charged insulating sphere ($r<R$), only the charge within radius $r$ is enclosed. For uniform volume charge density $\rho $:

$${Q}_{\text{enc}}=\rho \cdot \frac{4}{3}\pi {r}^{3}=Q\frac{{r}^{3}}{{R}^{3}}$$

Applying the same Gauss’s law steps:

$$E(4\pi {r}^{2})=\frac{Q{r}^{3}}{{\epsilon }_{0}{R}^{3}}$$

$$E=\frac{Qr}{4\pi {\epsilon }_{0}{R}^{3}}$$

Inside the sphere, $E$ is directly proportional to $r$. At the center, $E=0$. At the surface ($r=R$), the inside and outside expressions agree.

For a conducting sphere, all excess charge resides on the surface. Inside the conductor ($r<R$), ${Q}_{\text{enc}}=0$, so $E=0$. Outside, the field is $E=kQ/{r}^{2}$, the same as a point charge.

Worked Example: Electric Field Inside a Uniformly Charged Insulating Sphere

Scenario: A solid insulating sphere of radius $R=0.20$ m carries a total charge $Q=8.0\times {10}^{-6}$ C distributed uniformly. Determine the electric field at $r=0.10$ m from the center.

Step-by-step solution:

Equation used

$$E=\frac{Qr}{4\pi {\epsilon }_{0}{R}^{3}}$$

Given

$$Q=8.0\times {10}^{-6}\text{ C}$$

$$r=0.10\text{ m}$$

$$R=0.20\text{ m}$$

$$\frac{1}{4\pi {\epsilon }_{0}}=k=8.99\times {10}^{9}{\text{ N}\cdot\text{m}}^{2}/{\text{C}}^{2}$$

Working

$$E=\frac{kQr}{{R}^{3}}=\frac{(8.99\times {10}^{9})(8.0\times {10}^{-6})(0.10)}{(0.20{)}^{3}}$$

$$(0.20{)}^{3}=8.0\times {10}^{-3}$$

$$E=\frac{(8.99\times {10}^{9})(8.0\times {10}^{-7})}{8.0\times {10}^{-3}}$$

$$E=\frac{7.192\times {10}^{3}}{8.0\times {10}^{-3}}$$

Answer

$$E\approx 8.99\times {10}^{5}\text{ N/C}$$

Interpretation: At half the radius, the field is half the surface value. This confirms the linear proportionality inside the sphere.

Applying Gauss’s Law: Cylindrical Symmetry

For a long, straight charge distribution, cylindrical symmetry means the electric field depends only on the perpendicular distance $r$ from the axis and points radially outward (or inward) from that axis.

Derivation: Electric Field from an Infinite Line of Charge

Starting point: Gauss’s law.

$$\oint \vec{E}\cdot d\vec{A}=\frac{{Q}_{\text{enc}}}{{\epsilon }_{0}}$$

Step 1 — Choose the Gaussian surface

Draw a coaxial cylinder of radius $r$ and length $L$ centered on the line charge. This shape matches the cylindrical symmetry.

Step 2 — Apply the symmetry argument

On the curved surface: $\vec{E}$ is radial and constant in magnitude. $\vec{E}$ is parallel to $d\vec{A}$, so $\vec{E}\cdot d\vec{A}=E dA$. On each flat cap: $\vec{E}$ is perpendicular to $d\vec{A}$ (the cap normal is along the axis, but $\vec{E}$ is radial). So $\vec{E}\cdot d\vec{A}=0$ on the caps.

Step 3 — Simplify the integral

Only the curved surface contributes:

$$\oint \vec{E}\cdot d\vec{A}=E(2\pi rL)$$

Step 4 — Calculate the enclosed charge

For uniform linear charge density $\lambda $:

$${Q}_{\text{enc}}=\lambda L$$

Step 5 — Solve for $E$

$$E(2\pi rL)=\frac{\lambda L}{{\epsilon }_{0}}$$

Result

$$E=\frac{\lambda }{2\pi {\epsilon }_{0}r}$$

The electric field from an infinite line of charge is inversely proportional to $r$. Doubling the distance halves the field.

Examiner InsightA common FRQ asks for the field inside and outside a cylindrical conductor or insulating cylinder. Inside a solid insulating cylinder with uniform $\rho $: ${Q}_{\text{enc}}=\rho \pi {r}^{2}L$, giving $E=\rho r/(2{\epsilon }_{0})$. Inside a conducting cylinder: $E=0$.

Exam TipState the symmetry argument and the zero-flux caps for full credit.

Applying Gauss’s Law: Planar Symmetry

An infinite sheet of charge has planar symmetry: the electric field is uniform in magnitude and perpendicular to the sheet on both sides.

Derivation: Electric Field from an Infinite Sheet of Charge

Starting point: Gauss’s law.

$$\oint \vec{E}\cdot d\vec{A}=\frac{{Q}_{\text{enc}}}{{\epsilon }_{0}}$$

Step 1 — Choose the Gaussian surface

Draw a rectangular pillbox (a short cylinder or box) that straddles the sheet. Each flat face has area $A$ and is parallel to the sheet, one on each side.

Step 2 — Apply the symmetry argument

On each flat face: $\vec{E}$ is perpendicular to the sheet and therefore parallel to $d\vec{A}$ (outward normal). The field has the same magnitude $E$ on both faces by symmetry. On the side walls: $\vec{E}$ is parallel to the walls, so $\vec{E}\cdot d\vec{A}=0$.

Step 3 — Simplify the integral

$$\oint \vec{E}\cdot d\vec{A}=EA+EA=2EA$$

Step 4 — Calculate the enclosed charge

For surface charge density $\sigma $:

$${Q}_{\text{enc}}=\sigma A$$

Step 5 — Solve for $E$

$$2EA=\frac{\sigma A}{{\epsilon }_{0}}$$

Result

$$E=\frac{\sigma }{2{\epsilon }_{0}}$$

The electric field from an infinite sheet is constant — it does not depend on the distance from the sheet. This is a key result that distinguishes planar from spherical or cylindrical symmetry.

For two parallel sheets with equal and opposite surface charge densities ($+\sigma $ and $-\sigma $), the fields add between the plates and cancel outside. Between the plates: $E=\sigma /{\epsilon }_{0}$. Outside both plates: $E=0$. This is the basis of the parallel-plate capacitor.

Symmetry	Gaussian surface	Field result	Distance dependence
Spherical	Concentric sphere	$E=\frac{Q}{4\pi {\epsilon }_{0}{r}^{2}}$	$∝1/{r}^{2}$
Cylindrical	Coaxial cylinder	$E=\frac{\lambda }{2\pi {\epsilon }_{0}r}$	$∝1/r$
Planar	Pillbox	$E=\frac{\sigma }{2{\epsilon }_{0}}$	Independent of distance