Focus Question: What is the flux?

- Electric Flux – Electric flux (Φ) is a measure of the amount of electric field that penetrates a surface.
- Flux is a scalar quantity that is the dot product of electric field and the surface area the field passes through.

$

\begin{gathered}

\emptyset=E A_{\perp} \\

\emptyset=E \cdot A \\

\emptyset=E A \cos \theta

\end{gathered}

$

Flux is greater for:

Strong fields, bigger surfaces, more perpendicular surfaces.

*only the perpendicular component of the surface area will have flux going through it.

Calculate flux for a non-uniform electric field.

- For a non-uniform field, break the surface up into tiny elements, each of area dA:

**Example A:** E is uniform (constant) from left to right, and our surface is a cube with face area A. What’s the total flux through the (closed) cube shown?

**Answer/Explanation**

- Flux due to A1: $E \cdot A_1=E A \cos 180=-E A$
- Flux due to A2: $E \cdot A_2=E A \cos 180=E A$
- Flux due to $\mathrm{A} 3: E \cdot A_3=E A \cos 90=0 * \mathrm{E}$ is perpendicular to $\mathrm{dA}$ (the field does not pass through the surface.
- Flux due to A4: $E \cdot A_4=E A \cos 270=0$ The net flux is 0 .

Net flux will always be zero through a close surface if the field source is from outside of the surface since the field will enter one face of the surface and exit another.

*If a closed surface has no net charge, then the net flux through it is zero.

**Example B: Derivation of Gauss’s Law** – Consider a point charge +Q inside a spherical shell of radius R. Determine the flux through the sphere.

**Answer/Explanation**

<p*Every point on the sphere has E radially outward. dA is radially outward by definition, so $\theta=0$ everywhere.

$

\begin{aligned}

& d \phi=E \cdot d A \cos \theta \rightarrow d \phi=E \cdot d A \cos 0=E \cdot d A \\

& \phi=E \int d A

\end{aligned}

$

*Every point on the sphere is the same distance from the charge, so the field on the sphere is equal at all points to $E=\frac{k Q}{R^2} \iint d A$ is simply the surface area of the shell.

$

\begin{aligned}

\phi=\frac{k Q}{r^2}\left(4 \pi r^2\right) \rightarrow(k= & \left.\frac{1}{4 \pi \varepsilon_0}\right) \rightarrow \phi=\left(\frac{1}{4 \pi \varepsilon_0}\right) \frac{Q}{r^z}\left(4 \pi r^z\right) \\

& \rightarrow \phi=\frac{Q}{\varepsilon_0}

\end{aligned}

$

- Gauss’s Law – The total of the electric flux out of a closed surface is equal to the charge enclosed divided by the permittivity:

$

\phi=\oint \vec{E} \cdot \overrightarrow{d A} \cos \theta=\frac{Q_{\text {enclosed }}}{\epsilon_0}

$

*The $\cos \theta$ can be removed if the angle between the field vector and $\mathrm{dA}$ is constant.

Gauss’s Law is used to find the electric field due to various types charge distributions. When using Gauss’s Law, an imaginary surface is drawn in the region you want to find the electric field. As long as the Gaussian surface has appropriate symmetry in regard to the field, properties of the flux through that surface can be used to calculate the field.

**Example C: Thin Hollow Shell** – a) Find the electric field inside and outside a hollow shell of uniform charge +Q and b) sketch a graph of electric field from distance of the center of the shell.

**Answer/Explanation**

**Inside the Shell:**The electric field is to be found inside the shell, so a Gaussian surface is drawn as shown inside the sphere. The Gaussian surface needs to be symmetric with respect to the sphere, so a Gaussian sphere of variable radius r, where r<R is used.

$\emptyset=\oint \vec{E} \cdot \overrightarrow{d A} \cos \theta=\frac{Q_{e n c}}{\epsilon_0}$

${ }^* \boldsymbol{Q}_{\text {enc }}$ is the charge enclosed by the Gaussian surface. All the charge is outside of the surface $(>\mathrm{r})$, so the charge enclosed is zero, making the field zero as well.

$

\rightarrow E=0

$

**Outside the Shell:** The electric field is to be found outside the shell, so a Gaussian surface is drawn as shown outside the sphere. The Gaussian surface needs radially outward (spherical) symmetry, so a Gaussian sphere of variable radius R, where r>R is used. *the Gaussian surface will always be a sphere for a sphere of charge.

$

\emptyset=\oint \overrightarrow{\boldsymbol{E}} \cdot \overrightarrow{\boldsymbol{d A}}=\frac{Q_{e n c}}{\epsilon_0}

$

$*$ The angle between $\mathrm{E}$ and $\mathrm{dA}$ is always zero, and $\mathrm{E}$ is constant along the Gaussian surface, so $\oint \vec{E} \cdot \overrightarrow{d A} \cos \theta$ is simply EA here, where $\mathrm{A}$ is the surface are of the Gaussian surface. $(\oint \vec{E} \cdot \overrightarrow{d A} \cos \theta$ will usually be EA or zero when using Gauss’s law in Physics C).

For this problem, the Gaussian surface encloses all the charge, so $Q_{e n c}=Q$.

$

\begin{aligned}

\phi=E \int d A=\frac{Q}{\varepsilon_0} & \rightarrow E\left(4 \pi r^2\right)=\frac{Q}{\varepsilon_0} \rightarrow E=\frac{Q}{4 \pi \varepsilon_0 r^2} \\

& \rightarrow \boldsymbol{E}=\frac{\boldsymbol{K Q}}{\boldsymbol{r}^2}

\end{aligned}

$