Gauss’ law

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Gauss’ law relates the electric field through a closed surface with the charge contained within that surface.

Electric Flux

Electric flux is the amount of electric field passing through a surface.

Open surface

Bring an open surface in an electric field. The normal vector \(\hat n\) is perpendicular to the tiny surface \(dA\) and chosen to be upwards. The electric field at \(dA\) is \(\vec E\)

The electric flux \(d\phi\) that goes through \(dA\) is defined as the scalar $$ \shaded{ \begin{align*} d\phi = \vec E \cdot \hat n\, dA = \vec E \cdot d\vec A = E\,dA\,\cos\theta \end{align*} } \nonumber $$

The flux for the whole surface \(A\) follows from the \(\int dA\).

Closed surface

By convention, the normal \(\hat n\) of a closed surface is from the inside to the outside of the surface.

The total flux through region \(R\) follows from the integral $$ \newcommand{oiint}{\subset\!\!\supset\kern-1.65em\iint} \shaded{ \phi = \iint_R \vec E \cdot d\vec A } \quad \left[\frac{Nm^2}{C}\right] \tag{flux} $$

Notes

• The circle in the integral sign reminds us that it is a closed surface.
• If the flux is \(0\), then what flows in also flows out.
• If more flows out than in, the flux is positive.

Example

A point charge \(+Q\) in a sphere of radius \(R\)

On the sphere, the flux is the same everywhere. Both \(d\vec A\) and \(\vec E\) are radial (\(\newcommand{\parallelsum}{\mathbin{\!/\mkern-5mu/\!}} d\vec A\parallelsum\vec E\)), so the dot-product becomes a multiplication of scalars \(|\vec E|\,|d\vec dA|\,\cos 0=E.dA.1\) $$ \newcommand{oiint}{\subset\!\!\supset\kern-1.65em\iint} \phi = \oiint \vec E \cdot dA = E \underbrace{\oiint\,dA}_\text{surface} = 4\pi R^2 \, E \nonumber $$

The electric field \(E_R\) at distance \(R\) $$ \vec E = \frac{Q}{4\pi\varepsilon_0 R^2}\,\hat r \nonumber $$

Substituting \(E_R\) in the expression for \(\phi\) $$ \phi = \bcancel{4\pi R^2} \, \frac{Q}{\bcancel{4\pi R^2}\,\varepsilon_0} = \frac{Q}{\varepsilon_0} \nonumber $$

Note:

• This flux \(\phi\) is independent of the distance \(R\). This is not surprise, because if you think of it as air flowing out, then all the air has to come out somehow no matter how big the sphere is. This implies that we could have taken any kind of surface around the point charge, and have found the same result.
• The same relation holds for any number of charges inside the surface.

Gauss’ law

Gauss’ law relates the electric field through a closed surface with the charge contained within that surface.

Definition

The law was first formulated by Joseph-Louis Lagrange, an Italian mathematician and astronomer (1773), followed by Carl Friedrich Gauss, a German mathematician and physicist (1835).

This brings us to Gauss’ law, also known as Maxwell’s first equation $$ \newcommand{oiint}{\subset\!\!\supset\kern-1.65em\iint} \shaded{ \phi = \oiint_S \vec E \cdot d\vec A = \frac{\sum_i Q_i}{\varepsilon_0} } \tag{Gauss’ law} \label{eq:gauss} $$

Where

• \(Q_i\) is any charge inside any closed surface \(S\)
• \(\varepsilon_0\) is the permittivity of free space.

Note

• If the flux is \(0\), it means there is no net charge inside the surface.
• Gauss’ law always holds, no matter how weird the charge distribution inside the surface is, or how weird the shape of the surface.
• Both Coulomb’s and Gauss’ law make the electric field with a charge \(Q\).

Gauss’ law holds, but will not help you calculate \(\vec E\) when the charges are not distributed symmetrically. Here we will use spherical symmetry, cylindrical symmetry and flat planes with uniform charge distributions.

Example: using spherical symmetry

Consider a sphere with radius \(R\) and charge \(Q\) uniformly distributed over the surface. We want to know the electric field at distance \(r\) inside and outside the sphere.

The key is choosing the Gaussian surface right. We choose a sphere with radius \(r\).

Symmetry arguments

1. Because of the symmetry the electric field is the same anywhere on the Gaussian surface.
2. The electric field is either pointing radially outwards or radially inwards \(\Longrightarrow\) \(d\vec A\) and \(d \vec E\) are parallel or anti-parallel
\(\Longrightarrow\) so the dot-product becomes a multiplication of scalars.

The surface area of the Gaussian surface is \(4\pi r^2\).

Inside the charged sphere, for \(r\lt R\)

The enclosed charge is \(0\). Applying equation \(\eqref{eq:gauss}\) $$ \newcommand{oiint}{\subset\!\!\supset\kern-1.65em\iint} \begin{align*} \oiint_S \vec E \cdot d\vec A &= \frac{\sum_i Q_i}{\varepsilon_0} \\ \Rightarrow E \oiint_S dA &= \frac{0}{\varepsilon_0} \\ \Rightarrow 4\pi r^2 E &= 0 \\ \Rightarrow E &= 0 \end{align*} $$

That means that anywhere inside the sphere, there is no electric field! This is non-trivial. All distributed charges cancel out on the inside.

Outside the charged sphere, for \(r\gt R\)

The enclosed charge is \(Q\). Applying equation \(\eqref{eq:gauss}\) $$ \newcommand{oiint}{\subset\!\!\supset\kern-1.65em\iint} \begin{align*} \oiint_S \vec E \cdot d\vec A &= \frac{\sum_i Q_i}{\varepsilon_0} \\ \Rightarrow E \oiint_S dA &= \frac{Q}{\varepsilon_0} \\ \Rightarrow E &= \frac{Q}{4\pi r^2\varepsilon_0} \end{align*} $$

Note:

• The direction is radially outwards is \(Q\) is positive, and inwards otherwise.
• This is non-trivial as well, because it behaves as a point charge at the center of the sphere as we have seen in the first example.

Conclusion

The electric field

• inside the sphere is \(0\),
• outside the sphere, it is the same as if there was a point charge at the center of the sphere.

Plot of the electric field, \(E(r)\)

Key to Gauss’ law that the electric field falls of at \(\frac{1}{r^2}\). A gravitational field also falls of at \(\frac{1}{r^2}\) \(\Longrightarrow\) if there was an hollow spherical planet, there would be no gravitational force inside.

Example: using plane symmetry

Express the electric field anywhere in space for a very large plane with a uniformly distributed charge \(Q\), with charge density \(\sigma\) on area \(A\) $$ \sigma = \frac{Q}{A} \quad \left[\frac{C}{m^2}\right] \nonumber $$

If the very large plane is \(\newcommand{\parallelsum}{\mathbin{\!/\mkern-5mu/\!}} \parallelsum\) \(xy\)-plane, there is symmetry along the \(x,y\)-axis \(\Longrightarrow\) physical characteristics will not depend on \(x\) or \(y\) \(\Longrightarrow \vec E\) can only have a \(z\)-component \(\Longrightarrow\) \(\vec E \perp\) the plane.

As the Gaussian surface, we choose a closed cylinder through the plane

The Gaussian surface takes advantage of symmetry

1. The rounded wall is \(\perp\) to \(\vec E \Longrightarrow \phi_\text{wall}=0\).
2. The flat end plates are \(\newcommand{\parallelsum}{\mathbin{\!/\mkern-5mu/\!}} \parallelsum\) to \(\vec E\).
3. Both end plates are distance \(d\) from the plane.

For both end plates \(\vec E\) is pointed away from the plate. Let the surface of each end plate be \(A\) $$ \newcommand{oiint}{\subset\!\!\supset\kern-1.65em\iint} \begin{align*} \phi &= \phi_\text{wall} + \phi_\text{top} + \phi_\text{bottom} \\ &= 0 + 2 \oiint_S \vec E \cdot d\vec A = 2 \oiint_S E\,dA \\ &= 2EA \end{align*} $$

Apply \(\eqref{eq:gauss}\) $$ \newcommand{oiint}{\subset\!\!\supset\kern-1.65em\iint} \begin{align*} \phi &= \frac{\sum_i Q_i}{\varepsilon_0} \quad \Rightarrow \quad 2E\bcancel{A} = \frac{\sigma \bcancel{A}}{\varepsilon_0} \end{align*} $$

So, the electric field is independent of distance \(d\) $$ \shaded{ E = \frac{\sigma}{2\varepsilon_0} } \tag{E-field plate} $$

While this is for an infinitely large plane, but also holds if the distance \(d\) is small compared to the linear size of the plane.

Example: two charged planes

Two very large plates, with surface charge densities (\(Q/A\)), \(+\sigma\) and \(-\sigma\), a distance \(d\) apart

Sum the electric field vectors using the superposition principle $$ \shaded{ E = \frac{\sigma}{\varepsilon_0} } \tag{E-field plates} $$

Picture

Note that, when you get to the edge of the plates, the symmetry argument is no longer true.

Example: using cylindrical symmetry (lec.12)

A very long cylinder with radius \(R\) with uniform charge distribution throughout the whole cylinder with positive density \(\rho\). What is the electric field inside and outside the cylinder?

Let’s start with outside the cylinder, \(r\leq R\). The gaussian surface is a cylinder with radius \(r\) and length \(l\). The ends of the gaussian surface are flat perpendicular to the axis of symmetry

Symmetry arguments:

1. On the curved gaussian surface, the distance \(r\) is the same everywhere, so the electric field is the same.
2. Given that the curved gaussian surface is a cylinder, the electric field must everywhere be radial, perpendicular to the axis of symmetry.
3. The electric flux through the flat end surfaces must be zero, because \(\vec E\perp d\vec A\)
4. On the curved gaussian surface, \(\newcommand{\parallelsum}{\mathbin{\!/\mkern-5mu/\!}} \vec E\parallelsum d\vec A\)

To apply Gauss’ law, we only have to take the curved surface into account. The curved area is of the gaussian surface is \(l\,2\pi\,r\) and the volume of the enclosed charge is \(l\,\pi R^2\) $$ \newcommand{oiint}{\subset\!\!\supset\kern-1.65em\iint} \newcommand{\parallelsum}{\mathbin{\!/\mkern-5mu/\!}} \begin{align*} \oiint_S \vec E \cdot d\vec A &= \frac{\sum_i Q_i}{\varepsilon_0} & \left(\vec E \parallelsum d\vec A\right) \\ \Rightarrow E\,A_\rm{curved} &= \frac{V_\rm{encl.}\,\rho}{\varepsilon_0} \\ \Rightarrow E\,\bcancel{l}\,2\cancel{\pi}\,r &= \frac{\bcancel{l}\,\cancel{\pi} R^2\,\rho}{\varepsilon_0} \\ \Rightarrow \vec E &= \frac{R^2\,\rho}{2\,\varepsilon_0\,r} \hat r \end{align*} $$

Inside the cylinder, \(r\leq R\), a similar gaussian surface, but inside the cylinder

In applying Gauss’ law, the first term is the same, but and the volume of the enclosed charge is \(l\,\pi r^2\) $$ \newcommand{oiint}{\subset\!\!\supset\kern-1.65em\iint} \newcommand{\parallelsum}{\mathbin{\!/\mkern-5mu/\!}} \begin{align*} \oiint_S \vec E \cdot d\vec A &= \frac{\sum_i Q_i}{\varepsilon_0} & \left(\vec E \parallelsum d\vec A\right) \\ \Rightarrow E\,A_\rm{curved} &= \frac{V_\rm{encl.}\,\rho}{\varepsilon_0} \\ \Rightarrow E\,\bcancel{l}\,2\cancel{\pi}\,\bcancel{r} &= \frac{\bcancel{l}\,\cancel{\pi} r^\bcancel{2}\,\rho}{\varepsilon_0} \\ \Rightarrow \vec E &= \frac{r\,\rho}{2\,\varepsilon_0} \hat r \end{align*} $$

Plotting \(E(r)\)

If it were a solid conductor, the charge inside would be zero