Gauss law.

Gauss law allows easy (compared with using Coulomb formula) calculation of static electric fields for simple configurations.

Remainder: flux and electric field.

Lets consider closed surface \( S\) and charge \(Q\). Charge can be either inside the volume enclosed by surface or outside of it. Let us calculate flux of electric field through the surface. We introduced concept of flux earlier: \[ \Phi = \int_S \vec{E} \cdot \vec{dS} \] where \(dS\) is normal to local surface area. Now we require that surface \(S\) is closed one.

Electric field is: \[ \vec{E}= \frac{1}{4 \pi \epsilon_0} \frac{\vec{r}_o - \vec{r}_s}{ |\vec{r}_o - \vec{r}_s|^3 } \] That is a result of experiment, do don't have to prove it.
Because this field has some specific properties:

Field is "radial"
Amplitude reduces as \( 1/r^2 \)
Fields from several sources are added together.

These properties impose some limitations on flux \( \Phi\) of electric field \( \vec{E} \).

We take intuitive approach first and the follow with strict derivation.

Charge is outside \(S\)

Solid angle

A solid angle is a measure of how large an object appears to an observer from a particular point in three-dimensional space. It's the 3D analog of a regular (plane) angle. Instead of measuring the length of an arc (like in 2D angles), a solid angle measures the area that an object subtends on the surface of a unit sphere centered at the observation point. Its unit is the steradian (symbol: sr), which is dimensionless. Mathematically: \[ \Omega = \frac{A}{r^2} \] where \( A \) is area of the object and \( r \) is distance from observation point to the object.
Taking angle of area into account, in vector form: \[ d \Omega = \frac{ | \vec{dS} | }{ r^2}\] i.e. take the area you see when you look at the patch from the charge and divide by the square of the distance; that ratio is independent of the distance and is measured in steradians.
If you look at the area from the edge then \( \vec{r} \cdot \vec{dS} \) is 0, and solid angle is zero.

Using concept of solid angle applied to static E field flux.

Every infinitesimal cone of angle \( d \Omega \) carries the same amount of flux because

\( E \propto 1/r^2 \) but the area of any cut of the cone grows as \(r^2\), so \[ \int_S \vec{E} \cdot \vec{dS} \] stays constant.
There is no "spillage" from cone to neighbouring cones - static electric field is radial.

If charge is outside the surface \(S\) then cone of solid angle \(d \Omega \) crosses surface twice. When entering the surface, contribution to total flux is negative ( angle between \( \vec{dS}\) and \( \vec{E}\)) is more than 90 degrees. When exiting the surface, contribution of of same amount, but negative.
We can conclude that if charge is outside the closed surface - total flux is zero.

Charge inside the volume.

Looking at picture above we can see that if charge is inside the volume then vector \( \vec{E} \) is mostly in the same direction as \( \vec{dS} \). We can expect flux to be not zero. How can we calculate exact value?
It does not matter what is the shape of the surface.

We can select surface which makes calculation of surface integral simple. So we do integration over spherical surface with charge at its centre.

Surface element of the sphere is: \[ | \vec{dS} | = R^2 sin(\theta) \partial \theta \partial \phi \] In this case angle between vectors \( \vec{E} \) and \( \vec{dS}\) is zero - they are parallel.
\[ \oint_S \vec{E} \cdot \vec{dS} = \int_0^{\pi} \int_0^{2\pi} \frac{Q}{4 \pi \epsilon_0 R^2} R^2 sin(\theta) \partial \theta \partial \phi\] \[ \oint_S \vec{E} \cdot \vec{dS} = \frac{Q }{ 4 \pi \epsilon_0} \int_0^{\pi} \int_0^{2\pi} sin(\theta) \partial \theta \partial \phi = \frac{Q}{\epsilon_0} \] Distance from the charge to the surface cancels out - field strength goes down as \( 1/R^2 \) but surface area of the sphere grows as \( R^2 \). Result is important - flux of static electric field through any closed surface equals charge inside the surface. That is Gauss law. \[ \oint_S {\vec{E} \cdot \vec{dS}} = \frac{Q}{ \epsilon_0} \]

Differential version of Gauss law.

Divergence theorem states that flux through the surface S equals flow from sources/sinks inside enclosed volume \( V \): \[ \oint_S \vec{E} dS = \int_V (\nabla \cdot \vec{E}) dV \] and we can find flux of electric field through S if we find divergence of \( \vec{E} \) and integrate it over volume. What is divergence of static electric field?

Divergence of static electric field.

We take naive approach and calculate all derivatives: \[ \nabla \cdot \vec{E} = \frac{\partial E_x}{\partial x} + \frac{\partial E_y}{\partial y} + \frac{\partial E_z}{\partial z} \] and we get

: \[ \nabla \cdot \vec{E} = -\frac{3}{|r|^3} + \frac{(x_o - x_s)^2}{|r|^5} + \frac{(y_o - y_s)^2}{|r|^5} + \frac{(z_o - z_s)^2}{|r|^5} = -\frac{3}{|r|^3} + \frac{| \vec{r} |^2 }{|r|^5} \] If we look at this expression we can see singularity when \( |\vec{r}| = 0\). At all other points divergence is zero. \[ \nabla \cdot \vec{E} = \begin{cases} 0 & \text{if } |\vec{r}| \ne 0 \\ undefined & \text{if } |\vec{r}| = 0 \end{cases} \] How we can deal with singularity?

We replace point charge with "cloud" of charges (volume charge density is \( \rho_V \) ) and then make this cloud shrink to 0.
From Stoke's theorem (whatever flows through closed surface originates from interior of this surface) we can write \[ \oint_S {\vec{E} \cdot \vec{dS}} = \int_V { \nabla \cdot \vec{E} {dV}} \] On another hand we can write that total charge inside volume \(V\) encloded by surface \( S \) is sum/integral \[ Q = \int_V {\rho_V dV }\] Then we can write Gauss law as: \[ \int_V { \nabla \cdot \vec{E} {dV}} = \frac{1}{\epsilon_0} \int_V {\rho_V dV }\] Volume \(V\) can be anything so integrands should be equal. Thus we have Gauss law \[ \nabla \cdot \vec{E} = \frac{1}{\epsilon_0} \rho_V \]

Using Gauss law

Gauss law in integral form is: \[ \oint_S { \vec{E} \cdot \vec{dS}} = \frac{Q}{4 \pi \epsilon_0} \] If we have a scenario where \( \vec{E} \) is constant on surface \(S\) then it can be taken out of integral and we can obtain expression for \( \vec{E} \). It requires symmetry of charge distribution (not object itself).
If \( E \) is constant on surface \( S \) then: \[ \oint_S { \vec{E} \cdot \vec{dS}} = E \oint_S { dS} = E A = \frac{Q}{ \epsilon_0} \] where \(A\) is the area of surface \(S\).
Then we can calculate electric field as: \[ E = \frac{Q}{ \epsilon_0 A} \] and direction of the field \( \vec{E} \) is same as direction of surface element \( \vec{dS} \).

Hollow sphere of charges

Consider charges distributed over hollow sphere of radius \(R_s\). There are no charges inside the sphere.

It does not matter from which direction you look at this sphere - it looks same from any direction. There is spherical symmetry. Field does not change no matter what the direction is and \( \vec{E} \) does not change with direction.

We choose sphere ( radius \(r \) ) as Gaussian surface.

If sphere \( r < R_s \) then there are no charges enclosed in Gaussian sphere. Electric field is zero.
If sphere \( R > R_s \) enclosed charge is \(Q\) (total charge of hollow sphere) and electric field is: \[ E = \frac{Q}{A \epsilon_0} = \frac{Q}{4 \pi r^2 \epsilon_0} \] Area of the sphere radius \(R\) is \( 4 pi R^2 \).

It is so simple that we don't need to write any program.
This expression is same as expression for field of the charge located at (0,0,0) point. For Gaussian surface, once whole charge is enclosed, there is no difference.

Cylinder (for coaxial cable)

Lets calculate electric field inside coaxial cable: two cylinders, one hollow.
Charges are distributed over outer surface of smaller cylinder and over inner surface of larger cylinder. Charges are same on both surfaces. Surface charge density is \( \r ho \).

Due to cylindrical symmetry we choose cylinder as gaussian surface. Fields in different regions of coaxial cable are:

\( r \lt a \): There is no charge enclosed inside gaussian surface, field is zero.
\( r \gt a \) and \( \rho \lt b \): Surface area of inner cylinder is: \[ A = 2 \pi a dl \], where \(a\) is radius of inner cylinder and \( dl\) is length of coaxial cable.
If charge surface density (on surface of inner cylinder ) is \( \rho \), then enclosed charge (cable length is \( dl\) ) is: \[ Q = A \rho \] and by Gauss theorem: \[ E = \frac{Q }{ A \epsilon_0 } = \frac{A \rho }{ 2 \pi a dl\pi \epsilon_0} \] Field in space between cylinders: \[ E = \frac{\rho}{ 2 \pi \epsilon_0 r} \] It drops as \( \propto 1/r\) just as field of infinite line of charge does.
\( r \gt b \): opposite charges compensate each other, field is zero.

Charged volume - electron cloud

Now lets consider a bit different charge distribution - uniformly charged sphere. Charge is not confined to the surface but is distributed all over the volume of the sphere.
Again, we take advantage of its spherical symmetry - field can not change with \( \theta \) and \( \phi \).

Electric Field Inside the Sphere ( \( r < R_s \) Using Gauss’s Law: \[ \oint \vec{E} \cdot d\vec{A} = \frac{Q_{\text{enc}}}{\varepsilon_0} \] Gaussian surface: sphere of radius \(r\). Enclosed charge: \[ Q_{enc} = \frac{4}{3} \pi r^3 \] Surface area: \[ A = 4 \pi r^2 \] So: \[ E(r) \cdot 4\pi r^2 = \frac{\frac{4}{3}\pi r^3 \rho}{\varepsilon_0} \] Solving: \[ E(r) = \frac{\rho r}{3\varepsilon_0} \] The field increases as you move outward from the center.
Electric Field Outside the Sphere ( \( r>R_s \) ) Now the entire charge \(Q\) is enclosed: \[ E(r) \cdot 4\pi r^2 = \frac{Q}{\varepsilon_0} \] So: \[ E(r) = \frac{1}{4\pi \varepsilon_0} \cdot \frac{Q}{r^2} \] This is identical to the field of a point charge at the center.

Gauss law is always true but not always useful. It is only useful when field \( \vec{E} \) is constant on surface \(S \), so that it can be taken out of integral \( \int_S \vec{E} \cdot \vec{dS} \)