Discontinuity in electric field

[amjad-sh]

I was recently reading jackson's book of electrodynamics regarding surface charge distributions and discontinuity in electric field.
I reached that part and almost didn't understand anything.
I will write the whole text:

One of the common problems in electrostatics is the determination of electric field or potential due to a given surface distribution of charges.
Gauss law allows us to write a partial result directly. If a surface S with a unit normal $\vec n$ has a surface charge density of $\sigma(x)$ and electric field $\vec E_{1}$ and $\vec E_{2}$ on either side of the surface, then gauss law tells us immediately that $(\vec E_{1}-\vec E_{2})\cdot \vec n=\frac{\sigma}{\varepsilon_{0}} ...*$
This does not determine $\vec E_{1}$ and $\vec E_{2}$ if there are no other sources of field and the geometry and form $\sigma$ are especially simple.All that (*) says is that there is a discontinuity of $\frac{\sigma}{\varepsilon_{0}}$ in the normal component of electric field in crossing a surface with a surface charge density $\sigma$ the crossing being made from the "inner" to the "outer" side of the surface.
The tangential component of the electric field can be shown to be continuous across a boundary surface by using $\oint \vec E \cdot \vec dl =0$ for the line integral of $\vec E$ along a closed path. It is only necessarily to take a rectangular path with negligible ends and one side of either side of the boundary.
A general result for the potential (and hence, the field by diffrentiation) at any point in space not just at the surface, can be obtained from $\phi(x)=\int {\frac{\sigma(x')}{|x-x'|}}\ d^3 x'$ by replacing $\rho d^3x$ by$\sigma da$ :$$\phi(x)=\int {\frac{\sigma(x')}{|x-x'|}}\ da',$$

My Questions are:
a) I didn't get that part:"If a surface S with a unit normal $\vec n$ has a surface charge density of $\sigma(x)$ and electric field $\vec E_{1}$ and $\vec E_{2}$ on either side of the surface, then gauss law tells us immediately that $(\vec E_{1}-\vec E_{2})\cdot \vec n=\frac{\sigma}{\varepsilon_{0}} ...*$ ". E1 and E2 represent what? and where should I start to get this result.

b)what does he mean by this:"This does not determine $\vec E_{1}$ and $\vec E_{2}$ if there are no other sources of field and the geometry and form $\sigma$ are especially simple".

c)What did he mean by this, where is the discontinuity?:"All that (*) says is that there is a discontinuity of $\frac{\sigma}{\varepsilon_{0}}$ in the normal component of electric field in crossing a surface with a surface charge density $\sigma$ the crossing being made from the "inner" to the "outer" side of the surface".

d) can anybody explain this more deeply to me:"The tangential component of the electric field can be shown to be continuous across a boundary surface by using $\oint \vec E \cdot \vec dl =0$ for the line integral of $\vec E$ along a closed path. It is only necessarily to take a rectangular path with negligible ends and one side of either side of the boundary".Thanks

 

[jtbell]

E1 and E2 represent what?

$\vec E_1$ is the electric field on one side of the surface, and $\vec E_2$ is the electric field on the other side of the surface.

where should I start to get this result.

Start by drawing a Gaussian surface as shown in the diagram here: http://hyperphysics.phy-astr.gsu.edu/hbase/electric/elesht.html

b)what does he mean by this:"This does not determine ⃗E1\vec E_{1} and ⃗E2\vec E_{2} if there are no other sources of field [...]

If you know only the difference between two numbers, it does not tell you the numbers themselves. For example, if a - b = 5, then you might have a = 10 and b = 5; or a = 11 and b = 6; or a = 2 and b = -3; etc. In order to find a and b, you need more information or assumptions.

where is the discontinuity?

The discontinuity is at the surface (where the charge density $\sigma$ is located). As you move across the surface, the normal component of $\vec E$ changes abruptly.

can anybody explain this more deeply to me:"The tangential component of the electric field can be shown to be continuous across a boundary surface by using ∮⃗E⋅⃗dl=0\oint \vec E \cdot \vec dl =0 for the line integral of ⃗E\vec E along a closed path. It is only necessarily to take a rectangular path with negligible ends and one side of either side of the boundary".

See for example the left part of the diagram at the top of this page: http://farside.ph.utexas.edu/teaching/em/lectures/node59.html

Jackson is a graduate-level (master's / PhD) textbook. Students who take a course using it have normally already completed an undergraduate-level course in electromagnetism using e.g. Griffiths, which covers material such as the above. If you're trying to learn electromagnetism from the beginning, or if it's been some time since you studied it at an intermediate undergraduate level, you might want to supplement Jackson with Griffiths or a similar textbook, which is more "approachable."