36. Electric field of a point charge (Электрическое поле точечного заряда)

37. Electric field of a uniformly charged sphere (Электрическое поле равномерно заряженной сферы)

38. Flat capacitor. Field. Surface charge. Capacity. (Плоский конденсатор. Поле. Поверхностный заряд. Вместимость.)

Assuming the plates are large enough so that the E field between them is uniform and directed perpendicular, then applying Gauss’s Law over surface S we find:

,

where S is the area of surface perpendicular to the E field and σ is the surface charge density on the plate, Therefore,

Everywhere between the plates

Potential difference between the plates can be found from:

We integrate along an E field line, d is the plate separation.

The capacity can be defined as:

Energy can be found as:

W’ volume energy density – объемная плотность энергии

V volume between plates – объем между обкладками

39.1 Mutual inductance between a cylindrical coil with rectangular cross section and a wire passing along the axis of the coil (Взаимная индуктивность между цилиндрической катушкой прямоугольного поперечного сечения и проводом, проходящим вдоль оси катушки)

L et’s define flux of the magnetic flux density B into the fields of a conductor through the surface S bounded by one coil winding. We know that the B lines in this case have the shape of concentric circles and pass the surface of each winding along the normal to it. B depends only on the distance r. Let’s divide the surface of the coil into small elements, within each of which the B does not change. These are narrow strips, parallel to the conductor, with an area dS = h ∙ dr. The desired flux Ф₂₁ is getting by integrating over a surface S bounded by a contour:

Let's find the magnetic flux density B created by an infinite conductor of radius R.

From Ampere's law, if r ≤ R then:	From Ampere's law, if r ≥ R then

So, magnetic flux density B created by an infinite conductor equal:

Then,

The total flux Ф₂₁ through all windings of the coil is w times greater, so the desired mutual inductance is equal to:

39.2 Inductance of a cylindrical coil with the rectangular cross section(Индуктивность цилиндрической катушки прямоугольного сечения).

4 0.1 Electric field induced by charged line placed above conducting surface (Электрическое поле, создаваемое заряженной линией, помещенной над проводящей поверхностью).

На картинке длины векторов D₁ и D₂ должны быть равны (обязательно) потому что у нас симметричны заряженные провода в методе изображений, одинаковые расстояния до точки от проводов, значит одинаковые модули напряжённости и электрического смещения.

because of the symmetry. So .

The electric displacement of a charged wire has the formula:

All displacement vectors are directed downwards, so the sign of the projections is negative.

So, the modules of electric field intensity and electric displacement are simply calculated according to the following formulas:

4 0.2. Magnetic field induced by the line with a current placed above a ferromagnetic surface with infinitely high magnetic permeability

In short: here the inductance of a two-wire line is calculated using the images method.

We can represent the inductance as the sum of the internal inductance (of the wire itself) and the inductance from the other wire and the images (of these wires).

The internal inductance of one wire is , but since we are looking for the inductance of the system of two wires, so we multiply by 2. And we get .

Now we consider the external inductance. The formula is as follows: the flux linkage by current. We find the flux linkage of each wire and images.

First Ф₁ and Ф₂. The formula for the flux is the integral of BdS. And B = µ₀H, where .Substituting the value of B and finding of the integration limits (from the edge of the wire, to the middle of the other wire), we get the value for Ф₁.

And Ф₂ = Ф ₁, because we integrate in the opposite direction, but similarly (this is one "-" sign) and then the current itself is negative; it turns out minus by minus +.

Futher, we find the images flux by integrating from the point of the first wire to the point of the second.

And we substitute everything into the formulas above