Angles (углы)

Now let's continue the radial line outside the cylinder. Angle α is the same, because we know that the angle CPQ=180° and it consists of angles α, β and APB. On the other hand, we can consider this triangle ABP. Sum of angles α, β and APB also equal to 180°. That is a property of any kind triangle. We shall compare both relations:

And conclude:

That is a red line, which connects points B and P, now continue this line, and draw it outside the cylinder. There will be point O. After connecting this point with point A, we shall have here a triangle AOB of 90°.

Evidently

From this red triangle:

On the other hand, we have also another right triangle, which consists of a points vertices A O and P. And from this triangle we can say A and O is equal to line b. Angle between red and black lines (OPA) consists of angle α and angle that identical to β, because this angle was built by crossing two lines, so we can say:

So, if we compare AB of two relations, we can get

Let's look a cylinder. P is the arbitrary point, B is inverse point, C is a center. Let's look at PQ. It is a vertical line, which is normal to horizontal line.

Inside a triangle a sum of angles is 180°:

If we compare these expressions, we can get:

Combining all these relations:

Trigonometric relations (тригонометрические соотношения)

We just have got two relations:

Let us combine them:

Taking into account:

We shall get:

If this formula is right, then this boundary conditions will be satisfied.

Field induced by the line sources (поле, индуцированное линейными источниками)

The normal component of the field intensity induced by the surface charges may be expressed as

The normal component of the field intensity induced by the charged line in the cylinder center:

The normal component of the field intensity induced by the charged line in the inverse point:

The first term corresponds to field induced by the charged line in the inverse point. The second term corresponds to field induced by the charged line in the central point:

The field sources for the external domain (источники полей для внешней области)

Everywhere the dielectric permittivity is ε₀ and that is true only for the external space. We can calculate electric field in every point in this space as a superposition of 3 fields. First of them primary source which have charge density of τ. First image which is placed in the center of cylinder and has the linear charge density of . And now finally the charge, which is placed in point B, the inverse point with the special properties, which has a linear charge density .

The field sources for the internal domain (источники полей для внутренней области)

Честно, я не выкупил, почему здесь везде ε₀.

The inner space which is field with the dielectric. If we want to calculate the electric field, which is induced inside this area, we have to consider the only one primary source with such charge density . If we are looking for the field, which is inside this cylinder, we can not place images inside itself otherwise because in this case there would be some irregular points where the electric field which will come to infinity.

Application of the Images Method for calculating magnetic fields in the presence of cylindrical objects (Применение метода изображений для расчёта магнитных полей в присутствии цилиндрических объектов)