Electrostatics
Milevskiy Vladislav
1. Gauss's theorem
1.1. Flux of electric field strength
Let us introduce a new physical quantity - the flux of the electric field strength [2,3,4].
The field strength characterizes the electric field at a point in space. The flux of intensity does
not depend on the value of the intensity at a given point, but on the distribution of the field over
the surface of a particular area. It is for this quantity that the Gauss theorem is formulated. Let
us single out an element with area ΔS in the field. It must be so small that the electric field
strength at all its points can be considered the same. Draw a normal n to the element. The
direction of this normal is chosen arbitrarily (Fig. 1). The angle between the vectors E and n
will be denoted by α. Then, by definition, the flux of the electric field strength E is the product
of the area ΔS of the surface and the projection of the electric field strength on the normal to
the element:
∆𝑁 = 𝐸⃑ ∙ 𝑛⃑∆𝑆 = 𝐸∆𝑆𝑐𝑜𝑠(𝛼)
The flux can be positive or negative depending on the value of the angle α.
Fig.1. Definition of vectors.

Fig.2. Homogeneous field.
In reality, the field strength flux can be interpreted as a value proportional to the number
of field lines penetrating this element. The lines penetrating the element ΔS also penetrate the
element ΔS0, which is the projection of ΔS onto the plane perpendicular to the vector E (Fig.
2). The tension flux can be written in the form:
∆𝑁 = 𝐸∆𝑆𝑐𝑜𝑠(𝛼) = 𝐸∆𝑆0
(2)
If the field is inhomogeneous and the surface is arbitrary, then the flow is defined as
follows. The entire surface must be divided into small elements with an area of ΔSi, the tension
fluxes through each of these elements must be calculated, and then the fluxes through all
elements should be summed (Fig. 3)
Fig. 3. Inhomogeneous field.
𝑁 = ∑ 𝐸 ∙ 𝑛 ∆𝑆𝑖
(3)
The flow through a closed surface is also defined. The positive normal to any element
of a closed surface is taken to be the outer normal, i.e., the normal directed not inside the surface,
but outward.
1.2. The Gauss theorem for a point charge
The Gauss theorem establishes a connection between the flow of the electric field
strength through a closed surface and the charge inside this surface [4, 5]. Let us first consider
a simple special case [3]. Let us calculate the flow of the vector E in a homogeneous medium
through a spherical surface, in the center of which there is a point charge q. The field strength
at each point on the surface of the sphere is the same in absolute value, and the projection is:
𝐸𝑛 = 𝑘
𝑞
(4)
𝜀𝑟 2
The flow of the vector E through the surface of the sphere is:
𝐸𝑛 = 4𝜋𝑘
𝑞
𝜀
(5)
This result, as expected, is also valid for any closed surface containing charge q.
1.3. Generalization of the Gauss theorem
The Gauss theorem can be easily generalized to the case of any number of point charges.
The flux of tension through the surface area S for each charge is determined by the formula (5)
of fluxes from all charges. Therefore, summing the expressions () for all charges, we find:
𝑁 = 4𝜋𝑘
∑𝑞
𝜀
(6)
If the algebraic sum of charges inside the surface is zero, then N = 0. Gauss's theorem
can also be generalized for the case when the charge is distributed in space continuously. The
coefficient k in formula (6) is equal to unity in the absolute system of units and 1/4πε0 in SI.
Therefore, the Gauss theorem in SI does not contain the factor 4π:
𝑁=
𝑄
𝜀𝜀0
(7)
Summary: The Gauss theorem relates the electric field strength flow through a closed
surface to the total charge inside this surface.
2. Conductor
2.1. Conductor in an electrostatic field
The word "conductor" in physics refers to a conducting body of any size and shape.
It can be not only a metal, but also an electrolyte and, in general, any body containing
free charges, such as electrons or ions. For definiteness, we will understand the conductor below
as a metal body.
Consider the process of introducing a conductor into an electrostatic field (Fig. 2).
Fig. 4. Conductor in a uniform electrostatic field.
As soon as the conductor enters the field, the electrons in it begin to move towards the
positively charged plate. On the part of the surface of the conductor facing this plate, a negative
charge appears, so that the surface region is enriched with electrons. On the opposite side of the
surface, there are somewhat fewer electrons than needed to neutralize the positive ionic charge
of the crystal lattice, and this part of the surface turns out to be positively charged. The positive
and negative parts of the surface create their own field, opposite in direction to the external one.
If the conductor has been in the field for a long time) and the current no longer flows in it, then
both fields - the external E and the field of the conductor's own surface charges (Eint) - exactly
compensate each other at all points inside and on the surface. Indeed, if in some area of the
conductor the fields E and Einternal did not compensate each other, then in this area
a force would act on the electrons and a current would flow. No current means full field
compensation.
2.2. Electrostatic field at the surface of a conductor
As stated above, the electrostatic field inside the conductor is zero. Let us find out what
the field will be outside the conductor, near its surface. First of all, it is clear that the intensity
vector is normal to the surface of the conductor, since it is one of the equipotential surfaces.
The absolute value of E is determined only by the charge density c at a given location and, as
we will show below, is equal to 4πσ in a vacuum.
Indeed, in the immediate vicinity of the selected small area of the surface - ΔS (Fig. 3)
it can be considered an infinite plane.
Fig. 5. Field at the surface of the conductor
Its field is equal to 2πσ and is directed (if σ>0) from the section AC in both directions
(bold arrows on the left). The field created in the same place by the rest of the surface must
cancel the field of the AC section inside the conductor, so it must also be equal to 2πσ (arrows
on the right). Therefore, outside the conductor, both strengths add up and give 4πσ (in the SI
system, σ/ε0).
We have previously proved this for the special case of a conducting plate, i.e. small
parallelepiped. The second example is the field of a surface-charged sphere, equal to E=q/R2
on its surface. Expressing q in terms of the surface charge density σ, we again obtain
E = 4πσ
(8)
We emphasize once again that this is true, strictly speaking, only for a conductor in an
electrostatic field. If the field is not electrostatic, i.e. the current in the conductor is not equal to
zero, then there must be a component of the intensity along the surface, which ensures the
existence of the current. In addition, in this case, the field inside the conductor is not equal to
zero either - after all, to prove this position, we assumed that there is no current.
3. List of sources used
1. Zaitsev. Electrical Engineering 2000
2. A. Zilberman. Electricity and Magnetism 1970
3. G. Myakishev. PHYSICS, Electrodynamics 1998
4. I. Saveliev Physics Course. Electricity. 2000.
5. O. Troitsky. Lightning is a weapon of the gods of 1998.
4. List of illustrations
Pic. 1 Definition of vectors 3
Pic. 2 Homogeneous field 4
Pic. 3 Heterogeneous field 4
Pic. 4 Explorer in a homogeneous electrostatic field 6
Pic. 5 Field at the surface of the conductor 7
5. Index
flow of tension................................ 3, 4, 5
conductor........................................ 6, 8
Gauss theorem................................ 3, 5