Electrostatics in the language of power lines

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Electrostatics on the tongue
power lines

Generally, electrostatics is studied this way: they write Coulomb’s law for the force of interaction between two exact charges $q_1$ and $q_2$: $F = \frac{q_1 q_2}{r^2}$ (r is the distance between the charges) - and then, with the help of sometimes simple and sometimes complex mathematical calculations, they obtain all sorts of consequences of this law, considering that the charge of any body can be represented as the perfection of properly distributed exact charges.

But often an accurate mathematical analysis of the electric field and charge distribution turns out to be difficult. Then simple messages come to the rescue - symmetry and patterns of field lines. Of course, they cannot give as complete a picture as mathematical calculations, but they are visual and help to “see” and predict what features the electric field will have and how charges will interact.

What are field lines

Let’s take the exact charge $q$. This charge creates an electric field in the surrounding space, which is completely determined if at each point in space the direction of the field is given - the force that would act on an exact unit positive charge if it were placed at that point. In order to visually represent the electric field throughout space, you can try to represent it by a variety of vectors, for example, at different points in space so that each of them shows the direction of the field (its magnitude and direction) at the point from which it emerges. However, it immediately becomes obvious that this method of depicting the field is unsuccessful: individual arrows, overlapping each other, create a confusing picture, which will need to be understood.

Faraday proposed that the field be represented by lines of force. These are lines, the direction of which at each point coincides with the direction of the field at this point. To accurately charge, drawing field lines is quite simple. These are straight lines (rays) passing through the point at which the charge is located. If the charge is serial, then the lines of force are

Fig. 1.

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Considering figure 3 and 4, one can notice that with opposite charges, the drain lines seem to connect the charges with each other, and in the case of identical charges, the drain lines seem to push apart. If we attribute to the plum lines the properties of contracting along their length and repelling in the transverse direction, then we could say that the plum lines connecting dissimilar charges bind them to each other, and the confluent lines of dissimilar charges, repelling, repel the charges themselves from each other. Of course, one cannot assume that confluent field lines really exist, but with this interpretation, often just a glance at the field picture allows one to predict the result of the interaction of charges. These same “properties” of plum lines, as it is not difficult to report, can also speak about something more - about the magnitude of the direction of the field.

Gauss’s theorem

So, the plum line pattern shows the direction of the field direction vector at each point. But that’s not all. It turns out that these pictures can talk about something more - about the magnitude of the directionality of the field.

Let’s take a point positive charge $q$. Let us select a cone with the apex at the point at which the charge is located, and follow the plum lines passing through an area with area $S_1$, located at a distance $r_1$ from the charge (Fig. 5). At a distance of $r_2$, these drain lines will pass through the area with a larger area of ​​$S_2$. The ratio of the diameters of the pads is equal to the ratio $\frac{r_2}{r_1}$, therefore $\frac{S_2}{S_1} = \left( \frac{r_2}{r_1} \right)^2$.

But this means that the density of line plums $n = \frac{N}{S}$ (the number of line plums passing through an area of ​​unit area) is inversely proportional

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is national to the square of the distance to the charge. Since $\frac{n_1}{n_2} = \frac{r_2^2}{r_1^2}$, then $n \sim \frac{1}{r^2}$.

The directivity of the field of the exact charge changes in the same way: $E = \frac{q}{r^2}$.

Thus, the density of field lines is proportional to the direction of the field: $n \sim E$. It is only necessary that the power lines are not interrupted anywhere. For an exact positive charge, the lines of force must begin at the charge and go to infinity. In the case of a negative charge, the lines of force must go from infinity and end at the charge.

It turns out that it is always possible to draw a picture of the field lines so that their density is proportional to the direction of the field. To do this, the lines of force must be driven out on positive charges and end on negative ones (or go to infinity), and the number of lines of force leaving the charge or going to it must be proportional to the magnitude of this charge.

Let’s mentally draw a sphere with the center at the point where we find the charge. If we agree to assume that $4\pi q$ lines of force emerge from each positive charge (and end at each negative charge), then their density on the spherical surface will be equal to $\frac{4\pi q}{4\pi r^2} = \frac{q}{r^2}$, that is, it will be equal to the direction of the field. The total number of lines of force intersecting the surface of the sphere does not depend on its radius and is equal to $4\pi q$. It is clear that the same will be the number of lines of force coming out of any closed surface inside which we find a charge (or going into it), no matter what shape this surface has.

This statement is called Gauss’s theorem. It is true, of course, for any charge distribution. It can be shown that for any system of charges the number of force

lines emerging from a closed surface of arbitrary shape, minus the number of force lines entering it, is equal to $4\pi Q$, where $Q$ is the total charge that is located in the volume limited by this surface. If the charge $Q$ is negative, then the number of lines that leave the volume bounded by the surface is less than the number of lines of force entering it. If there are no charges in the volume bounded by the surface, then the same number of lines of force come out of it as they enter.

Potential and equipotential surfaces

Before we begin to implement our knowledge - solving specific problems, let us remember one more characteristic of the field - potential. The fact is that for it you can find a geometric picture that is closely related to the picture of lines of force. This connection often helps to draw and analyze a picture of field lines.

Let us first recall what an electric field potential is.

If, by moving a charge in the field, we return to the starting point, then the total work that is done when crossing an exact charge from one point in the field to another does not depend on the shape of the path. Otherwise, it would be possible to first follow the path where it is smaller, and return along the path where it is greater. The total work would not be zero. Due to the fact that the work does not depend on the shape of the path, the concept of potential can be introduced. The field potential at a given point is equal to the work done by the field when crossing a unit exact positive charge from a given point in the field to infinity.

Potential can be either positive or negative. The field potential of a positive charge is positive, and the field potential of a negative charge is negative. In the first case, the force acting

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on the transferred charge from the side of the field, is directed in the same direction in which the charge moves, and in the second direction - against the movement (the field does negative work). A potential is a scalar that has no direction in space. The exact charge $q$ at a distance $r$ from it has a potential equal to $\frac{q}{r}$. If we need to find the field potential of several charges, then we must add up the field potentials created by individual charges, placed taking into account their signs. How is potential related to the pattern of field lines? Let’s check that the potential is constant everywhere. Such probabilities are called equipotential. For an exact charge, the equipotentials are spheres centered on the point where the charge is located. It is easy to see that the field lines must always be perpendicular to the equipotentials. If this were not so, then there would be a component of directional veracity, and therefore work would be done when moving charges along it. The potential would change along the veracity, and it could not be equipotential. One can see another important connection between equipotential tests and field lines. If two close equipotential checks are taken, then the difference in their potentials is equal to the work of transferring an exact unit positive charge from one check to the other. Since the force acting on such a charge is equal to the direction of the field, then $\Delta \varphi = E \Delta r$, or $E = \frac{\Delta \varphi}{\Delta r}$. This means that equipotential probabilities are denser where the field is stronger, and less frequent where the field is weaker: with the same potential difference, the distance $\Delta r$ between the probabilities is greater where the field is weaker, and vice versa.

Now we can move on to solving specific problems.

Field of a uniform charged hollow sphere

We can immediately assert that, due to the symmetry of the field, the field lines are directed everywhere along the radius and the density of the field lines is the same in all directions. Inside the sphere the field is very zero. Otherwise, the field lines inside the sphere were directed towards the center or away from the center of the sphere. This means that there is a precise charge at the center of the sphere.

In order for the electron field to be outside the sphere, remember that the direction of the field is equal to the density of the field lines, unless the total number of field lines emerging from the charge $Q$ is equal to $4\pi Q$. Let us draw a sphere of radius $R$ with its center at the center of the charged sphere. The number of lines of force crossing the sphere we tested is equal to the direction of the field multiplied by the area of ​​the sphere being tested: $N = E \cdot S = E \cdot 4\pi R^2$. According to Gauss’s theorem

$$N = 4\pi Q,$$ from here $E = \frac{Q}{R^2}$.

The field strength outside the sphere is the same as that of an exact charge located at the center of the sphere. It is clear that the potential at any point in the field outside the sphere will be the same as the potential of the field of the exact charge. The potential of all points lying on the surface of the sphere is the same and is equal to $\frac{Q}{r}$, where $r$ is the radius of the sphere. The potential of any point inside the sphere will be the same: when moving a charge to this point from the surface of the sphere, we do not do any work (inside the sphere, the directed field, and therefore the force acting on the charge, are equal to zero). The dependence of the direction of the electron field of a uniform charged sphere on the distance to the center of the sphere is shown in figure 6, 6.

The absence of a field inside a charged sphere can, of course, be shown

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and non-predestination from Coulomb’s law.
Let us take an arbitrary point $A$ (Fig. 7) inside the sphere. Let’s imagine a narrow cone that comes from point $A$, cutting out two small spherical areas $\Delta S_1$ and $\Delta S_2$ on the surface of the sphere. If $r_1$ and $r_2$ are the distances from point $A$ to these sections, then $\frac{\Delta S_1}{\Delta S_2} = \frac{r_1^2}{r_2^2}$.

Since the sphere is charged uniformly, the charges of the areas are proportional to their areas. Therefore, $\frac{q_1}{q_2} = \frac{r_1^2}{r_2^2}$, and the fields created at point $A$ by these sections cancel each other out:

$$\frac{E_1}{E_2} = \frac{\frac{q_1}{r_1^2}}{\frac{q_2}{r_2^2}} = 1.$$

In this way, the entire sphere can be divided into pairs of sections. Therefore, the field at point $A$ is zero.

The idea of ​​plum lines made it possible to obtain the same result easier and faster.

Uniformly charged plane and parallel-plate capacitor

Let us now consider the field of an infinite (more correctly said, very large) uniformly charged plane, per unit area of ​​which there is a charge $σ$.

It is clear that the plum lines of the field created by the plane are perpendicular to the plane and their density is the same (Fig. 8): if the plum lines were inclined in one direction, this would mean that one direction along the plane is different from the others; if the density of the plumbing lines were different at different points of the plane’s approach, then this would mean that one point, which, of course, cannot be, differs in some way from the others.

Let us draw a cylindrical surface with bases parallel to the plane and equal in area to one. Drain field lines do not intersect the side surface

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cylinder, and the number of lines of force passing through the base of the cylinder, according to Gauss’s theorem, is equal to 4lσ. This means that in each direction (to the left of the plane and to the right of it) there are 2σ lines of force passing through a unit area (Fig. 8). Thus, the direction of the field of a uniformly charged infinite plate is the same everywhere, does not depend on the distance to the plate and is equal to

$$E = 2\pi\sigma.$$

If we now take two oppositely charged planes (Fig. 9), then on both sides of the capacitor formed by these planes, the fields cancel each other. Inside, between the plates, the fields are directed equally and, adding up, give the direction of the field

$$E = 4\pi\sigma.$$

Although the field at the edges of the capacitor will be distorted, in the middle part we can consider it to be the same in both magnitude and direction. Such a field is called homogeneous.

Conductor in an electric field

For up to six years now we have considered the position of the charges given, and all we had to do was find the electric field that these charges create. The problem becomes more complicated when we have conductors through which charges can move freely. If, for example, you bring a charge (Fig. 10) to the conductor, then the electrins in the conductor begin to move under the influence of the field of this charge. In this case, charges will appear on the surface of the conductor, the field of which inside the conductor is directed opposite to the external field. This will continue until the field inside the conductor becomes zero. In this case, the charges that appear on the surface of the conductor will, of course, change the field outside the conductor.

The field inside the conductor is always