For a more visual graphic representation of fields, in addition to lines of tension, surfaces of equal potential or equipotential surfaces are used. As the name suggests, an equipotential surface is a surface on which all points have the same potential. If the potential is given as a function of x, y, z, then the equation of the equipotential surface has the form:
Field strength lines are perpendicular to equipotential surfaces.
Let's prove this statement.
Let the line and the line of force make a certain angle (Fig. 1.5).
Let's move a test charge from point 1 to point 2 along the line. In this case, the field forces do work:
. (1.5)
That is, the work done by moving the test charge along the equipotential surface is zero. The same work can be defined in another way - as the product of the charge by the modulus of the field strength acting on the test charge, by the amount of displacement and by the cosine of the angle between the vector and the displacement vector, i.e. cosine of the angle (see Fig. 1.5):
.
The amount of work does not depend on the method of its calculation; according to (1.5), it is equal to zero. It follows from this that and, accordingly, which is what needed to be proved.
The equipotential surface can be drawn through any point in the field. Consequently, an infinite number of such surfaces can be constructed. It was agreed, however, to draw the surfaces in such a way that the potential difference for two adjacent surfaces would be the same everywhere. Then, by the density of the equipotential surfaces, one can judge the magnitude of the field strength. Indeed, the denser the equipotential surfaces are, the faster the potential changes when moving along the normal to the surface.
Figure 1.6a shows equipotential surfaces (more precisely, their intersections with the drawing plane) for the field point charge. In accordance with the nature of the change, the equipotential surfaces become denser as they approach the charge. Figure 1.6b shows equipotential surfaces and tension lines for the dipole field. From Fig. 1.6 it is clear that with the simultaneous use of equipotential surfaces and tension lines, the field picture is especially clear.
For uniform field equipotential surfaces obviously represent a system of planes equidistant from each other, perpendicular to the direction of the field strength.
1.8. Relationship between field strength and potential
(potential gradient)
Let there be an arbitrary electrostatic field. In this field we draw two equipotential surfaces in such a way that they differ from each other in potential by the amount dφ(Fig. 1.7)
The tension vector is directed normal to the surface. The normal direction is the same as the x-axis direction. Axis x drawn from point 1 intersects the surface 
at point 2.
Line segment dx represents the shortest distance between points 1 and 2. The work done when moving a charge along this segment:
On the other hand, the same work can be written as:
Equating these two expressions, we get:
where the partial derivative symbol emphasizes that differentiation is carried out only with respect to x. Repeating similar reasoning for axes y And z, we can find the vector:
, (1.7)
where are the unit vectors of the coordinate axes x, y, z.
The vector defined by expression (1.7) is called the gradient of the scalar φ . For it, along with the designation, the designation is also used. ("nabla") means a symbolic vector called the Hamiltonian operator
For a visual representation of vector fields, a picture of field lines is used. A line of force is an imaginary mathematical
a curve in space, the direction of the tangent to which at each
the point through which it passes coincides with the direction of the vector
fields at the same point(Fig. 1.17).
Rice. 1.17:
The condition of parallelism of the vector E → and the tangent can be written as the equality to zero of the vector product E → and the arc element d r → of the field line:
Equipotential is the surface on which for which the electric potential is constantϕ. In the field of a point charge, as shown in Fig. , spherical surfaces with centers at the location of the charge are equipotential; this can be seen from the equation ϕ = q ∕ r = const.
By analyzing the geometry of electric field lines and equipotential surfaces, it is possible to indicate a number of general properties of the geometry of the electrostatic field.
Firstly, lines of force begin at charges. They either go to infinity or end on other charges, as in Fig. .
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Secondly, in a potential field, field lines cannot be closed. Otherwise, it would be possible to specify a closed loop such that the work electric field when a charge moves along this contour, it is not equal to zero.
Thirdly, the lines of force intersect any equipotential normal to it. Indeed, the electric field is everywhere directed towards the fastest decrease in the potential, and on the equipotential surface the potential is constant by definition (Fig. ). 
Rice. 1.20:
And finally, the field lines do not intersect anywhere except at points where E → = 0. The intersection of field lines means that the field at the intersection point is an ambiguous function of coordinates, and the vector E → does not have a specific direction. The only vector that has this property is the zero vector. The structure of the electric field near the zero point will be analyzed in problems for ??
. The field line method is, of course, applicable to the graphical representation of any vector fields. So, in the chapter ?? we will meet the concept of magnetic lines of force. However, the geometry
magnetic field
completely different from the geometry of the electric field. Rice. 1.21: The idea of lines of force is closely related to the concept of a force tube. Let's take any arbitrary closed loop L and through each point we draw an electric line of force (Fig. ). These lines form the power tube. Let us consider an arbitrary section of the tube with surface S. We draw the positive normal in the same direction in which the field lines are directed. Let N be the flow of the vector E → through the section S. It is easy to see that if there are no electric charges inside the tube, then the flow N remains the same along the entire length of the tube. To prove it, we need to take another cross section S ′. According to Gauss's theorem, the electric field flux through a closed surface limited by the side surface of the tube and sections S, S′ is equal to zero, since there are no electric charges inside the power tube. Flow through
lateral surface
The result is a complete analogy with the flow of an incompressible fluid. In those places where the tube is thinner, the field E → is stronger. In those places where it is wider, the field E → is stronger. Consequently, the density of the field lines can be used to judge the electric field strength.
Before the invention of computers, for experimental reproduction of power lines they took glass vessel with a flat bottom and poured into it a liquid that does not conduct electrical current, for example, castor oil or glycerin. Powdered crystals of gypsum, asbestos or some other oblong particles were evenly stirred into the liquid. Metal electrodes were immersed in the liquid. When connected to sources of electricity, the electrodes excited an electric field. In this field, the particles are electrified and, attracted to each other by oppositely electrified ends, are arranged in the form of chains along the lines of force. The picture of field lines is distorted by fluid flows caused by forces acting on it in a non-uniform electric field.
To Be Done Yet
Rice. 1.22:
The best results are obtained from the method used by Robert W. Pohl (1884-1976). Staniol electrodes are glued onto a glass plate, between which an electrical voltage is created. Then, by lightly tapping it, oblong particles, for example, gypsum crystals, are poured onto the plate. They are located along it along the lines of force. In Fig. ??
The picture of field lines obtained in this way between two oppositely charged circles of staniol is depicted.
▸ Problem 9.1 Write down the equation of field lines in arbitrary orthogonal
coordinates
▸ Problem 9.2
Write down the equation of field lines in spherical coordinates. Let's find the relationship between tension electrostatic field , which is his power characteristics, and potential - energy characteristic of the field. Moving work single point positive charge from one point of the field to another along the axis X = provided that the points are located infinitely close to each other and x 1 – x 2 , dx . equal to E x dx . The same work is equal to j 1 -j 2 = dj
Equating both expressions, we can write where the partial derivative symbol emphasizes that differentiation is performed only with respect to X. , Repeating similar reasoning for the y and z axes
we can find vector E:
where i, j, k are unit vectors of the coordinate axes x, y, z.
i.e. the field strength E is equal to the potential gradient with a minus sign. The minus sign is determined by the fact that the field strength vector E is directed towards descending side potential.
To graphically depict the distribution of the potential of an electrostatic field, as in the case of the gravitational field (see § 25), equipotential surfaces are used - surfaces at all points of which the potential has the same value.
If the field is created by a point charge, then its potential, according to (84.5),
Thus, the equipotential surfaces in in this case- concentric spheres. On the other hand, the tension lines in the case of a point charge are radial straight lines. Consequently, the tension lines in the case of a point charge perpendicular equipotential surfaces.
Tension lines always normal to equipotential surfaces. Indeed, all points of the equipotential surface have the same potential, so the work done to move a charge along this surface is zero, i.e., the electrostatic forces acting on the charge are Always directed along the normals to equipotential surfaces. Therefore, vector E always normal to equipotential surfaces, and therefore the lines of the vector E are orthogonal to these surfaces.
An infinite number of equipotential surfaces can be drawn around each charge and each system of charges. However, they are usually carried out so that the potential differences between any two adjacent equipotential surfaces are the same. Then the density of equipotential surfaces clearly characterizes the field strength at different points. Where these surfaces are denser, the field strength is greater.
So, knowing the location of the electrostatic field strength lines, it is possible to construct equipotential surfaces and, conversely, from the known location of equipotential surfaces, the magnitude and direction of the field strength can be determined at each point in the field. In Fig. 133 shows, as an example, the form of tension lines (dashed lines) and equipotential surfaces (solid lines) of the fields of a positive point charge (a) and a charged metal cylinder having a protrusion at one end and a depression at the other (b).
A graphical representation of fields can be made not only with tension lines, but also with the help of potential differences. If we combine points with equal potentials in an electric field, we get surfaces of equal potential or, as they are also called, equipotential surfaces. At the intersection with the drawing plane, equipotential surfaces give equipotential lines. Drawing equipotential lines that correspond to different meanings potential, we get a visual picture that reflects how the potential of a particular field changes. Moving along the equipotential surface of a charge does not require work, since all field points along such a surface have equal potential and the force that acts on the charge is always perpendicular to the movement.
Consequently, tension lines are always perpendicular to surfaces with equal potentials.
The most clear picture of the field will be presented if we depict equipotential lines with equal potential changes, for example, 10 V, 20 V, 30 V, etc. In this case, the rate of change of potential will be inversely proportional to the distance between adjacent equipotential lines. That is, the density of equipotential lines is proportional to the field strength (the higher the field strength, the closer the lines are drawn). Knowing the equipotential lines, it is possible to construct the intensity lines of the field under consideration and vice versa.
Consequently, images of fields using equipotential lines and tension lines are equivalent.
Numbering of equipotential lines in the drawing
Quite often, equipotential lines in the drawing are numbered. In order to indicate the potential difference in the drawing, an arbitrary line is designated by the number 0, next to all other lines the numbers 1,2,3, etc. are placed. These numbers indicate the potential difference in volts between the selected equipotential line and the line that was selected as zero. At the same time, we note that the choice of the zero line is not important, since physical meaning has only the potential difference for the two surfaces, and it does not depend on the choice of zero.
Point charge field with positive charge
Let us consider as an example the field of a point charge, which has a positive charge. The field lines of a point charge are radial straight lines, therefore, equipotential surfaces are a system concentric spheres. The field lines are perpendicular to the surfaces of the spheres at each point of the field. Concentric circles serve as equipotential lines. For a positive charge, Figure 1 represents the equipotential lines. For a negative charge, Figure 2 represents equipotential lines.
This is obvious from the formula that determines the field potential of a point charge when the potential is normalized to infinity ($\varphi \left(\infty \right)=0$):
\[\varphi =\frac(1)(4\pi \varepsilon (\varepsilon )_0)\frac(q)(r)\left(1\right).\]
System parallel planes, which are at equal distances from each other, are equipotential surfaces of a uniform electric field.
Example 1
Assignment: The field potential created by a system of charges has the form:
\[\varphi =a\left(x^2+y^2\right)+bz^2,\]
where $a,b$ are constants Above zero. What shape do equipotential surfaces have?
Equipotential surfaces, as we know, are surfaces in which the potentials are equal at any points. Knowing the above, let us study the equation that is proposed in the conditions of the problem. Divide the right and left sides of the equation $=a\left(x^2+y^2\right)+bz^2,$ by $\varphi $, we get:
\[(\frac(a)(\varphi )x)^2+(\frac(a)(\varphi )y)^2+\frac(b)(\varphi )z^2=1\ \left( 1.1\right).\]
Let us write equation (1.1) in canonical form:
\[\frac(x^2)((\left(\sqrt(\frac(\varphi )(a))\right))^2)+\frac(y^2)((\left(\sqrt( \frac(\varphi )(a))\right))^2)+\frac(z^2)((\left(\sqrt(\frac(\varphi )(b))\right))^2) =1\ (1.2)\]
From equation $(1.2)\ $ it is clear that the given figure is an ellipsoid of revolution. Its axle shafts
\[\sqrt(\frac(\varphi )(a)),\ \sqrt(\frac(\varphi)(a)),\ \sqrt(\frac(\varphi )(b)).\]
Answer: The equipotential surface of a given field is an ellipsoid of revolution with semi-axes ($\sqrt(\frac(\varphi )(a)),\ \sqrt(\frac(\varphi )(a)),\ \sqrt(\frac( \varphi )(b))$).
Example 2
Assignment: The field potential has the form:
\[\varphi =a\left(x^2+y^2\right)-bz^2,\]
where $a,b$ -- $const$ is greater than zero. What are equipotential surfaces?
Let's consider the case for $\varphi >0$. Let's bring the equation specified in the conditions of the problem to canonical form; to do this, we divide both sides of the equation by $\varphi , $ we get:
\[\frac(a)(\varphi )x^2+(\frac(a)(\varphi )y)^2-\frac(b)(\varphi )z^2=1\ \left(2.1\ right).\]
\[\frac(x^2)(\frac(\varphi )(a))+\frac(y^2)(\frac(\varphi )(a))-\frac(z^2)(\frac (\varphi )(b))=1\ \left(2.2\right).\]
In (2.2) we got canonical equation single-sheet hyperboloid. Its semi-axes are equal to ($\sqrt(\frac(\varphi )(a))\left(real\ semi-axis\right),\ \sqrt(\frac(\varphi )(a))\left(real\ semi-axis\right ),\ \sqrt(\frac(\varphi )(b))(imaginary\semi-axis)$).
Consider the case when $\varphi
Let's imagine $\varphi =-\left|\varphi \right|$ Let's bring the equation specified in the conditions of the problem to canonical form, to do this we divide both sides of the equation by minus modulus $\varphi ,$ we get:
\[-\frac(a)(\left|\varphi \right|)x^2-(\frac(a)(\left|\varphi \right|)y)^2+\frac(b)(\ left|\varphi \right|)z^2=1\ \left(2.3\right).\]
Let us rewrite equation (1.1) in the form:
\[-\frac(x^2)(\frac(\left|\varphi \right|)(a))-\frac(y^2)(\frac(\left|\varphi \right|)(a ))+\frac(z^2)(\frac(\left|\varphi \right|)(b))=1\ \left(2.4\right).\]
We have obtained the canonical equation of a two-sheet hyperboloid, its semi-axes:
($\sqrt(\frac(\left|\varphi \right|)(a))\left(imaginary\semi-axis\right),\ \sqrt(\frac(\left|\varphi \right|)(a) )\left(imaginary\ semi-axis\right),\ \sqrt(\frac(\left|\varphi \right|)(b))(\real\ semi-axis)$).
Let's consider the case when $\varphi =0.$ Then the field equation has the form:
Let us rewrite equation (2.5) in the form:
\[\frac(x^2)((\left(\frac(1)(\sqrt(a))\right))^2)+\frac(y^2)((\left(\frac(1 )(\sqrt(a))\right))^2)-\frac(z^2)((\left(\frac(1)(\sqrt(b))\right))^2)=0\ left(2.6\right).\]
We have obtained the canonical equation of a right circular cone, which rests on an ellipse with semi-axes $(\frac(\sqrt(b))(\sqrt(a))$;$\ \frac(\sqrt(b))(\sqrt(a ))$).
Answer: As equipotential surfaces for a given potential equation, we obtained: for $\varphi >0$ - a one-sheet hyperboloid, for $\varphi