Equipotential surfaces and lines of force of the electrostatic field.

I would like to be able to visualize the electrostatic field. The scalar potential field can be geometrically represented as a set equipotential surfaces ( in the flat case - lines), or level surfaces, as mathematicians call them:

For each such surface the condition holds (by definition!):

(*)

Let's present this condition in equivalent notation:

Here the vector perpendicular to the surface element belongs to the surface under consideration (the scalar product of non-zero vectors is equal to zero precisely under this condition). We have the opportunity to determine the unit normal vector to the surface element in question:

If we return to physics, we conclude that the vector of the electrostatic field strength is perpendicular to the equipotential surface of this field!

Mathematical content of the concept of "scalar field gradient":

The direction of the vector is the direction in which the function increases most rapidly;

This is the increment of a function per unit length along the direction of maximum increase.

How to construct an equipotential surface?

Let the equipotential surface given by equation (*) pass through a point in space with coordinates ( x,y,z). Let us set arbitrarily small displacements of two coordinates, for example x=>x+dx And y=>y+dy. From equation (*) we determine the required displacement dz, such that the end point remains on the equipotential surface under consideration. In this way you can “get” to the desired point on the surface.

Vector field line.

Definition. The tangent to the field line coincides in direction with the vector defining the vector field under consideration.

A vector and a vector are the same in direction (i.e. parallel to each other) if

In coordinate notation form we have:

It is easy to see that the following relations are valid:

The same result can be reached if we write down the condition for the parallelism of two vectors using their vector product:

So, we have a vector field. Consider the elementary vector as a force line element of a vector field.

In accordance with the definition of a power line, the following relations must be satisfied:

(**)

This is what the differential equations of a field line look like. It is possible to obtain an analytical solution to this system of equations in very rare cases (field of a point charge, constant field, etc.). But it is not difficult to graphically construct a family of force lines.

Let the field line pass through the point with coordinates ( x,y,z). We know the values of the projections of the tension vector onto the coordinate directions at this point. Let us choose an arbitrarily small mixing, for example, x=>x+dx. Using equations (**) we determine the required displacements dy And dz. So we moved to the neighboring point of the force line. The construction process can be continued.

NB! (Nota Bene!). The power line does not completely determine the tension vector. If a positive direction is specified on a power line, the voltage vector can be directed either in a positive or negative direction (but along the line!). The field line does not determine the vector modulus (i.e., its magnitude) of the vector field under consideration.

Properties of the entered geometric objects:

The relationship between tension and potential.

For a potential field, there is a relationship between potential (conservative) force and potential energy

where ("nabla") is the Hamiltonian operator.

Because the That

The minus sign shows that vector E is directed towards decreasing potential.

To graphically display the potential distribution, equipotential surfaces are used - surfaces at all points of which the potential has the same value.

Equipotential surfaces are usually drawn so that the potential differences between 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. In the figure, the dotted line shows the lines of force, the solid lines show sections of equipotential surfaces for: a positive point charge (a), a dipole (b), two charges of the same name (c), a charged metal conductor of complex configuration (d).

For a point charge the potential therefore equipotential surfaces are concentric spheres. On the other hand, tension lines are radial straight lines. Consequently, the tension lines are perpendicular to the equipotential surfaces.

It can be shown that in all cases the vector E is perpendicular to the equipotential surfaces and is always directed in the direction of decreasing potential.

Examples of calculations of the most important symmetrical electrostatic fields in vacuum.

1. Electrostatic field of an electric dipole in a vacuum.

An electric dipole (or double electric pole) is a system of two equal in magnitude opposite point charges (+q,-q), the distance l between which is significantly less than the distance to the field points under consideration (l<< r).

The dipole arm l is a vector directed along the dipole axis from the negative to the positive charge and equal to the distance between them.

The electric moment of the dipole re is a vector coinciding in direction with the dipole arm and equal to the product of the charge modulus |q| on shoulder I:

Let r be the distance to point A from the middle of the dipole axis. Then, given that

2) Field strength at point B on the perpendicular restored to the dipole axis from its center at

Point B is equidistant from the +q and -q charges of the dipole, so the field potential at point B is zero. Vector Ёв is directed opposite to vector l.

3) In an external electric field, a pair of forces acts on the ends of the dipole, which tends to rotate the dipole in such a way that the electric moment re of the dipole turns along the direction of the field E (Fig. (a)).

In an external uniform field, the moment of a pair of forces is equal to M = qElsin a or In an external inhomogeneous field (Fig. (c)), the forces acting on the ends of the dipole are not identical and their resultant tends to move the dipole to a field region with higher intensity - the dipole is pulled into a region of a stronger field.

2. Field of a uniformly charged infinite plane.

An infinite plane is charged with constant surface density The tension lines are perpendicular to the plane under consideration and directed from it in both directions.

As a Gaussian surface, we take the surface of a cylinder, the generators of which are perpendicular to the charged plane, and the bases are parallel to the charged plane and lie on opposite sides of it at equal distances.

Since the generators of the cylinder are parallel to the tension lines, the flux of the tension vector through the side surface of the cylinder is zero, and the total flux through the cylinder is equal to the sum of the fluxes through its bases 2ES. The charge contained inside the cylinder is equal to . By Gauss's theorem where:

E does not depend on the length of the cylinder, i.e. The field strength at any distance is the same in magnitude. Such a field is called homogeneous.

The potential difference between points lying at distances x1 and x2 from the plane is equal to

3. The field of two infinite parallel oppositely charged planes with equal absolute value surface charge densities σ>0 and - σ.

From the previous example it follows that the tension vectors E 1 and E 2 of the first and second planes are equal in magnitude and are everywhere directed perpendicular to the planes. Therefore, in the space outside the planes they compensate each other, and in the space between the planes the total tension . Therefore, between the planes

(in dielectric.).

The field between the planes is uniform. Potential difference between planes.
(in dielectric ).

4.Field of a uniformly charged spherical surface.

A spherical surface of radius R with a total charge q is charged uniformly with surface density

Since the system of charges and, consequently, the field itself is centrally symmetrical relative to the center of the sphere, the lines of tension are directed radially.

As a Gaussian surface, we choose a sphere of radius r that has a common center with the charged sphere. If r>R, then the entire charge q gets inside the surface. By Gauss's theorem, whence

At r<=R замкнутая поверхность не содержит внутри зарядов, поэтому внутри равномерно заряженной сферы Е = 0.

Potential difference between two points lying at distances r 1 and r 2 from the center of the sphere

(r1 >R,r2 >R), is equal to

Outside the charged sphere, the field is the same as the field of a point charge q located at the center of the sphere. There is no field inside the charged sphere, so the potential is the same everywhere and the same as on the surface

THEORETICAL BASIS OF WORK.

There is an integral and differential relationship between the electric voltage and the electric potential:

j 1 - j 2 = ∫ E dl (1)

E = -grad j (2)

The electric field can be represented graphically in two ways, complementary to each other: using equipotential surfaces and lines of force (field lines).

A surface where all points have the same potential is called an equipotential surface. The line of its intersection with the drawing plane is called equipotential. Field lines are lines whose tangents at each point coincide with the direction of the vector E . In Figure 1, dotted lines show equipotentials, solid lines show electric field lines.

Fig.1

The potential difference between points 1 and 2 is 0, since they are at the same equipotential. In this case from (1):

∫E dl = 0 or ∫E dlcos ( Edl ) = 0 (3)

Because the E And dl in expression (3) are not equal to 0, then cos ( Edl ) = 0 . Therefore, the angle between the equipotential and the field line is p/2.

From differential connection (2) it follows that the lines of force are always directed in the direction of decreasing potential.

The magnitude of the electric field strength is determined by the “density” of the field lines. The denser the field lines, the smaller the distance between the equipotentials, so that the field lines and equipotentials form “curvilinear squares”. Based on these principles, it is possible to construct a picture of field lines, having a picture of equipotentials, and vice versa.

A fairly complete picture of field equipotentials makes it possible to calculate the value of the projection of the intensity vector at different points E to the chosen direction X , averaged over a certain coordinate interval ∆х :

E avg. ∆х = - ∆ j /∆х,

Where ∆х - increment of coordinate when moving from one equipotential to another,

∆ j - the corresponding potential increment,

E avg. ∆х - average value E x between two potentials.

DESCRIPTION OF THE INSTALLATION AND MEASUREMENT TECHNIQUE.

To model the electric field, it is convenient to use the analogy that exists between the electric field created by charged bodies and the electric field of direct current flowing through a conducting film with uniform conductivity. In this case, the location of the electric field lines turns out to be similar to the location of the electric current lines.

The same statement is true for potentials. The distribution of field potentials in a conducting film is the same as in an electric field in a vacuum.

Electrically conductive paper with conductivity equal in all directions is used as a conductive film.

Electrodes are installed on the paper so that good contact is ensured between each electrode and the conductive paper.

The working diagram of the installation is shown in Figure 2. The installation consists of module II, remote element I, indicator III, power supply IV. The module is used to connect all used devices. The remote element is a dielectric panel 1, on which a sheet of white paper 2 is placed, on top of it is a sheet of copy paper 3, then a sheet of electrically conductive paper 4, on which electrodes 5 are attached. Voltage is supplied to the electrodes from module II using connecting wires. Indicator III and probe 6 are used to determine the potentials of points on the surface of electrically conductive paper.

A wire with a plug at the end is used as a probe. Potential j probe is equal to the potential of the point on the surface of the electrically conductive paper that it touches. The set of field points with the same potential is the image of the field equipotential. The TEC-42 power supply is used as the IV power source, which is connected to the module using a plug connector on the rear wall of the module. A V7-38 voltmeter is used as an indicator Ш.

PROCEDURE FOR PERFORMANCE OF THE WORK.

1. Place 1 sheet of white paper 2 on the panel. Place carbon paper 3 and a sheet of electrically conductive paper 4 on it (Fig. 2).

2. Install electrodes 5 on electrically conductive paper and secure with nuts.

3. Connect power supply IV (TEC – 42) to the module using the plug connector on the rear wall of the module.

4. Using two conductors, connect indicator III (voltmeter B7 – 38) to the “PV” sockets on the front panel of the module. Press the corresponding button on the voltmeter to measure DC voltage (Fig. 2).

5. Using two conductors, connect electrodes 5 to module P.

6. Connect the probe (wire with two plugs) to the socket on the front panel of the module.

7. Connect the stand to a 220 V network. Turn on the general power supply of the stand.

Let us find the relationship between the electrostatic field strength, which is its 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.

From the definition of gradient (12.4) and (12.6). follows that

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 decreasing 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 this case are 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. By depicting equipotential lines that correspond to different potential values, 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 only the potential difference for two surfaces has a physical meaning, 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 of 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 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 ($\varphi \left(\infty \right)=0$):

\[\varphi =\frac(1)(4\pi \varepsilon (\varepsilon )_0)\frac(q)(r)\left(1\right).\]

A system of 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 greater than 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 obtained the canonical equation of a one-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:

Let us rewrite equation (1.1) in the form:

We have obtained the canonical equation of a two-sheet hyperboloid, its semi-axes:

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

Equipotential lines. Equipotential surfaces

Equipotential surfaces and lines of force of the electrostatic field.

Fig.1

Numbering of equipotential lines in the drawing

Point charge field with positive charge