Example 1.Determine the strength and potential of the field of an infinite plane, evenly charged with surface density Charge s. The dielectric constant ambient e.

From the symmetry task it is clear that the lines of tension electric field Must be parallel straight, perpendicular to the charged surface, as shown in Ri C.1.2. Consequently, the field of the charged plane is homogeneous.

Choose a closed surface in the form of a direct cylinder length. l. With an area of \u200b\u200bfoundation S., as it shown on the picture. Due to the homogeneity of the field of tension on the lower and upper bases of the cylinder the same in size and coincide in the direction with positive standards to these surfaces. In addition, we take into account that the stream of the vector of tension through the side surface is absent. therefore

A full charge in a volume of limited cylindrical surface is equal to q \u003d SS.. Then and.

Potential of the plane field.

In the resulting expression FROM - permanent integration, and the axis Oh Perpendicular to the plane. Taking the potential on the surface of the plane equal to zero, we get C \u003d.0.

Finally.

Example 2.Determine the strength and potential of the electrical field of an endless thread, evenly charged with a linear density of charge t. Dielectric permeability of the environment E.

From the axial symmetry of the problem, it follows that the filaments of the thread field lines are a system of radial rays emerging from the points belonging to the axis of symmetry. Then at points equidistant from the thread, the field strength is the same in size. Choose a straight cylinder length as a closed surface l.with a radius of foundation r.whose axis coincides with the thread (see Fig.1.3). In this case, the stream of tension vector through the base of the cylinder is absent, and the full flow through the closed surface is equal to a stream through the side surface of the cylinder s.:

Because At each point of the side surface of the cylinder, the direction coincides with the direction of normal at this point, and E (R) \u003d ConstT.

Full charge prisoner in the volume of the cylinder q \u003d TL. Then get

Potential of the field of the thread

gleie is permanent FROM determined by the choice of point of zero potential

Example 3. The Radius RF with a spherical cavity of the radius A in the center is charged with a charge density variable by linear law R \u003d A (R - A), where a is a constant value, and R is the distance from the center of the ball. Determine the tension and potential of the electrical field of the ball. Dielectric constant Bowl and environment to take equal to 1.

From spherical symmetry task it follows that the electric field strength depends only on distance r. from the center of the ball, i.e. E \u003d E (R), And the lines of the vector are radial rays emerging from the beginning of the coordinates combined with the center of the ball. At the same time on any spherical surface r \u003d Const. The field strength is equal in the same one at each point of this surface, and the direction coincides with the direction of normal to the surface. Therefore, when using the Gauss Theorem as a closed surface, it is convenient to choose the surface r \u003d Const..

Consider the flow of the electric field strength vector through a spherical surface. r.< a . Because There is no charge in the cavity, the flow of the electric field strength vector through a spherical surface r.< a equal to 0, and therefore electric field There is no cavity.

Let the surface radius, through which the stream of tension vector is calculated, satisfies inequality a.< r < R . Then the stream of tension vector

On the other hand, this flow is accurate e 0 equal to the charge concluded in the volume of a spherical layer limited by the spheres of radii but and r.:

In this way,

And for a.< r < R .

Consider the stream of the vector of electric field strength through the surface of the sphere r\u003e R.. Here is still N. \u003d, and the charge enclosed in the volume under consideration is equal to the total charge of the ball, i.e.

Therefore, when r\u003e R..

Calculate the potential of the ball. Because and for r.< a Е \u003d 0, then j 1 \u003d C 1,

moreover With 1. - constant. In area a.< r < R

Question from Oroksa

Two electric field strength lines cannot intersect at one point. Is it possible to argue that two equipotential surfaces can also be crossed at one point?

Answer:Two equipotential surfaces can intersect at one point, provided that the field strength at this point is zero. In fact, the vector of tension determines the direction of normal to each equipotential surface. Since the surfaces intersect, at the point of intersection, their normal does not coincide, and therefore the electric field cannot have a certain direction. This is possible only with a zero value of the field value.

Computer model demonstrates picture silest lines and equipotential surfaces Point charge and systems of two point charges. You can change the values \u200b\u200bof charges and their signs, as well as the distance between charges. When installing the cursor at any point and press the left mouse button, the computer highlights the display of the module of the electric field strength vector and the potential φ at this point. It should be noted that spherical equipotential surfaces are possible in the electric field of two point charges.

2.7. Examples of potential calculation

Consider the calculation of the potential of the field created by some charged bodies.

Fig.3. The location of the observation point A with respect to the point source of the Q field

1 .Example Point charge. Using relations (1) and (4) and combining point 1 with observation point BUT (Fig. 3), and point 2 removing on infinity, we get

Comment. The concept of vector field was still used - fields of vectors (each point of space is put in line with vector). The performance of the potential as the function of the space point allows you to enter the concept of a scalar field - the fields of numbers (each point of space is put in accordance with an accuracy of an arbitrary additive constant, number). Therefore, in the future, in addition to the concepts, the electrostatic field and the field will use the concept of the field.

2. Example. We find the potential of the endless uniformly charged with a linear filament density.

Our task is to find as a function.

As already shown for infinitely long threads:.

Choose somewhere the point of which we start, for example at the point, then

.

We agree that at the start point, the potential is zero:, then

.

This is with such a normalization. If in such a task is to normalize, then we get, that is, all potentials at all points are infinitely large. Who needs such a normalization? If you are interested in some particular spatial range, then it is necessary to normalize somewhere near this area, then all the potentials will be finite pleasant numbers. From infinite thread, it is impossible to escape from infinity, for the same reason you should not expect a good result from as the amount of potentials of point fragments of the thread, each of which is normalized by the condition.

3. Example. Potential of a spot dipole field (first method).

So, we consider the point dipole and the observation point is far enough from this our dumbbell.

Let's look for the potential of this observation point.

First, the principle of superposition.

Secondly, as you know

Taking into account that there is a projection of the vector on and the fact that the distance to the dipole is very large, then

Potential field of a point dipole (second method).

We use the finished formula for field strength:

. Here is such an expression, we at one time brought. I've been on it.

Let's choose such a direction so that it is convenient for us to calculate this integral, as the field is potential, the result will not depend on the type of trajectory. The orientation of the dipole, the direction of movement and the vector is shown in the figure. Taking into account some relations (5) will take the form.

Consider a field created by charge, evenly distributed over the infinite thread. We decided this task at the last lecture, using the principle of superposition of electric fields (see 1.11).

Now we will show, it is somewhat easier to calculate this field using the Gauss theorem.

Determine the field strength at a distance r.from the thread charged with constant linear density:

, [CL / M] (2.10)

Surrounding the thread with a closed cylindrical surface (Fig. 2.7.). Cylinder height - h., and the radius of its foundation - r..

Fig. 2.7.

The field created by the charged thread has a cylindrical symmetry. In this regard, the voltage vectors at all points of the side surface of the cylinder will be the same by the module and are directed radially, that is, perpendicular to the side surface of the cylinder. On the bases of the cylinder vectors directed by radially, "slide" on the basis, forming a straight angle with the normal .

Calculate flow vector through the surface of the selected cylinder. A full flow through this closed "Gaussian" the surface is made of a stream through the side surface of the cylinder and through two bases:

The last two integrals are zero, since the "sliding" on the bases of the cylinder vectors do not permeate them and do not create any stream. Formally, these two integrals are zero, because between vectors and straight angle I.

. In this way

At all points of the side surface of the cylinder E.=E. r. \u003d cons

.

Therefore, the flow through the side surface of the cylinder is equal

(2.11)

This is the flow of the electric field strength vector calculated by definition of the flow.

Now we use the Gauss theorem, noting the previous thing that "the charge prisonered inside the Gaussian surface" in this case is concentrated on the stretch of the thread h.- on the axis of the cylinder:

In this way

(2.12)

From here is now easy to get familiar to us a hyperbolic dependence of the intensity of the field from the distance to the thread - r.(See 1.11).

(2.13)

1. Field of an infinite uniformly charged plane. Flat condenser field

Let the electric field are created by charge, evenly distributed over the surface of an infinite plane, with a surface density  (Fig. 2.8.)

Fig. 2.8.

From the symmetry of the task it follows that the field is sent everywhere perpendicular to the surface. We find out how the field strength changes as it removes from the charged plane.

As a Gaussian surface, it is convenient to choose a cylinder. The axis of the cylinder will send a perpendicular plane, its base is positioned at a distance H.symmetrically on both sides of the surface.

Calculate the strength of the voltage vector through the side surface and base of the cylinder. As follows from fig. 2.8., Stroy vector stream through the side surface of the cylinder is zero, since here the tension vectors "slide" on the surface and

.

Then the full flow through a closed cylindrical surface can be written as a stream through two bases of the cylinder.

This is the value calculated to determine the flow.

Now we use the Gauss theorem, noting that the charge q., "Located inside the Gaussian surface", in this case focuses on the site S.=S. Osn, "cut" cylinder on an infinite plane

(2.15)

We combine the results (2.15) and (2.14) to the Gauss equation:

Where follows

(2.16)

Output. A field created by an infinite uniformly charged plane, uniformly. It does not change with a distance from the charged surface, nor in the direction.

Now consider another important example. Let the field be created two infinite planes, variemented, but with the same over the size of the surface density of the charge (Fig. 2.9.). This is an important idealization of electrostatics - flat condenser. Each resting of this capacitor creates a homogeneous field, the tension of which we have just installed (2.16):

.

Fig. 2.9.

Power lines of the field of a positively charged plane are directed from it, and negative to the plane. When these fields are addition, the tension of the resulting field outside the condenser turns out to be zero, and inside the condenser, where these fields coincide in the direction - the field doubles:

. (2.17)

Example 1. Thin, infinitely long thread is charged homogeneously with linear charge density λ . Find tensions electrostatic field E.(r.) at random r. From the thread.

Let's make a drawing:

Analysis:

Because The thread carries point charge, Apply the di method. Highlight the infinitely small element of the length of the conductor dLwhich will contain a charge dQ.=dlλ.. Calculate the intensity of the field created by each element of the conductor in an arbitrary point A from the thread at a distance but. The vector will be directed along a straight line connecting the point charge with the observation point. The resulting field is obtained by normal to the thread along the x axis. It is necessary to find the quantity dE X.: dE X \u003d.dE.cOSα. .

A-priory:

.

Value dL, r., change consistently when changing the position of the element dL. Express them through the value of α:

where dα. - infinitely small increment of angle α as a result of rotation of the radius-vector relative to the point A when moving on the thread on dL. Then dL \u003d.r. 2 dα / A.. When moving dL From to point about angle varies from 0 0 to π / 2.

Hence .

Checking dimension: [E] \u003d V / m \u003d kgm / MFM \u003d CLV / CLM \u003d V / m;

Answer:.

Method 2.

By virtue of the axial symmetry of the charge distribution, all points are located at an equal distance from the threads, equivalent and the intensity of the field in them is the same, i.e. E.(r.) \u003d const, where r.- distance from the point of observation to the thread. Direction E. At these points always coincides with the direction of normal to the thread. On the Gauss Theorem; Where Q.- A chance embraced by the surface - S 'through which the flow is calculated, select the radius A in the form of a cylinder and forming the thread. Considering that the side surface of the cylinder is normal, we obtain for stream:

T. K. E.\u003d const.

S. Bok.pov. \u003d. On the2π .

On the other hand E.2ïan \u003d Q / ε 0 ,

where λn \u003d Q..

Answer:E.=λ /4πε 0 but.

Example 2. . Calculate the tension of a uniformly charged endless plane with surface density of charges σ .

Tension lines are perpendicular and directed in both sides of the plane. As a closed surface, we choose the surface of the cylinder, the base of which is parallel to the plane, and the axis of the cylinder is perpendicular to the plane. Because The forming cylinders are parallel to the lines of tension (α \u003d 0, cos α \u003d 1 ), That flow vector strength through the side surface is zero, and a complete stream through a closed cylindrical surface is equal to the amount of flow through its base. The charge enclosed inside the closed surface is Σ S. OSN. , then:

F E \u003d 2 E.S. OUND or F E \u003d \u003d, then E \u003d \u003d

Answer:E \u003d, does not depend on the length of the cylinder and at any distance from the plane the same in the module. Field of a uniformly charged plane uniformly.

Example 3. . Calculate the field of two infinitely charged planes, with a surface density of + σ and -σ, respectively.

E \u003d E \u003d 0; E \u003d E + + E - \u003d.

Answer: The resulting field strength in the region between the planes is equal to E \u003d, and outside the volume bounded by the planes is zero.

Example 4. . Calculate the field strength uniformly charged with the surface density of charge + σ spherical surface of the radius R..

Then, and

if R.< R , то внутри замкнутой поверхности нет зарядов и электростатическое поле отсутствует (Е=0).

Answer:.

Example 5. . Calculate tension volume-charged with bulk density ρ , ball radius R..

In the form of a closed surface, take the sphere.

If a r. ≥R. , then \u003d 4πr 2 E; E \u003d.

if R.< R , то сфера радиусом r., covers the charge Q "equal to q" \u003d (as the charges are treated as volumes, and volumes like cubes of radii)

Then in TGAUS

Answer:; Inside a uniformly charged bowl, tension grows linearly with distance r.from his center, and outside - decreases back proportionally r. 2 .

Example number 6. . Calculate the intensity of the field of an endless, round cylinder charged with a linear charge density λ , radius R..

Thread vector stream Through the cylinder ends is 0, and through the side surface:

Because , or ,

then (if R\u003e R)

if λ\u003e 0, Е\u003e 0, the vector ē is directed from the cylinder,

if λ.< 0, Е < 0 , вектор Ē направлен к цилиндру.

If R.< R, то замкнутая поверхность зарядов внутри не содержит, поэтому в этой области Е = 0

Answer: (R\u003e R); E \u003d 0 (R\u003e R). Inside uniformly charged over the surface of an endless, round cylinder, there is no field.

Example 7. . The electric field was created by two infinitely long parallel planes with surface planes of charges 2 nkl / m 2 and 4ken / m 2. Determine the intensity of the field in the regions І, ІІ, ІІІ. Build a graph of addiction Ē (r.) .

Planes divide space on 3 areas

Direction ē The resulting field towards more.

In the projection on r.:

; «–»; ;

; «–»; ;

; «+»; .

Schedule Ē (r.)

Select scale: E. 2 =2 E. 1

E 1 \u003d 1; E 2 \u003d 2

Answer:E. І \u003d -345 V / m; E. І i \u003d -172 v / m; E. І ii \u003d 345 V / m.

Example number 8. . Ebonite solid ball radius R. \u003d 5 cm is charged, evenly distributed with volumetric density ρ \u003d 10 NL / M 3. Determine the electric field strength at points: 1) at a distance r. 1 \u003d 3 cm from the center of the sphere; 2) on the surface of the sphere; 3) at a distance r. 2 \u003d 10 cm from the center of the sphere.

Consider a field created by charge, evenly distributed over the infinite thread. We decided this task at the last lecture, using the principle of superposition of electric fields (see 1.11).

Now we will show, it is somewhat easier to calculate this field using the Gauss theorem.

Determine the field strength at a distance r. From the thread charged with constant linear density:

, [CL / M] (2.10)

Surrounding the thread with a closed cylindrical surface (Fig. 2.7.). Cylinder height - h., and the radius of its foundation - r..

The field created by the charged thread has a cylindrical symmetry. In this regard, the voltage vectors at all points of the side surface of the cylinder will be the same by the module and are directed radially, that is, perpendicular to the side surface of the cylinder. On the bases of the cylinder, the vectors aimed at the same radially, "slide" on the basis, forming a straight angle with the normal.

Calculate the vector stream through the surface of the selected cylinder. A full flow through this closed "Gaussian" the surface is made of a stream through the side surface of the cylinder and through two bases:

The last two integrals are zero, since the "sliding" for the bases of the cylinder vectors do not permeate them and do not create any stream. Formally, these two integrals are zero, because between vectors and straight angle and. In this way

At all points of the side surface of the cylinder E. =E R. \u003d const and.

Therefore, the flow through the side surface of the cylinder is equal

This is the flow of the electric field strength vector calculated by definition of the flow.

Now we use the Gauss theorem, noting the previous thing that "the charge prisonered inside the Gaussian surface" in this case is concentrated on the stretch of the thread h. - on the axis of the cylinder:

In this way

(2.12)

From here is now easy to get familiar to us a hyperbolic dependence of the intensity of the field from the distance to the thread - r. (See 1.11).

1. Field of an infinite uniformly charged plane. Flat condenser field

Let the electric field are created by charge, evenly distributed over the surface of an infinite plane, with a surface density S (Fig. 2.8.)

From the symmetry of the task it follows that the field is sent everywhere perpendicular to the surface. We find out how the field strength changes as it removes from the charged plane.

As a Gaussian surface, it is convenient to choose a cylinder. The axis of the cylinder will send a perpendicular plane, its base is positioned at a distance H. Symmetrically on both sides of the surface.

Calculate the strength of the voltage vector through the side surface and base of the cylinder. As follows from fig. 2.8., Stroy vector stream through the side surface of the cylinder is zero, since here the tension vectors "slide" on the surface and.

Then the full flow through a closed cylindrical surface can be written as a stream through two bases of the cylinder.

This is the value calculated to determine the flow.

Now we use the Gauss theorem, noting that the charge q., "Located inside the Gaussian surface", in this case focuses on the site S. = S. Osn, "cut" cylinder on an infinite plane

We combine the results (2.15) and (2.14) to the Gauss equation:

Where follows

Output. A field created by an infinite uniformly charged plane, uniformly. It does not change with a distance from the charged surface, nor in the direction.

Now consider another important example. Let the field are created by two infinite planes, charged differently, but with the same size of the charge density (Fig. 2.9). This is an important idealization of electrostatics - flat condenser. Each resting of this capacitor creates a homogeneous field, the tension of which we have just installed (2.16):

.

Power lines of the field of a positively charged plane are directed from it, and negative to the plane. In addition to these fields, the tension of the resulting field outside the condenser turns out to be zero, and inside the condenser, where these fields are coincided in the direction, the field doubles.