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Among school problems in physics, those that are connected with concentric conducting spheres stand alone. These spheres can be charged, grounded, can be in the field of external charges, etc., there are many variations. In the school course of physics, these tasks are among the most difficult. Last but not least, of course, the misunderstanding of this material is due to the inability of teachers to explain it correctly and accessible. So, let's try to figure out what kind of conducting spheres and what they eat.
Potential inner sphere φ 2 is determined by the known relation:
Then the total potential φ in on the surface of the inner sphere is:
The potential on the surface of the outer sphere also consists of two potentials: the inner sphere φ’ 1 and the actual outer sphere φ’ 2 .
The potential of the inner sphere φ’ 1 at a distance R from its center is determined by the well-known relation:
The formula that determines the potential of the outer sphere φ’ 2 on its surface is also well known:
Then the total potential on the surface of the outer sphere is equal to:
Decision. Before joining spheres with a conductor, the charge of the first was:
After the connection, part of the charge from the inner sphere flowed onto the outer sphere. The current stopped at the moment when the potential of the ball became equal to the potential of the outer shell. It is more convenient therefore to seek not the potential of the sphere, but the equal potential of the outer shell. In accordance with the results obtained in the previous problem, this potential is determined by the expression:
where q 1 and q 2 - charges of the ball and outer shell after connecting them with a conductor, respectively. By law of conservation of charge q = q 1 + q 2. After simple transformations, we get:
Let's start with the picture for the solution of the problem:
After grounding the conductive shell, the entire positive charge formed on it due to the phenomenon of electrostatic induction flows to the ground. It only has a negative charge, since it is attracted to the positive charge of the inner sphere
Decision. Knowing the potential of the ball at the initial time and its radius, you can find the charge on it:
Due to the phenomenon electrostatic induction on the outer shell there must be a separation of charge. The negative charge will flow to the inner surface of the shell, the positive charge will flow to the outer surface of the shell (see the figure). The same phenomenon arose in the previous problems, but we did not take it into account. Why? In the condition of the tasks it was indicated that the shell is thin, and such a "run-up" of charges did not lead to any significant change in the configuration of the electrostatic field.
In this task, taking this phenomenon into account is important, since the shell is grounded. After grounding, the positive charge from the sheath drains to the ground, leaving only a negative q 2, since it is attracted to a positive charge q 1 of the inner sphere. The potential of a grounded sheath becomes equal to the ground potential, that is, zero. In this connection and in accordance with the result obtained in solving the first problem, we obtain the equality:
Using the expression for calculating the potential of the inner sphere of a similar system, obtained in the first problem, we find the finally required potential of the ball:
Experience shows that very few people understand the solution of these problems in all the details from the first time. Usually it takes a long time and persistently to explain to students all those little things, without the realization of which the solution is reduced to empty transformations of letter expressions in order to obtain the answer given at the end of the textbook. To understand the physical essence of these tasks and learn how to apply the knowledge gained in the future is not easy. However, this is the main methodological value of this topic in the school physics course. The best assistant in its study will certainly be a professional tutor, a competent tutor who can come up with an understandable explanation for you and answer all the questions that arise. By the way, if there are any, you can ask them below in the comments.
Sergey Valerevich
Electrostatics
1
(CT 2001, Test 9. A19). Conductive ball of radius R
has a positive charge +
q
. If at a distance of 2 R from the center of the ball put a point negative charge -2 q, then the potential at the center of the sphere
A typical mistakewhen the solution comes from an incorrect interpretation of the wording: "the field inside the conducting ball is missing." From this statement an erroneous conclusion is drawn: both characteristics of the field: both the strength and the potential are zero. In reality, in this case, only the field strength is zero; free charges stop moving along the surface of the conductor when the vector 
at any point of the surface is perpendicular to it. The surface of the conductor in this case is equipotential. The work of moving a test charge in a volume bounded by a surface is zero, because the force acting on the charge is zero; from this it follows that the potential energy of the charge when moving from point to point does not change: the electric potential in the volume bounded by the conducting surface is constant and equal to the potential on the surface itself. The change in the strength and potential of a charged sphere with the distance from the charge center can be illustrated by graphs (see Fig.).
Decision
The potential created spherically symmetrically distributed charge q in the center of the ball, the same as on its surface and is equal to φ 1 = q/4 πε 0 R ; the potential created at the same point by the point charge -2 q , located outside the sphere at a distance 2 R from its center, is equal to φ 2 = – 2 q/4 πε 0 2 R; The potential in the center of the sphere is the result of the superposition of two fields, i.e. φ = φ 1 + φ 2 = q/4 πε 0 R–2 q/4 πε 0 2 R= 0 .
When running test tasks, it is convenient to record as short as possible. For example, in this case one general equation φ = q/4 πε 0 r.
Because the point negative charge is twice as large as the charge distributed on the sphere and is twice as far from the center of the sphere, from the recorded equation it is seen that the potentials of both charges are equal in magnitude and opposite in sign, therefore, the resultant potential is 0.
2 (CT 2001 Test 11. A 19).
Inside the spherical metal layer, the inner and outer radii of which are correspondingly equal R
and 2
R
,
on distance R /
2
q
.
Potential in the center of the sphere is ...
Decision
Difficultiescauses the construction of a picture of the distribution of charges on the surfaces of the spherical layer.
Thanks to electrostatic induction on inner surface sphere appears charge –
q
, and on the outer + q
.
Potential in the center of the sphere 
;
; 
; 
– answer 1.
A similar task - A 19 in the test number 12, 2001. :
Inside the spherical metal layer, the inner and outer radii of which are respectively 2 R and 4 R , on distance R from the center there is a point positive charge q . The potential in the center of the sphere is ....
Answer: 
3 (CT 2000, Test ... A19). Metal sphere of radius R 1 , having a potential φ 1 , surrounded by an uncharged spherical conductive shell of radius R 2 . Find the potential of the ball after it is for some time connected to the shell?
Decision
Potential of a charged sphere 
.
If a charged sphere touches the inner surface of the shell, the charges, aiming to settle at as large as possible from each other, pass to the shell. The field strength inside the shell becomes equal to 0, the field potential at the shell points and inside it is equal to 
(but not 0! See the previous problem), where 
. Thus, the potential of the shell and the ball inside and connected to it is equal to 
– answer.
Correct answer -1.
4 (CT 2000, Test ... A19). Metal sphere of radius R 1 , having a potential φ 1 , are surrounded by a spherical conductive shell of radius R 2 . What will be the potential of the ball if the shell is grounded?
Decision
The most frequent errorconsists in the fact that the potential of the field created by charges induced on the shell when it is grounded is not taken into account.
First option solutions. Potential on the surface of the ball created by the charge of the ball 
, is 
. After the sheath is grounded, an induced charge appears on it 
, which creates a potential on the shell and inside it 
.
The superposition of the initial field of the ball and the field created by the induced shell charge gives the potential on the surface of the sphere
– answer.
The second option solutions. It is possible to proceed from the argument that the potential of the shell after grounding is 0,as is accepted in the art (a grounded shell is taken as the reference point of the potential energy), from this condition is the magnitude and sign of the induced charge. Shell potential φ 2 is added from the potential due to the charge induced on it q 2 and the potential of the field of the sphere φ 1 :
, 
q 1
= –
q 2.
.
We find the potential of the ball after the earthing of the shell:
(1).
Of 
we find 
, we substitute in (1), we obtain answer: 
.
This solution has some drawbacks:
- in previously solved problems, the potential (potential energy) was measured from a point infinitely removed from the charge, which has a clear physical meaning; it is always logical to use the same origin;
- the solution turned out to be more cumbersome.
5
(CT 2001 Test 3. A19). Thin fixed radius ring R is uniformly charged so that per unit length of the ring there is a charge + γ .
In a vacuum on the axis of the ring at a distance l from its center is placed a small ball, which has a charge of + q. If the ball is released, then in the process of motion it will acquire the maximum kinetic energy equal to
| 1)  | 2)  | 3)  |
| 4)  | 5)  |
R 
resolution
By the law of conservation of energy E = U, where U Is the interaction energy of a point charge and a ring.
A typical error:when solving this problem, it is believed that, because of the symmetry of the charge distribution over the ring, it is possible to "pull" it to the center of the ring and find the potential of the field created by the ring at the location of the charge as the potential of the point charge field: 
. However, this can not be done, since the symmetry of the charge is not spatial, but plane. In fact, it is necessary to break the ring into small elements, which can be considered as material points, to determine the field potential of each such point charge at the location of the charge + q
and sum the results:
The energy of interaction of a ring with a charge and the maximum kinetic energy of a charge are equal, according to the law of conservation of energy: 
- the answer.
Similarly, the potential is found in solving problem A 19 from test No. 8 of the 2001 CT: a uniformly distributed charge of 10 -9 cells is uniformly distributed over a thin wire ring of radius 3 cm. Determine the potential difference between the center of the ring and the point located on the axis of the ring at a distance of 4 cm from the center. Answer: 120 V.
6
(CG 2000, Test 3. A 20). If the metal sphere of radius R
1
, charged to potential φ
1
, connect the thin
wire with an uncharged metal sphere of radius R
2, the total bond potential is equal to
| 1)  | 2)  | 3)  |
| 2)  | 3)  |
|
| 4)  | 5)  |
7
(CT 2001. Test 2. A 
19). What kind of work must be done to make three identical point positive charges q, which are in a vacuum along one straight line at a distance a
from each other, placed in the vertices of an equilateral triangle with a side a/2.
7.1. There are two electrodes in the form of concentric spheres with radii a (internal) and b (external). Such a system is called a ball condenser. Find the potential of any point of the field between the electrodes.
7.2. Calculate the potential electric field dipole.
7.3. Find the field potential of a system of charges in a volume with linear dimensions l, at distances.
7.4. Draw a potential diagram of a system of two charged spheres.
7.5. Calculate the potential of the field of a sphere with a radius a, uniformly charged in volume: a) inside the ball; b) outside the sphere. Draw a graph 
, where r is the distance from the center of the sphere. Solve the problem by integrating the Poisson equation in spherical coordinates, and also using the relationship between the field strength and the potential.
7.6. On a thin wire ring of radius R the charge is uniformly distributed q. Investigate the dependence of the potential of the electric field on the axis of the ring from the distance to its center. Find the tension as a gradient of the potential.
7.7. Sphere of radius 
, uniformly charged by a charge 
, is surrounded by a thin concentric sphere of radius 
. What a charge 
it is necessary to inform the external sphere that the potential of the inner sphere with respect to infinity should vanish? Charge also uniformly distributed over its surface.
Problem Solving
and, consequently, varies in space in the same way as in the case of a point charge field, whence it follows that the potential difference between the inner sphere and some point of the field remote from the distance r from the center of the capacitor is
.
Potential difference 
between the electrodes (spheres) will be
.
Of these two formulas follows
.
Measuring between the electrodes, we can calculate the potential of any point of the field.
7.2. Let the system of charges 
is in the volume with linear dimensions l. Let us find the potential of the field created by this system of charges at distances r large compared to l. We choose the origin O inside the volume occupied by the system of charges, and determine the position of the charges by means of radii-vectors 
(in the figure ___ one of the radius vectors is shown 
charge). Potential at a point determined by the radius vector 
, is
.
As 
, then we can assume that
(symbol 
we have denoted the unit vector). Then
.
We use the formula
at 
.
Now we can write down
The first term of this expression is the potential of the point charge field by the quantity 
. The second term is of the same form as the expression that determines the dipole field potential. The role of the electric moment of a dipole is played by the quantity
,
which is called the dipole electric moment of the charge system.
7.3. If the field is created by several charges, then the potential of this field is equal to the sum of the potentials of the fields created by separate charges
Here 
- potential of the resulting field at the point under consideration with respect to infinity, 
- the distance from this point to charge, and the summation is over all point charges.
The field under consideration has axial symmetry, so the picture of the field in any plane passing through the dipole axis will be the same, and the vector 
lies in this plane. The position of the point M with respect to the dipole will be characterized by the radius vector , or with the help of polar coordinates r and 
. Charge position 
with respect to the center of the dipole is defined by the vector 
, and charge 
- vector 
. It's obvious that 
, where 
- Dipole shoulder. Distances from charges and up to the selected point M will be denoted respectively 
and 
. As 
, then we can assume that
Potential at a point determined by the radius vector , is
.
Composition 
, the difference 
. Consequently,
,
where 
Is the electric moment of the dipole.
It is seen from this formula that the potential of the dipole field is determined by its electric moment. Comparing the potential of the dipole field with the potential of the field of the point charge, it is seen that the dipole field potential decreases with distance faster 
, than the potential of the point charge field 
.
From figure ___ it is visible, that 
. therefore
.
7.4. Let the inner sphere, the radius of which , has a positive charge , and the outer one with a radius - negative charge 
, and 
.
Outside the spheres, the potential will be
,
since it is created jointly by both spheres (the potential is the work of external forces performed when moving a unit positive charge from infinity to a given point of the field). The work on the displacement of a single positive charge from infinity to the region between spheres will be equal to the sum of two works: 
(work against forces acting from the outer sphere on the way from infinity to its surface) and 
(work against the field of the inner sphere), i.e.
.
Within the smaller sphere, the potential will be constant and equal to
.
The graph, constructed according to the first and second formulas, is shown in Figure ____1.
If the charges of the spheres are equal in magnitude and opposite in sign, i.e. 
(such a system is called a spherical condenser), then the potential in the outer region vanishes, and between the plates is equal to
.
The graph shown in figure ___2 is obtained.
If the inner sphere has a negative charge, and the outer one - a positive one, then the graph turns over and looks like it is depicted in figure ____3.
7.5. Integration of the Poisson equation in spherical coordinates. We introduce a spherical coordinate system 
,
,, taking the center of the ball as the origin. The Gaussian equation in differential form (Poisson's equation), which determines the field potential, takes the form
where 
.
Due to the spherically symmetric charge distribution, the potential depends only on the distance r and is independent of the angles and i.e. . Therefore, the Poisson equation simplifies and takes the form
.
Here through 
the potential inside the sphere is denoted, and 
- outside the ball. Integrating these equations, we find
,
.
The constants A, B, C, D must be determined from the following boundary conditions.
1) The potential 
it is necessary to remain finite when 
, from which it immediately follows that 
.
2)
at 
, from which it follows that 
.
3) The potential of the electrostatic field is a continuous function of the coordinates, so it is necessary that 
.
4) The normal component of the vector it should not experience a jump as it passes through the surface of the sphere, i.e. 
at 
, since the surface charge density on the surface of the sphere is zero. The last condition is equivalent to the requirement
.
From the last two conditions we find
,
.
The required potentials are finally written in the form
,
.
From these formulas it is seen that outside the sphere the field potential is analogous to the field of a point charge.
Draw a graph .
Let 
,
. Let us find the volume charge density
and attitude
.
Now we can write down that the potential inside the sphere
Let's make the table
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Potential outside the sphere 
.
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Now we are plotting .
Relationship between tension and potential.
The dependence of the intensity of the electrostatic field on the distance to the center of the sphere inside the sphere has the form (see the solution of the problem 1.5.4.)
,
those. Within the sphere, the field strength increases linearly with the distance from the center. When 
,
, with it reaches a maximum and becomes equal to
.
When 
the field strength depends on the distance as the field strength of the point charge.
Changing potential in the field of a charged sphere
The potential of the field inside the sphere
,
where - the potential of a point on the surface of the sphere (the potential of the field of a point charge), equal to
.
Finally we get
.
Taking into account that the volume charge density
,
can be written
,
those. we arrived at the same formula as for solving the problem by integrating the Poisson equation.
1.7.6. The potential of the resulting field at point A
,
there is a potential field created by the charge 
element of the ring 
.
is the linear charge density, r is the distance from the element up to the specified point. From the last two formulas we have
.
Resultant potential
From geometric considerations it follows that
.
Consequently,
.
Field strength
.
Expression Analysis 
and 
shows that in the center of the ring ( 
) the potential has a maximum value, and the field strength vanishes.
When 
and the potential and tension tend to zero.
When 
derivative 
vanishes, therefore, at this point the field strength is maximal, and on the graph 
(see Fig. ____) there will be an inflection point. Schedule is located in the 1 st and 3 rd quarters, i.е. 
,
. This means that when passing through the center of the ring ( ) vector changes its direction to the opposite.
Schedule is located in the 1 st and 2 nd quarters, i.е. on both sides of the ring at points lying on its axis, the potential is positive.
Using the example of solving this problem, we can verify that when the origin of the potential is changed, the potential difference between any two points does not change. The whole character of the dependence of the potential on the distance does not change either. For example, if you select the origin in the center of the ring, i.e. if we assume that 
, then the potential of any point lying on the axis of the ring is equal to
.
This formula can be easily obtained on the basis of the superposition principle.
If the origin of the potential is chosen in the center of the ring, then the potential of the field created by the elementary charge at the point A, can be represented in the form
.
Integrating this expression over the entire ring, we obtain the formula
.
Dependency graph , without changing its character, is shifted down parallel to itself by an amount 
(dotted line in figure _____2). When potential tends to value 
.
7.7. The potential is numerically equal to the work done by the forces of the electric field when one positive charge is transferred from a given point of the field (in our case, from the surface of the inner sphere) to infinity, i.e.
,
where - the resultant field strength at all points of the integration interval.
In the interval 
the field is created only by the charge of the inner sphere. Vector , regardless of the magnitude and sign of the charge , is directed along the radius from the center. When moving a unit positive charge from before the forces of the field are doing a positive job. When 
i.e. outside the second sphere the work of the field forces is negative and, consequently, the vector it is directed along the radius to the center of the sphere. At points field is determined by the algebraic sum of the charges on both spheres. Charge must be negative and should be larger in magnitude than the charge . Since the vectors and 
are collinear (or anticollinear with ), then the scalar product 
can be replaced by a product 
(in the case when the two vectors are directed in the opposite direction, the field strength must be regarded as negative). In the formula integrand 
suffers a break in the point 
. Therefore, the integral must be divided into two integrals ranging from before and from before 
:
.
When tensions
,
while 
.
Substituting these expressions into the corresponding integrals, we obtain
.
Integrating and deriving similar terms, we obtain
.
Since by the hypothesis of the problem 
, then
.
Dependency graph 
is shown in figure ____.
Let us analyze the obtained graph.
According to the condition of the problem, for a given value potential on the surface of the inner sphere . When 
the potential is constant and equal to the potential on the surface, hence, the plot on the section from before 
is a straight line coinciding with the axis of abscissae. When vector suffers a break. As 
, then the dot (as well as the point ) represent singular points. Location on vector is directed along the radius vector . Therefore, as the distance from the surface of the inner sphere decreases, the potential decreases to a certain value 
. Location on vector is directed towards the radius vector , therefore, as the distance from the surface of the outer sphere increases, the potential increases, and when 
. Despite the fact that in points and vector suffers a gap, function is continuous.
