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)
Electric field can be represented graphically in two ways, complementary to each other: using equipotential surfaces and lines of tension (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 field lines electric field.
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 thicker power 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. ∆x = - ∆ 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 direct current, flowing through a conductive 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.
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
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 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 ($\varphi \left(\infty \right)=0$):
\[\varphi =\frac(1)(4\pi \varepsilon (\varepsilon )_0)\frac(q)(r)\left(1\right).\]
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
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 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).
For greater clarity, the electric field is often depicted using field lines and equipotential surfaces.
Power lines – these are continuous lines, the tangents to which at each point through which they pass coincide with the electric field strength vector (Fig. 1.5). The density of field lines (the number of field lines passing through a unit area) is proportional to the electric field strength.
Equipotential surfaces (equipotentials) – surfaces of equal potential. These are surfaces (lines) on which the potential does not change when moving. Otherwise, the potential difference between any two equipotential points is zero. The lines of force are perpendicular to the equipotentials and directed in the direction of decreasing potential. This follows from equation (1.10).
Let us consider as an example an electric field created at a distance 
from a point charge. According to (1.11,b) the intensity vector coincides with the direction of the vector 
, if the charge is positive, and opposite to it if the charge is negative. Consequently, the field lines diverge radially from the charge (Fig. 1.6, a, b). The density of field lines, like tension, is inversely proportional to the square of the distance ( 
) to charge. The electric field equipotentials of a point charge are spheres centered at the location of the charge.
1.6. Gauss's theorem for electric field in vacuum
The main task of electrostatics is the problem of finding the intensity and potential of the electric field at each point in space. In section 1.4, we solved the problem of the field of a point charge, and also considered the field of a system of point charges. In this section we will talk about a theorem that allows you to calculate the electric field of more complex charged objects. For example, a charged long thread (straight), a charged plane, a charged sphere and others. Having calculated the electric field strength at each point in space using equations (1.12) and (1.13), we can calculate the potential at each point or the potential difference between any two points, i.e. solve the basic problem of electrostatics.
For a mathematical description, we introduce the concept of intensity vector flow or electric field flow. Flux (F) vector 
electric field through a flat surface area 
the quantity is called:
,
(1.16)
– electric field strength, which is assumed to be constant within the site 
;
– angle between vector direction 
and unit normal vector 
to the site 
(Fig. 1.8). Formula (1.16) can be written using the concept of a scalar product of vectors:
. (1.15,a)
In the case when the surface 
not flat, to calculate the flow it must be divided into small parts 
, which can be approximately considered flat, and then write down expression (1.16) or (1.16,a) for each piece of surface and add them. In the limit when the surface S i very small ( 
), such a sum is called a surface integral and is denoted 
. Thus, the flow of the electric field strength vector through an arbitrary surface 
is determined by the expression:
.
(1.17)
As an example, consider a sphere of radius 
, the center of which is a positive point charge 
, and determine the electric field flow through the surface of this sphere. The lines of force (see, for example, Fig. 1.6, a) emerging from the charge are perpendicular to the surface of the sphere, and at each point of the sphere the modulus of the field strength is the same
.
Area of a sphere 
,
Then 
.
Magnitude 
and represents the flow of electric field through the surface of the sphere. Thus, we get 
. It can be seen that the flow of electric field through the surface of the sphere does not depend on the radius of the sphere, but depends only on the charge itself 
. Therefore, if you draw a series of concentric spheres, then the flow of the electric field through all these spheres will be the same. Obviously, the number of field lines crossing these spheres will also be the same. It was agreed that the number of lines of force emerging from the charge should be equal to the flow of the electric field: 
.
If the sphere is replaced by any other closed surface, then the electric field flux and the number of field lines crossing it will not change. In addition, the electric field flux through a closed surface, and hence the number of field lines penetrating this surface, is equal to 
not only for the field of a point charge, but also for the field created by any collection of point charges, in particular by a charged body. Then the value 
should be considered as the algebraic sum of the entire set of charges located inside a closed surface. This is the essence of Gauss's theorem, which is formulated as follows.
The flux of the electric field strength vector through an arbitrary closed surface, inside of which there is a system of charges, is equal to 
, Where 
is the algebraic sum of these charges.
Mathematically, the theorem can be written as
.
(1.18)
Note that if on some surface S vector 
constant and parallel to the vector 
, then the flow through such a surface. Transforming the first integral, we first took advantage of the fact that the vectors 
And 
parallel, which means 
. Then they took out the value 
for the sign of the integral due to the fact that it is constant at any point on the sphere 
. When applying Gauss's theorem to solve specific problems, they specifically try to choose a surface for which the conditions described above are satisfied as an arbitrary closed surface.
Let us give several examples of the application of Gauss's theorem.
Example 1.2. Calculate the electric field strength of a uniformly charged endless thread. Determine the potential difference between two points in such a field.Solution. Let us assume for definiteness that the thread is positively charged. Due to the symmetry of the problem, it can be argued that the lines of force will be straight lines radiating from the axis of the thread (Fig. 1.9), the density of which decreases according to some law as they move away from the thread. According to the same law, the magnitude of the electric field will also decrease 
. Equipotential surfaces will be cylindrical surfaces with an axis coinciding with the thread.
Let the charge per unit length of the thread be equal to 
. This quantity is called linear charge density and is measured in SI units [C/m]. To calculate the field strength, we apply the Gauss theorem. To do this, as an arbitrary closed surface 
choose a cylinder of radius 
and length 
, the axis of which coincides with the thread (Fig. 1.9). Let us calculate the electric field flux through the surface area of the cylinder. The total flow is the sum of the flow through lateral surface cylinder and flow through the bases
However, 
, since at any point on the bases of the cylinder 
. It means that 
at these points. Flow through the side surface 
. According to Gauss' theorem, this total flux is 
. Thus, we got
.
The sum of the charges located inside the cylinder can be expressed through the linear charge density 
:
. Considering that 
, we get
,
,
(1.19)
those. the intensity and density of the electric field lines of a uniformly charged endless thread decreases in inverse proportion to the distance ( 
).
Let's find the potential difference between points located at distances 
And 
from the thread (belonging to equipotential cylindrical surfaces with radii 
And 
). To do this, we use the connection between the electric field strength and the potential in the form (1.9, c): 
. Taking into account expression (1.19), we obtain a differential equation with separable variables:
.
Solution. The electric field of a uniformly charged plane is shown in Fig. 1.10. Due to symmetry, the lines of force must be perpendicular to the plane. Therefore, we can immediately conclude that the density of the lines, and, consequently, the electric field strength will not change with distance from the plane. Equipotential surfaces are planes parallel to a given charged plane. Let the charge per unit area of the plane be 
. This quantity is called surface charge density and is measured in SI units [C/m2].
Let's apply Gauss's theorem. To do this, as an arbitrary closed surface 
choose a cylinder of length 
, the axis of which is perpendicular to the plane, and the bases are equidistant from it (Fig. 1.10). Total electric field flux 
. The flux through the side surface is zero. The flux through each of the bases is 
, That's why 
. By Gauss's theorem we get:
.
The sum of the charges inside the cylinder 
, we find through the surface charge density 
:
. Then where from:
.
(1.20)
From the resulting formula it is clear that the field strength of a uniformly charged plane does not depend on the distance to the charged plane, i.e. at any point in space (in one half-plane) is the same both in magnitude and direction. This field is called homogeneous. The lines of force of a uniform field are parallel, their density does not change.
Let us find the potential difference between two points of a uniform field (belonging to equipotential planes 
And 
, lying in the same half-plane relative to the charged plane (Fig. 1.10)). Let's direct the axis 
vertically upward, then the projection of the tension vector onto this axis is equal to the modulus of the tension vector 
. Let's use equation (1.9):
.
Constant value 
(the field is homogeneous) can be taken out from under the integral sign: 
. Integrating, we get: . So, the potential of a uniform field depends linearly on the coordinate.
The potential difference between two points of the electric field is the voltage between these points ( 
). Let us denote the distance between equipotential planes 
. Then we can write that in a uniform electric field:
.
(1.21)
Let us emphasize once again that when using formula (1.21), we must remember that the quantity 
- not the distance between points 1 and 2, but the distance between the equipotential planes to which these points belong.
Example 1.4. Calculate the electric field strength of two parallel planes uniformly charged with surface charge densities 
And 
.
Solution. Let's use the result of Example 1.3 and the principle of superposition. According to this principle, the resulting electric field at any point in space 
, Where 
And 
- electric field strengths of the first and second planes. In the space between vector planes 
And 
are directed in one direction, so the modulus of the resulting field strength. In external space vector 
And 
directed in different directions, therefore (Fig. 1.11). Thus, the electric field exists only in the space between the planes. It is homogeneous, since it is the sum of two homogeneous fields.
, and the radius of the sphere is 
.Solution. Due to the symmetry of the charge distribution, the field lines should be directed along the radii of the sphere.
Consider an area inside a sphere. As an arbitrary surface 
choose a sphere of radius 
, the center of which coincides with the center of the charged sphere. Then the electric field flow through the sphere S:
. Sum of charges inside the sphere 
radius 
is equal to zero, since all charges are located on the surface of a sphere of radius 
. Then, by Gauss's theorem: 
. Because the 
, That 
. Thus, there is no field inside a uniformly charged sphere.
Let's consider a region outside the sphere. As an arbitrary surface 
choose a sphere of radius 
, the center of which coincides with the center of the charged sphere. Electric field flow through a sphere 
:
. The sum of the charges inside the sphere is equal to the total charge 
charged sphere radius 
. Then, by Gauss's theorem: 
. Considering that 
, we get:
.
Let's calculate the electric field potential. It is more convenient to start from the outer area 
, since we know that at an infinite distance from the center of the sphere the potential is taken equal to zero. Using equation (1.11,a) we obtain a differential equation with separable variables:
.
Constant 
, because the 
at 
. Thus, in external space ( 
):
.
Points on the surface of a charged sphere ( 
) will have potential 
.
Consider the area 
. In this area 
, therefore from equation (1.11,a) we obtain: 
. Due to the continuity of the function 
constant 
should be equal to the potential value on the surface of the charged sphere: 
. Thus, the potential at all points inside the sphere is: 
.