Potential difference between two capacitors. Capacitors

Electrical capacity

When a charge is imparted to a conductor, a potential φ appears on its surface, but if the same charge is communicated to another conductor, then the potential will be different. It depends on the geometric parameters of the conductor. But in any case, the potential φ is proportional to the charge q.

The SI unit of capacitance is farad. 1 F = 1Cl / 1V.

If the potential of the surface of the ball

(5.4.3)

(5.4.4)

More often in practice, smaller units of capacitance are used: 1 nF (nanofarad) = 10 –9 F and 1pcF (picofarad) = 10 –12 F.

There is a need for devices that store charge, and solitary conductors have a small capacity. Empirically, it was found that the electrical capacity of a conductor increases if another conductor is brought to it - due to electrostatic induction phenomena.

Capacitor Are two conductors called covers located close to each other .

The design is such that the external bodies surrounding the capacitor do not affect its electrical capacity. This will be done if the electrostatic field is concentrated inside the capacitor, between the plates.

Capacitors are available in flat, cylindrical and spherical capacitors.

Since the electrostatic field is inside the capacitor, the lines of electrical displacement start at the positive plate, end at the negative plate, and do not disappear anywhere. Consequently, the charges on the plates opposite in sign, but equal in magnitude.

The capacitance of a capacitor is equal to the ratio of the charge to the potential difference between the capacitor plates:

(5.4.5)

In addition to capacitance, each capacitor is characterized by U slave (or U etc . ) - maximum allowable voltage, above which breakdown occurs between the capacitor plates.

Connecting capacitors

Capacitive batteries- combinations of parallel and series connections of capacitors.

1) Parallel connection of capacitors (fig.5.9):

V in this case common is tension U:

Total charge:

Resulting capacity:

Compare with parallel connection of resistances R:

Thus, when capacitors are connected in parallel, the total capacitance is

The total capacity is greater than the largest capacity in the battery.

2) Series connection of capacitors (fig.5.10):

Common is charge q.

Or , from here

(5.4.6)

Compare with serial connection R:

Thus, when capacitors are connected in series, the total capacity is less than the smallest capacity included in the battery:

Calculation of capacities of various capacitors

1.Capacity flat capacitor

Field strength inside the capacitor (Figure 5.11):

Voltage between plates:

where is the distance between the plates.

Since the charge, then

(5.4.7)

As you can see from the formula, the dielectric constant of a substance has a very strong effect on the capacitance of a capacitor. This can be seen experimentally: we charge the electroscope, bring a metal plate to it - we got a capacitor (due to electrostatic induction, the potential has increased). If a dielectric with ε is introduced between the plates, more than that of air, then the capacitance of the capacitor will increase.

From (5.4.6) it is possible to obtain the units of measurement ε 0:

(5.4.8)

2. Cylindrical capacitor capacity

The potential difference between the plates of the cylindrical capacitor shown in Figure 5.12 can be calculated using the formula:

A large number of capacitors that are used in technology are close in type to a flat capacitor. This is a capacitor, which consists of two parallel conducting planes (plates), which are separated by a small gap filled with a dielectric. Charges of equal magnitude and opposite in sign are concentrated on the plates.

Electric capacity of a flat capacitor

The electrical capacitance of a flat capacitor is very simply expressed through the parameters of its parts. By changing the area of the capacitor plates and the distance between them, it is easy to make sure that the electric capacitance of a flat capacitor is directly proportional to the area of its plates (S) and inversely proportional to the distance between them (d):

The formula for calculating the capacity of a flat capacitor is easy to obtain using theoretical calculations.

Let us assume that the distance between the capacitor plates is much less than their linear dimensions. Then the edge effects can be neglected, and the electric field between the plates can be considered uniform. The field (E), which is created by two infinite planes carrying the same modulus and opposite charge, separated by a dielectric with a dielectric constant, can be determined using the formula:

where is the density of charge distribution over the plate surface. The potential difference between the considered capacitor plates located at a distance d will be equal to:

Substitute right side expressions (3) instead of the potential difference in (1) taking into account that, we have:

The energy of the field of a flat capacitor and the force of interaction of its plates

The formula for the field energy of a flat capacitor is written as:

where is the volume of the condenser; E is the field strength of the capacitor. Formula (5) connects the energy of the capacitor with the charge on its plates and the field strength.

The mechanical (pondemotor) force with which the plates of a flat capacitor interact with each other can be found using the formula:

In expression (6), minus indicates that the capacitor plates are attracted to each other.

Examples of problem solving

EXAMPLE 1

Exercise

What is the distance between the plates of a flat capacitor, if at a potential difference B, the charge on the capacitor plate is equal to C? The area of the plates, the dielectric in it is mica ().

Solution

The capacitance of a capacitor is calculated using the formula:

From this expression we obtain the distance between the plates:

The capacity of any capacitor is determined by the formula:

where U is the potential difference between the capacitor plates. Substituting the right-hand side of expression (1.3) instead of capacity into formula (1.2), we have:

Let's calculate the distance between the plates ():

Answer

EXAMPLE 2

Exercise

The potential difference between the plates of a flat air condenser is V. The area of the plates is equal, the distance between them is

m. What is the energy of the capacitor and what will it be equal to if the plates are moved apart to a distance

m. Please note that the voltage source is not switched off when the plates are extended.

Solution

Let's make a drawing.

The energy of the electric field of a capacitor can be found using the expression:

Since the capacitor is flat, its electrical capacitance can be calculated as:

7.6. Capacitors

7.6.3. Change in electrical capacity capacitor and capacitor bank

The capacitance of a capacitor can be changed by increasing or decreasing the distance between its plates, replacing the dielectric in the space between them, etc. In this case, it turns out to be decisive whether the capacitor is disconnected or connected to the voltage source.

If a capacitor (or capacitor bank):

connected to a voltage source, then the potential difference (voltage) between the plates of the capacitor remains unchanged and equal to the voltage at the poles of the source:

U = const;

disconnected from the voltage source, then the charge on the capacitor plates remains unchanged:

Q = const.

When connecting to each other eponymous covers two charged capacitors, their parallel connection.

U = Q total C total,

where Q total is the charge of the capacitor bank; C total - electrical capacity of the battery;

C total = C 1 + C 2,

where C 1 is the electrical capacity of the first capacitor; C 2 - electrical capacity of the second capacitor;

total charge

Q total = Q 1 + Q 2,

When connecting to each other dissimilar covers two charged capacitors take place (as in the case of connecting the plates of the same name) their parallel connection.

The parameters of such a capacitor bank are calculated as follows:

capacitor bank voltage

U = Q total C total,

where Q total is the charge of the capacitor bank; C total - battery capacity;

electrical capacity of the capacitor bank

C total = C 1 + C 2,

where C 1 - electrical capacity of the first capacitor; C 2 - electrical capacity of the second capacitor;

total charge

Q total = | Q 1 - Q 2 |,

where Q 1 is the initial charge of the first capacitor, Q 1 = C 1 U 1; U 1 - voltage (potential difference) between the plates of the first capacitor before connection; Q 2 - the initial charge of the second capacitor, Q 2 = C 2 U 2; U 2 - voltage (potential difference) between the plates of the second capacitor before connection.

Example 17. Two capacitors of the same electrical capacity are charged to a potential difference of 120 and 240 V, respectively, and then connected with the same charged plates. What will be the potential difference between the plates of the capacitors after the specified connection?

Solution . Before connecting the capacitor plates of the same name, each of them had a charge:

first capacitor -

second capacitor -

When connecting plates of the same name, we get a parallel connection of capacitors. The potential difference between the plates of the capacitor bank is determined by the formula

U = Q total C total,

The total charge of the battery of two capacitors, obtained by connecting their plates of the same name, is determined by the sum of the charges of each of them:

Q total = Q 1 + Q 2,

U = Q total C total = Q 1 + Q 2 2 C = C U 1 + C U 2 2 C = U 1 + U 2 2.

Let's calculate:

U = 120 + 240 2 = 180 V.

The potential difference between the plates of the capacitors after the specified connection will be 180 V.

Example 18. Two identical flat capacitors are charged to a potential difference of 200 and 300 V. Determine the potential difference between the plates of the capacitors after connecting their opposite plates.

Solution . Before connecting dissimilar capacitor plates, each of them had a charge:

first capacitor -

Q 1 = C 1 U 1 = CU 1,

where C 1 is the electrical capacity of the first capacitor, C 1 = C; U 1 is the potential difference between the plates of the first capacitor;

second capacitor -

Q 2 = C 2 U 2 = CU 2,

where C 2 is the electrical capacity of the second capacitor, C 2 = C; U 2 is the potential difference between the plates of the second capacitor.

When connecting opposite plates, we obtain a parallel connection of capacitors. The potential difference between the plates of the capacitor bank is determined by the formula

U = Q total C total,

where Q total is the total battery charge; C total - the total electrical capacity of the battery.

The total charge of the battery of two capacitors, obtained by connecting their opposite plates, is determined by the modulus of the difference in charges of each of them:

Q total = | Q 1 - Q 2 |,

and the total electrical capacity of a battery of two identical capacitors connected in parallel is

C total = C 1 + C 2 = 2C.

Therefore, the potential difference between the battery plates is determined by the expression

U = Q total C total = | Q 1 - Q 2 | 2 C = | C U 1 - C U 2 | 2 C = | U 1 - U 2 | 2.

Let's calculate:

U = | 200 - 300 | 2 = 50 V.

The potential difference between the plates of the capacitors after the specified connection will be 50 V.

Example 19. A flat air capacitor charged to 180 V and disconnected from the voltage source. An uncharged metal plate is introduced into the space between its plates, parallel to them, the thickness of which is 3 times less than the distance between the plates. Assuming that the metal plate is located symmetrically with respect to the capacitor plates, determine the potential difference that will be established between them.

Solution . When a metal plate is placed in a flat capacitor as shown in the figure, free electrons in the metal are redistributed:

the plane facing the positively charged plate of the capacitor receives an excess of electrons and is charged with a negative charge q 1 = −q;

the plane facing the negatively charged plate of the capacitor has a lack of electrons and is charged positive charge q 2 = + q.

As a result of charge redistribution, the plate remains neutral:

Q = q 1 + q 2 = −q + q = 0.

The redistribution of charge in the metal plate leads to the formation of a battery of two capacitors:

the positively charged plate of the capacitor and the negatively charged plane of the metal plate have the same modulo charges of the opposite sign; they can be considered as a capacitor with electrical capacity

C 1 = ε 0 S d 1,

where ε 0 is an electrical constant, ε 0 = 8.85 ⋅ 10 −12 Cl 2 / (N ⋅ m 2); S is the area of the capacitor plate; d 1 - the distance between the positively charged capacitor plate and the negatively charged plane of the metal plate;

the negatively charged plate of the capacitor and the positively charged plane of the metal plate also have the same modulo charges of the opposite sign; they can be considered as a capacitor with electrical capacity

C 2 = ε 0 S d 2,

where d 2 is the distance between the negatively charged capacitor plate and the positively charged plane of the metal plate.

Both capacitors have the same charge and form a series connection. The electric capacity of a battery of two capacitors in series connection is determined by the formula

1 C total = 1 C 1 + 1 C 2, or C total = C 1 C 2 C 1 + C 2.

With a symmetrical arrangement of the plate in the space between the capacitor plates (d 1 = d 2 = d), the capacitances of the capacitors are the same:

C 1 = C 2 = ε 0 S d,

the total electric capacity of the battery is given by the expression

C total = C 1 C 2 C 1 + C 2 = C 2 = ε 0 S 2 d,

where d = (d 0 - a) / 2; d 0 - the distance between the plates of the capacitor before the introduction of the plate; a is the thickness of the metal plate.

Potential difference between the battery plates

U = Q total C total = 2 d q ε 0 S = q (d 0 - a) ε 0 S,

where Q total is the battery charge of series-connected capacitors, Q total = q.

The initial potential difference is determined by the formula

U 0 = Q 0 C 0 = Q 0 d 0 ε 0 S,

where Q 0 is the charge of the capacitor before the introduction of the plate, Q 0 = q (the capacitor is disconnected from the voltage source); C 0 - electrical capacity of the capacitor before the plate is inserted.

The ratio of the potential difference before and after the introduction of the metal plate is determined by the expression

U U 0 = d 0 - a d 0.

From here we find the required potential difference

U = U 0 d 0 - a d 0.

Taking into account d 0 = 3a, the expression takes the form:

U = U 0 3 a - a 3 a = 2 3 U 0.

Let's calculate:

U = 2 3 ⋅ 180 = 120 V.

As a result of the introduction of a metal plate into the capacitor, the potential difference between its plates decreased and amounted to 120 V.

Example 20. A flat air capacitor is charged to 240 V and disconnected from the voltage source. It is vertically immersed in some liquid with a dielectric constant of 2.00 per one third of its volume. Find the potential difference that is established between the capacitor plates.

Solution . When a flat air capacitor is partially immersed in a liquid dielectric, as shown in the figure, free electrons on its plates are redistributed in such a way that:

part of the capacitor plates immersed in the dielectric has a charge q 1;
part of the capacitor plates remaining in the air has a charge q 2.

As a result of the redistribution of the charge over the area of the capacitor plates, a charge is established on its plates:

Q total = q 1 + q 2.

The area of the capacitor plates, when it is partially immersed in a liquid dielectric, is divided into two parts:

the part immersed in the dielectric has an area S 1; the corresponding part of the capacitor can be considered as a separate capacitor with electrical capacity

C 1 = ε 0 ε S 1 d,

where ε 0 is an electrical constant, ε 0 = 8.85 ⋅ 10 −12 Cl 2 / (N ⋅ m 2); ε is the dielectric constant of the capacitor; d is the distance between the capacitor plates;

the part remaining in the air has an area of S 2; the corresponding part of the capacitor can be considered as a separate capacitor with electrical capacity

C 2 = ε 0 S 2 d.

Both capacitors have the same potential difference between the plates and form a parallel connection. The electrical capacity of a battery of two capacitors in parallel connection is determined by the formula

C total = C 1 + C 2 = ε 0 ε S 1 d + ε 0 S 2 d = ε 0 d (ε S 1 + S 2),

and the charge on the battery plates is

Q total = C total U = ε 0 d (ε S 1 + S 2) U,

where U is the potential difference between the battery plates.

The electrical capacity of a capacitor before immersing it in a dielectric is determined by the expression

C 0 = ε 0 S 0 d,

and the charge on its plates is

Q 0 = C 0 U 0 = ε 0 S 0 d U 0,

where U 0 - the potential difference between the plates of the capacitor before the introduction of the plate; S 0 - the area of the plate.

The capacitor is disconnected from the voltage source, so its charge does not change after partial immersion in the dielectric:

Q 0 = Q total,

or, explicitly,

ε 0 S 0 d U 0 = ε 0 d (ε S 1 + S 2) U.

After simplification, we have:

S 0 U 0 = (εS 1 + S 2) U.

Hence it follows that the sought-for potential difference is determined by the expression

U = U 0 S 0 ε S 1 + S 2.

Taking into account the fact that part of the capacitor plates is immersed in the dielectric, i.e.

S 1 = ηS 0, S 2 = S 0 - S 1 = S 0 - ηS 0 = S 0 (1 - η), η = 1 3,

U = U 0 S 0 ε η S 0 + S 0 (1 - η) = U 0 ε η + 1 - η.

From here we find the required potential difference:

U = 240 2.00 ⋅ 1 3 + 1 - 1 3 = 180 V.

A physical quantity equal to the work that the field forces will perform by moving a charge from one point of the field to another is called tension between these points of the field.

Consider a uniform electrostatic field (such a field exists between the plates of a flat charged capacitor far from its edges):

During the movement of the charge, the field does work:

Conductor in an external electric field (one hundred happens, why is it induced)

Electrostatic induction,

guidance in conductors or dielectrics electric charges in a constant electric field.

V conductors mobile charged particles - electrons - move under the action external electric fields... The movement occurs until the charge is redistributed so that the electrical field inside conductor will fully compensate externalfield and the total electrical field inside conductor becomes zero. (If this did not happen, then inside a conductor placed in a constant electric field, electricity, which would contradict the law of conservation of energy.) As a result, equal in magnitude induced (induced) charges of opposite sign are formed on separate sections of the surface of the conductor (generally neutral).

In dielectrics placed in a constant electric field, polarization occurs, which consists either in a slight displacement of positive and negative charges inside the molecules in opposite directions, which leads to the formation of electric dipoles(with an electric moment proportional to the external field), or in a partial orientation of molecules with an electric moment in the direction of the field. In both cases, the electric dipole moment per unit volume of the dielectric becomes nonzero. Bound charges appear on the surface of the dielectric. If the polarization is inhomogeneous, then bound charges appear inside the dielectric. A polarized dielectric produces an electrostatic field that is added to the external field. (Cm. Dielectrics.)

Electric capacity, capacitor

Electrical capacity- a quantitative measure of the ability of a conductor to hold a charge.

The simplest ways to separate dissimilar electric charges - electrification and electrostatic induction - make it possible to obtain on the surface of bodies not a large number of free electric charges. For the accumulation of significant amounts of unlike electric charges, capacitors.

Capacitor Is a system of two conductors (plates) separated by a dielectric layer, the thickness of which is small compared to the dimensions of the conductors. So, for example, two flat metal plates located in parallel and separated by a dielectric layer form flat capacitor.

If the plates of a flat capacitor are told equal in magnitude charges of the opposite sign, then the electric field strength between the plates will be twice as high as the field strength of one plate. Outside the plates, the electric field strength is zero, since equal charges of opposite sign on two plates create electric fields outside the plates, the strengths of which are equal in magnitude, but opposite in direction.

Capacitor capacitance is called a physical quantity determined by the ratio of the charge of one of the plates to the voltage between the capacitor plates:

With a constant position of the plates, the electrical capacity of the capacitor is constant for any charge on the plates.

Farad is taken as the unit of electrical capacity in the SI system. 1 F is the electrical capacity of such a capacitor, the voltage between the plates of which is equal to 1 V when the plates are informed of opposite charges by 1 C.

The electrical capacity of a flat capacitor can be calculated by the formula:

, where

S - area of capacitor plates

d - distance between plates

- dielectric constant of the dielectric

The electrical capacity of the ball can be calculated by the formula:

Energy of a charged capacitor.

If inside the capacitor the field strength is E, then the field strength created by the charge of one of the plates is E / 2. In a uniform field of one plate, there is a charge distributed over the surface of the other plate. According to the formula for the potential energy of a charge in a uniform field, the energy of a capacitor is:

Using the formula for the capacitance of a capacitor
:

One of the most important parameters by which a capacitor is characterized is its electrical capacity (C). Physical quantity C, equal to:

called the capacitance of the capacitor. Where q is the magnitude of the charge of one of the capacitor plates, and is the potential difference between its plates. The capacitance of a capacitor is a value that depends on the size and design of the capacitor.

For capacitors with the same device and with equal charges on its plates, the potential difference of the air capacitor will be one times less than the potential difference between the plates of the capacitor, the space of which between the plates is filled with a dielectric with a dielectric constant. This means that the capacitance of a capacitor with a dielectric (C) is times greater than the electrical capacity of an air capacitor ():

where is the dielectric constant of the dielectric.

The unit of capacitance of a capacitor is the capacity of such a capacitor, which is charged with a unit charge (1 C) to a potential difference equal to one volt (in SI). The unit of capacitance of a capacitor (like any eclectic capacitance) in the International System of Units (SI) is the farad (F).

Electric capacity of a flat capacitor

In most cases, the field between the plates of a flat capacitor is considered uniform. The uniformity is broken only near the edges. When calculating the capacitance of a flat capacitor, these edge effects are usually neglected. This is possible if the distance between the plates is small in comparison with their linear dimensions. In this case, the capacitance of a flat capacitor is calculated as:

where is the electrical constant; S is the area of each (or the smallest) plate; d is the distance between the plates.

The electric capacitance of a flat capacitor, which contains N layers of dielectric, the thickness of each, the corresponding dielectric constant of the i-th layer, is equal to:

Electric capacity of a cylindrical capacitor

The design of a cylindrical capacitor includes two coaxial (coaxial) cylindrical conducting surfaces of different radii, the space between which is filled with a dielectric. The electric capacity of such a capacitor is found as:

where l is the height of the cylinders; - radius of the outer cover; - radius of the inner lining.

Capacities of a spherical capacitor

A spherical capacitor is a capacitor, the plates of which are two concentric spherical conducting surfaces, the space between them is filled with a dielectric. The capacity of such a capacitor is found as:

where are the radii of the capacitor plates.

Examples of problem solving

EXAMPLE 1

Exercise

The plates of a flat air condenser carry a charge that is evenly distributed with an area density. In this case, the distance between its plates is equal. How much will the potential difference on the plates of this capacitor change if its plates are moved apart to a distance?

Solution

Let's make a drawing.

In the problem, when the distance between the plates of the capacitor changes, the charge on its plates does not change, the capacitance and potential difference on the plates change. The capacity of a flat air condenser is:

where . The capacity of the same capacitor can be defined as: