Capacitors. Capacitors If the potential difference between the plates of a capacitor

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 that the determining factor is whether the capacitor is disconnected or connected to the voltage source.

If the 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, the charge on the capacitor plates remains unchanged:

Q = const.

When connected to each other covers of the same name two charged capacitors take place parallel connection.

U = Q total C total,

where Qtot is the charge of the capacitor bank; Ctot is the electrical capacity of the battery;

Ctot = 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

Qtot = Q 1 + Q 2,

When connected to each other different linings two charged capacitors takes place (as in the case of connecting plates of the same name) parallel connection.

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

• capacitor bank voltage

U = Q total C total,

where Qtot is the charge of the capacitor bank; Ctotal - battery capacity;

• electrical capacity of a capacitor bank

Ctot = 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 |,

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 - 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 by similarly charged plates. What will be the potential difference between the plates of the capacitors after the indicated 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 obtain a parallel connection of capacitors. The potential difference between the plates of a capacitor bank is determined by the formula

U = Q total C total,

The total charge of a 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:

Qtot = Q 1 + Q 2,

U = Q tot C tot = 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 capacitor plates after this 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 opposite plates of capacitors, 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 - 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 a 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 a battery of two capacitors obtained by connecting their opposite plates is determined by the modulus of the charge difference 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

Ctot = C1 + C2 = 2C.

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

U = Q tot C tot = |

Let's calculate:

Q 1 − Q 2 |

2 C = |

C U 1 − C U 2 |

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

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

• the plane facing the negatively charged capacitor plate has a lack of electrons and is charged with a 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 bank of two capacitors:

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

C 1 = ε 0 S d 1 ,

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

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

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 equal charges and form a series connection. The electrical capacity of a battery of two capacitors when connected in series 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 plates of the capacitor (d 1 = d 2 = d), the electrical capacitances of the capacitors are the same:

C 1 = C 2 = ε 0 S d ,

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

Ctot = 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 insertion of the plate; a is the thickness of the metal plate.

Potential difference between battery plates

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

where Qtot is the charge of a battery of series-connected capacitors, Qtot = 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 inserting the plate, Q 0 = q (the capacitor is disconnected from the voltage source); C 0 is the electrical capacity of the capacitor before inserting the plate.

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 introducing a metal plate into the capacitor, the potential difference between its plates decreased and amounted to 120 V.

Example 20. A flat-plate 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 the volume. Find the potential difference that will be established between the plates of the capacitor.

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;
• the part of the capacitor plates remaining in the air has a charge q 2.

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

Qtot = q 1 + q 2.

The area of ​​the capacitor plates when 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 capacitance

C 1 = ε 0 ε S 1 d,

where ε 0 is the electrical constant, ε 0 = 8.85 ⋅ 10 −12 C 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 S 2 ; the corresponding part of the capacitor can be considered as a separate capacitor with electrical capacitance

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 when connected in parallel 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 it is immersed 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 is the potential difference between the plates of the capacitor before the introduction of the plate; S 0 - lining area.

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 .

It follows that the desired 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 large number of capacitors that are used in technology are similar in type to a flat-plate 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 sign are concentrated on the plates.

Electric capacitance of a parallel plate 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 verify that the electrical 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 capacitance of a flat capacitor is easy to obtain using theoretical calculations.

Let us assume that the distance between the plates of the capacitor is much less than their linear dimensions. Then 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 a charge of the same magnitude and opposite in sign, separated by a dielectric with dielectric constant, can be determined using the formula:

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

Let's substitute the right side of expression (3) instead of the potential difference in (1), taking into account that , we have:

Field energy of a flat capacitor and the force of interaction between its plates

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

where is the volume of the capacitor; E is the field strength of the capacitor. Formula (5) relates the energy of a capacitor to the charge on its plates and the field strength.

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

In expression (6), the minus shows 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 the 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 side of expression (1.3) instead of capacity into formula (1.2), we have: Let's calculate the distance between the plates ():
Answer	m

EXAMPLE 2

Exercise	The potential difference between the plates of a flat air capacitor is equal to V. The area of ​​the plates is equal to , the distance between them 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 turned off when moving the plates apart.
Solution	Let's make a drawing. The energy of the electric field of the capacitor can be found using the expression: Since the capacitor is flat, its electrical capacitance can be calculated as:

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

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

While the charge is moving, the field does work:

1. A conductor in an external electric field (one hundred occurs, why is it induced)

Electrostatic induction,

induction of electric charges in conductors or dielectrics in a constant electric field.

IN conductors mobile charged particles - electrons - move under the influence external electrical fields. The movement occurs until the charge is redistributed so that the electrical energy created by it field inside will fully compensate externalThe movement occurs until the charge is redistributed so that the electrical energy created by it and total electrical The movement occurs until the charge is redistributed so that the electrical energy created by it field inside will become equal to zero. (If this had not happened, then inside a conductor placed in a constant electric field, an electric current would exist indefinitely, which would contradict the law of conservation of energy.) As a result, induced forces of equal magnitude ( induced) charges of opposite sign.

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

1. Dielectrics

Electrical capacity, capacitor Electrical capacity

– a quantitative measure of a conductor’s ability to hold a charge. The simplest methods of separating unlike electric charges - electrification and electrostatic induction - allow one to obtain a small amount of free electric charges on the surface of bodies. To accumulate significant quantities of opposite electrical charges, they are used.

capacitors Capacitor is a system of two conductors (plates) separated by a dielectric layer, the thickness of which is small compared to the size of the conductors. For example, two flat metal plates located parallel and separated by a dielectric layer form flat

capacitor.

If the plates of a flat capacitor are given charges of equal magnitude and opposite signs, then the electric field strength between the plates will be twice as strong as the field strength of one plate. Outside the plates, the electric field strength is zero, since equal charges of opposite signs on two plates create electric fields outside the plates, the strengths of which are equal in magnitude but opposite in direction. is a physical quantity determined by the ratio of the charge of one of the plates to the voltage between the plates of the capacitor:

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

The unit of electrical capacity in the SI system is Farad. 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 given opposite charges of 1 C each.

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

Energy of a charged capacitor.

If the field strength inside the capacitor is E, then the field strength created by the charge of one of the plates is E/2. In the 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 the capacitor is equal to:

Using the formula for the electrical capacity of a capacitor
:

Electrical capacity

When a charge is imparted to a conductor, a potential φ appears on its surface, but if the same charge is imparted to another conductor, the potential will be different. This 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 the farad. 1 F = 1 C/1 V.

If the potential of the sphere surface

(5.4.3)

(5.4.4)

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

There is a need for devices that accumulate charge, and isolated conductors have low capacity. It was experimentally discovered that the electrical capacity of a conductor increases if another conductor is brought close to it - due to electrostatic induction phenomena.

capacitors - these are two conductors called linings, 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 flat, cylindrical and spherical.

Since the electrostatic field is inside the capacitor, the electric displacement lines start on the positive plate, end on the negative plate, and do not disappear anywhere. Therefore, the charges on the plates

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

(5.4.5)

In addition to capacitance, each capacitor is characterized U slave (or U etc . ) – the maximum permissible voltage, above which a breakdown occurs between the plates of the capacitor.

Connection of capacitors

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

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

In this case, the common voltage is U:

Total charge:

Resulting capacity:

Compare with parallel connection of resistances R:

Thus, when connecting capacitors in parallel, the total capacitance

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

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

The common charge is q.

Or , from here

(5.4.6)

Compare with serial connection R:

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

Calculation of the capacitances of various capacitors

1.Capacitance of parallel plate capacitor

Field strength inside the capacitor (Fig. 5.11):

Voltage between plates:

where is the distance between the plates.

Since the charge is

.

(5.4.7)

As can be seen from the formula, the dielectric constant of a substance greatly affects the capacitance of the capacitor. This can also be seen experimentally: we charge the electroscope, bring a metal plate to it - we get a capacitor (due to electrostatic induction, the potential has increased). If you add a dielectric with ε greater than that of air between the plates, then the capacitance of the capacitor will increase.

From (5.4.6) we can obtain the units of measurement ε 0:

(5.4.8)

.

2. Capacitance of a cylindrical capacitor

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

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

is called the capacitance of the capacitor. Where q is the amount of charge on one of the plates of the capacitor, and is the potential difference between its plates. The electrical capacity 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 an air capacitor will be several times less than the potential difference between the plates of a capacitor, the space between the plates of which 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 capacitance of an air capacitor ():

where is the dielectric constant of the dielectric.

The unit of capacitor capacity is considered to be the capacitance of a capacitor that 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 (as well as any eclectic capacitance) in the International System of Units (SI) is the farad (F).

Electrical capacitance of a flat capacitor

The field between the plates of a flat-plate capacitor is in most cases considered uniform. Uniformity is only disrupted near the edges. When calculating the capacitance of a parallel-plate capacitor, these edge effects are usually neglected. This is possible if the distance between the plates is small compared to 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 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:

Electrical capacitance 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 electrical capacitance of such a capacitor is found as:

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

Capacitances of a spherical capacitor

A spherical capacitor is a capacitor whose plates are two concentric spherical conducting surfaces, the space between them is filled with a dielectric. The capacitance 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 planar air capacitor carry a charge that is uniformly distributed with a surface density of . In this case, the distance between its plates is equal to . By what amount 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 a 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 capacitor is: Where . The capacitance of the same capacitor can be determined as: where U is the potential difference across the capacitor plates. For the capacitor in the first case we have: For the same capacitor, but after the plates have been moved apart, we have: Using formula (1.3) and applying the relation: let's express the potential difference Therefore, for the capacitor in the second state we obtain: Let's find the change in potential difference:
Answer