Connection of Capacitors

Electromagnetism IChapter IIProf. Dr. T. Fahmy

CHAPTER II

Capacitors

After completing this chapter, the students will know:

The concept of the capacitor
The different types of capacitors connection.
The energy of charged capacitors.
The different types of material (insulator, semiconductor and conductor).
The dielectric constant.
The induced charge and induced electric field.

CHAPTER II

Capacitors

Definition of Capacitor:

Consider two conductors carrying charges of equal magnitude and opposite sign, as shown in Figure. Such a combination of two conductors is called a capacitor. The conductors are called plates. A potential difference V exists between the conductors due to the presence of the charges.

The capacitance:

The capacitance C of a capacitor is defined as the ratio of the magnitude of thecharge on either conductor to the magnitude of the potential difference betweenthe conductors:

Note that:

The SI unit of capacitance is the farad (F)

2.1: Connection of Capacitors

2.1.1: Capacitors in series connection

Note that,

In series connection

a)The potential difference V between the points A and B equals to the summation of potential difference on all capacitors, i.e.,

b)The capacitors have the same charge q, i.e.,

But as known,

Therefore,

Then:

The equation (2.5) can be written in general form as follow

2.1.2: Capacitors in parallel Connection

Note that,

In parallel connection

a)The total charge between points A and B

equals to the summation of charges on all

capacitors, i.e.,

b)The capacitors have the same voltage V, i.e.,

Also

Therefore,

and in general form, the total capacitance can be written as follow

Example (2.1):

You have three capacitors. How can you arrange them to give the greatestcapacitance?

Solution:

In Parallel

Example (2.2):

You have three capacitors. How can you arrange them to give the least capacitance?

Solution:

In Series

Example (2.3):

A 3.0 F and a 5.0 F capacitor are connected in parallel across a 12 V battery.

Find the charge on each capacitor.
Find the equivalent capacitance.

Solution:

Find the charge on each capacitor.

Find the equivalent capacitance.

Ceq = C1 + C2; Ceq = 3.0 F + 5.0 F; Ceq = 8.0 F

Example (2.4):

You have three capacitors: 1 mF, 2.0 mF and 3.0 mF. What are the maximum and minimum equivalent capacitances you can obtain by combining these three capacitors?

Solution:

The maximum equivalent capacitances will be in parallel connection:

The minimum equivalent capacitances will be in parallel connection:

Example (2.5):

In the following figure, calculate

1-The total capacitance of the system, Ctotal

2-The total charge of the system, Qtotal

3-The charge on each capacitor

4-The voltage on each capacitor

Solution:

1-The total capacitance of the system, Ctotal will be:

2-The total charge of the system, Qtotal will be:

3-Since, the capacitors are in series, therefore:

4-The voltage on each capacitor will be:

Example (2.6):

In the following figure, calculate

1-The total capacitance of the system, Ctotal

2-The total charge of the system, Qtotal

3-The voltage on each capacitor

4-The charge on each capacitor

Solution:

1-The total capacitance of the system, Ctotalis:

2-The total charge of the system, Qtotal will be:

3-Since, the capacitors are in series, therefore:

4-The charge on each capacitor will be:

Example (2.7):

In the following figure, calculate the equivalent capacitance between the points a and b.

Solution:

Since, C1 and C2 are in parallel, thus the equivalent capacitance is

C1,2 = C1 + C2

=2+4= 6 F

Since, C1,2 and C3 are in series, thus the equivalent capacitance is

Example (2.8):

In the circuit given below, C1=60µF, C2=20 µF, C3=9 µF and C4=12 µF. If the potential difference between points a and c Vac= 120V find the charge of the second capacitor

Solution:

Since, C1 and C2 are in series, thus the equivalent capacitance of C1 and C2 is

Since, C1,2 and C3 are in parallel, thus the equivalent capacitance of C1,2 and C3 is

Since, C1,2,3 and C4 are in series, thus the equivalent capacitance of C1,2,4 and C4 is

Then the total charge of the system will be:

Also,

The potential difference between the points b and c will be:

Therefore,

The potential difference between the points a and b will be:

Then:

Example (2.9):

Four capacitors are connected as shown

in the figure. Calculate the following:

The equivalent capacitance between

the points a and b.

Calculate the charge and voltage on each

capacitor if Vab=15 V.

Solution:

The equivalent capacitance can be calculated as follow:

Note that:

C1 and C2 are connected in series, so

Also, C1,2 and C3 are connected in parallel, so

Also, C1,2,3 and C4 are connected in parallel, so

The total charge can be calculated as follow:

Therefore,

The charge on 20 F capacitor = 6x10-5 C
and Voltage (V) on that capacitor is

Therefore, the voltage on the rest of the circuit is

15-3= 12 volt

Example (2.10):

In the following figure, calculate

1-The charge on each capacitor

2-Determine the voltage at the

point b

Solution:

Note that, C2 and C3 are connected in parallel. Therefore, the resultant capacitance of these capacitors is:

Also, the total capacitance of the circuit is

Therefore, the total charge of the circuit is

Since, C1 and C2,3 are connected in series, the voltage between the points a and b is

Also,

Also

Example (2.11):

Determine the equivalent capacitance between A and B for the group of capacitors in the drawing.

Solution:

The 24, 12, and 8.0-F capacitors are in series. We can find the equivalent capacitance for the three capacitors:

This 4.0F capacitance is in parallel with the 4.0F capacitance already shown in the text diagram. We find that the equivalent capacitance for the parallel group is

This 8.0F capacitance is between the 5.0 and the 6.0F capacitances and in series with them. The equivalent capacitance between A and B in the text diagram:

2.2: Energy of a Charged Capacitor

The work done to transport a charge of dq through a capacitor is expressed as follow

Where V is the potential difference

but, as known

Then

Therefore, the total work done (W) or the energy (U) of the charged capacitor is

Hence,

If the capacitor is considered as two parallel plates with cross-sectional area of A, the distance between the plates is d and the total surface charge density = Q/A. This leads to the capacitance of the capacitor can be expressed as follow:

But as known,

Then, from equation (2.17) in (2.18), we can obtain

Therefore, the energy density is u

Example (2.12):

For the system of capacitors shown in the figure,

Calculate the following:

The equivalent capacitance
The potential across each capacitor
The charge on each capacitor
The total energy stored by the group

Solution:

The equivalent capacitance can be calculated as follow:

Note that:

C1 and C2 are connected in parallel, so

Also,

C3 and C4 are connected in parallel, so

Therefore, the equivalent capacitance (Ceq) is:

The total charge can be estimated as follow:

The total energy stored by the group

Example (2.13):

Determine the energy of these two capacitors

before and after the connection?

Solution:

1-Before connection

Also,the total energy (U) can be calculated as follow

2-After connection

The total energy (U) can be calculated as follow

Example (2.14):

Squared parallel plate capacitor with capacitance of 4.65x10-10 F and having 0.4 mm gap between the plates is filled with a material of dielectric constant k= 2.1 and E= 6x107 V/m. Calculate the following:

The voltage (V)
The energy (U)

Solution:

The voltage (V)= E x d= 6x107 x0.4x10-3 = 2400 V
The energy (U) = ½ CV2 = ½ x 4.65x10-10 x (2400)2 = 0.13 J

Example (2.15):

Two capacitors C1 and C2 (where C1 ˃ C2) are charged to the same initial potential difference Vi. The charged capacitors are removed from the battery, and their plates are connected with opposite polarity as in the Figure. The switches S1 and S2 are then closed.

Find the final potential difference DVfbetween a and b after the switches are closed.
Find the total energy stored in the capacitors before and after the switches are closed and the ratio of the final energy to the initial energy.

Solution:

A)Note that:

The total charge Q in the system is

(1)

After the switches are closed, the total charge Q in thesystem remains the same but the charges on the individualcapacitors change to new values Q1fand Q2f. Because thesystem is isolated,

(2)

The charges redistribute until the potential difference is thesame across both capacitors, Vf . To satisfy this requirement,the charges on the capacitors after the switches are closed are:

Dividing the first equation by the second, we have

(3)

Combining Equations (2) and (3), we obtain:

(4)

Using Equations (3) and (4) to find Q1fin terms of Q, we have:

(5)

Finally,

(6)

(7)

As noted earlier, V1f=V2f =Vf.

To express Vfin terms of the given quantities C1, C2,and Vi, we substitute the value of Q from Equation (1) intoeither Equation (6) or (7) to obtain:

Before the switches are closed, the total energystored in the capacitors is:

After the switches are closed, the total energy stored in the capacitors is

Using the results of part (A), we can express this as:

Therefore, the ratio of the final energy stored to the initial energy stored is:

(8)

Example (2.16):

A parallel-plate capacitor with air between the plates has an area A = 2x10-4 m2 and a plate separation d =1 mm. Find its capacitance.

Solution:

Dielectric Materials

The materials can be classified according to its electrical conductivity into different types:

1-Insulators or Dielectrics

These materials are characterized by very low electrical conductivity and very high resistance. These materials do not have free electrons.

2-Semiconductors

Semiconductor materials are classified into two types as follows:

a)Intrinsic semiconductors

These materials are characterized by electrons and holes as charge carriers.

b)Extrinsic semiconductors

These materials are classified into two types

n- type

The majority of charge carriers are electrons in this type, whereas, the minority of charge carriers are holes.

p- type

The majority of charge carriers are holes in this type, whereas, the minority of charge carriers are electrons.

3-Conductor Materials

These materials are characterized by high electrical conductivity due to free electrons.

4-Superconductor Materials

These materials are characterized by very high electrical conductivity and resistance equals to zero.

2.4: Dielectric Constant

Dielectric constant is defined as the ratio between the capacitance of a capacitor (C) to the capacitance of the same capacitor when it contains a space between its plates (C0)

(2.20)

Also, as known

(2.21)

Substituting from equation (2.21) in equation (2.20)

Also,

Then from equation (2.23) in equation (2.22), we can get

Also, as we know

Then from equation (2.25) in equation (2.24), we can get

Finally, the dielectric constant can be expressed as follow:

2.5: The relation between the induced charge and dielectric constant

Suppose that, the electric field in case of space

between the plates of the capacitor is E0, whereas, will

be E in case of existence of a dielectric material between

the plates.

Therefore:

(2.27)

where, Ei is the induced electric field due to the induced charge qi inside the capacitor.

By substituting from equations (2.25) and (2.26) in equation (2.27), we can write the following equation:

(2.28)

But,

Where A is the area

Then

(2.29)

Example (2.17):

A capacitor with two parallel of charge equals to 30 C, the area of each plate is 1 m2. If a dielectric with permittivity () of 15x10-12 C2/N. m2 filled the region between the plates, calculate the following:

1-The resultant electric field of the dielectric ‘E’.

2-The density of the induced surface charge (i).

3-The resultant of the electric field due to the induced charge ‘Ei’.

Knowing that,  = 8.85x10-12 C2/N. m2

Solution:

q= 30 C

A= 1m2

= 15x10-12C2/N. m2

 = 8.85x10-12 C2/N. m2

1-As known,

Example (2.18):

An air filled capacitor consists of two parallel plates each with an area of 7.6 cm2, separated by a distance of 1.8 mm. If a 20 V potential difference is applied to these plates, calculate the following:

The electric field between the plates (E).
The surface charge density ()
The capacitance
The charge on each plate

Solution:

The electric field between the plates (E)

The surface charge density ()

The capacitance

The charge on each plate