# 4. Capacity

## Introduction

In 1745, Dutchman Pieter van Musschenbroek invented the first capacitor . While wearing a glass bottle to isolate a metal blade inside, he found that when holding the bottle in his hand, the electrical charge that could store the blade was much higher than when the bottle was on the table. The explanation is that the hand which is a conductor are induced signal charges which attract the opposite charges in the metal, allowing it easier to introduce the charge of the same sign. Placing a second metal blade out of the bottle, easy to charge input in the bottle can be stored very high charges. The capacitor van Musschenbroek became known as bottle of Leiden , which is the city where he worked. It is one of the most important inventions in the history of electricity, allowing higher charges accumulate, facilitating the achievement of electrostatic experiments.

## 4.1. Capacity of an insulated conductor

The potential in an isolated conductor is uniform across the driver and proportional to the total charge it. Defines the capability of the driver as the ratio between the charge and the potential on the conductor surface

(4.1)

The ability does not depend on the charge or the potential, because both increase in the same proportion; the ability depends solely on the shape and driver size. The potential $V_\mathrm{sup}$ really is the potential difference between the conductive surface and a point at infinity, which usually potential to arbitrate-zero.

In the international system of units, the unit for measuring capacity is the farad , named after the scientist Michael Faraday (1791-1867). The farad represented by the letter F, is the ability of a conductor with a charge of 1 C has a potential of 1 V:

(4.2)

A capacity of 1 F is very high and is common to find in practice skills in the order of 1 μF 1 nF or 1 pF.

## 4.2. conducting sphere isolated

An isolated conducting sphere (figure  4.1 ), all the charge builds up on the surface evenly due to the symmetry of the ball. In Appendix B it is shown how to calculate the electric field and the result is that, within the sphere the field is null and off the field is identical to the field of a point charge$Q$placed in the center of the sphere. The module of the electric field at a point which lies at a distance$r$ ball center is:

(4.3)

at where $k$is Coulomb's constant (9 × 10 9 N · m 2 / C 2 ) and$K$is the dielectric constant of the medium around the ball. Figure  4.2 shows the function module in the field of graphic distance to the center, in the case where the charge is positive.

To calculate $V_\mathrm{sup}$It integrates the tangential component of the electric field from the surface of the sphere to infinity along any path; a route that facilitates the calculation is in the direction and the radial direction which is the direction of the electric field lines:

(4.4)

The value of this integral is also equal to the shaded area in figure  4.2 . Using the expression obtained for$V_\mathrm{sup}$, Is the expression for the sphere of radius capacity $R$,

(4.5)

Note that if $Q$ is negative, $|Q|$ It should be replaced by $-Q$but the result of the integral in equation  4.4 is the same because in this case the direction of the field would be opposed to radial direction, introducing another negative sign. The potential on the surface in relation to infinity, is always the same charge signal, and as such, the capacity is always a positive number.

The larger the ball, the greater its capacity. It has been previously reported that the capacity does not depend on the charge stored on the ball, or the potential produced by such cargo. The ability depends only on the size and geometry of the conductor and the dielectric constant of the medium. In this case it is directly proportional to the radius of the sphere and the dielectric constant of the medium.

## 4.3. capacitors

In the opening chapter was mentioned the bottle of Leiden, which was the first capacitor built in history. The two conductors separated by an insulating (glass in this case) is designated plate . When there are charges a reinforcement opposite sign charges are induced in the other equipment, which decreases the potential of each plate in relation to the reference potential (the ground ). The reduction potential of the two armatures system compared with the potential that would have a single equipment with the same charge, implies a much higher capacity to the capacitor compared to the capacity of each one of the two plates separately.

If one of the plates have charge $Q$ the other has charge $-Q$. If Δ$V$ is the voltage between the two armatures, is defined as the capacity of the capacitor:

(4.6)

If between the two armatures an insulator is placed, Coulomb's constant, $k$, Which enters into the calculation of the potential difference Δ$V$From the force, it is replaced by $k/K$, at where $K$It is the constant dielectric insulator. As such, the insulator with the capacitance of the capacitor increases by a factor$K$. Thus, the Leyden jar glass bottle serves insulator and helps to increase the capacity. As the glass has a dielectric constant close to 6, the capacity of the glass bottle is approximately 6 times that obtained without glass between the plates.

The larger the capacity of a capacitor is easier to store charges therein. There are several different types of capacitors with different shapes and sizes (figure  4.3 ).

The insulator between the plates of a capacitor is also called dielectric . The dielectric also helps to increase the maximum potential difference that can exist between the plates. Each insulating material has a value of stiffness dielectric ($E_\mathrm{max}$) Which is the maximum electric field that the dielectric supports without producing breakage. The rupture of a dielectric occurs when the molecules or atoms are deformed to the point that some of them are ionized, forming cracks where the material is burned and becomes conductive because of the ions deposited in that slots. Figure  4.4 shows an acrylic block that was placed between two plate with a high voltage exceeding the dielectric strength of acrylic, producing its rupture. The slots where the dielectric was burnt create the so-called figure of Lichtenberg

The maximum potential difference that supports a capacitor dielectric with a thickness $d$ without burning it is then,

(4.7)

at where $E_\mathrm{max}$It is the dielectric rigidity. Different models of capacitors (figure  4.3 ) have different capacities and maximum potential differences depending on the size and dielectric used. In some applications it is also important that the dielectric response time is fast, since the charges are not induced in the molecules of the dielectric instantaneously. Table  4.1 indicates the dielectric constant and dielectric strength of various insulating materials.

Table 4.1: Constant and stiffness of some dielectrics.
Material  Dielectric constant, $K$ stiffness, $E_\mathrm{max}$ (KV / mm)
Water (20 ° C)80 -
Dry air1.000593
Oil2:2412
Paper3.716
Acrylic3.440
pyrex glass5.614
Porcelain75.7
Polyester2:5524
Paraffin2.1- 2.510

The dielectric strength of the dry air is 3 kV / mm. When the potential difference between two objects in the air exceeds 3000 V per millimeter of clearance, gives an abrupt electrical discharge of the objects. The high electrical forces ionize air molecules, and the discharge is the passage of positive and negative ions in the air between the two objects.

The clouds and the ground, which are conductive, act as plates of a capacitor with air dielectric. During a thunderstorm, the air humidity decreases the dielectric strength of the air and the maximum potential difference between the clouds and the earth decreases, with the possibility arise electrical discharges (Figure  4.5 ). The closer the clouds are objects on the floor, the greater the probability of being struck by lightning because $\Delta\,V_\mathrm{max} = E_\mathrm{max}\,d$ It is then smaller.

### 4.3.1. spherical capacitor

Figure  4.6 shows a spherical capacitor formed by two concentric spherical reinforcement, beams$R_1$ and $R_2$, Separated by a dielectric constant insulator $K$which occupies the space between the two spheres. The smallest sphere is attached to a wire that passes out of the larger ball without touching it in order to be able to store a charge$Q$ one of the balls and $-Q$ in the other.

The field produced by the two conductive spheres is given by the expression obtained in Appendix B , substituting $Q_1$ per $Q$ and $Q_2$ per $-Q$in equation  B.14 . As the sum of two charges is zero, the field outside of the two spheres is null. The electric field is confined to the region between the two spheres, where there is the dielectric that separates them. The expression of the field (assuming that$Q$ It is positive) is

(4.8)

And the potential difference between the spheres is the integral of the field in the radial direction between the two spheres:

(4.9)

Sharing the charge, $Q$, By the potential difference, $\Delta V$, One obtains the expression for the capacity of the spherical capacitor:

(4.10)

### 4.3.2. capacitor plan

A capacitor plane is formed by two flat armatures, area$A$Parallel and separated by a constant distance $d$As on the left side of the figure  4.7 . If the reinforcements are close enough to a good approximation is to consider the plan capacitor as a small part of a spherical capacitor with a very large radius, approaching infinity, and the two spheres about the same radius as shown in Figure  4.7 .

The charge and the area in the very large spherical capacitor approach is infinite, but the relationship between them, surface charge, remains finite. Equation  4.8 for the field inside the spherical capacitor must be written in function of the surface charge,$\sigma = Q/(4\,\pi\,R^2)$, and with $r$ equal to $R$for the field in the sphere of the neighborhood; the field inside the plane capacitor and then approximately:

(4.11)

And the potential difference between the plates is equal to

(4.12)

at where $d$ is the distance between the plates, $Q$ the charge on the armature and positive $A$the area of plate. From equation 4.6 we obtain the expression for the capacitor capacity plan:

(4.13)

The ability of a plane capacitor is directly proportional to dielectric constant and area of ​​the reinforcement and inversely proportional to the distance therebetween.

### example 4.1

A capacitor variable is constituted by two parallel flat plates in form of a circular sector angle of 80 ° and radius 5 cm, which can rotate around a common axis, as shown in Fig. If the distance between the plates is 0.5 cm, calculate the maximum capacity and the capacity when one of the plates 30 ° rotates from the position where the capacity is maximum.

Resolution . The maximum capacity is obtained when the two plates are completely superposed one above the other, so that the charge is distributed throughout the plate surface. The angle of 80 ° is equivalent to a fraction of 80/360 full circle; therefore, the area of reinforcement is:

The capacity is given by 4:13 with the dielectric constant of air,$K$ = 1:

When one of the plates rotates 30 °, the area in which the charge is distributed, corresponds only to the area of ​​the plate that is superimposed, i.e., a circular sector angle of 50 °. The area is then 5/8 of the total area of ​​reinforcement and capacity is directly proportional to the area, is 5/8 of the maximum capacity:

(4.14)

### 4.3.3. ultracapacitor

A capacitor can accomplish a battery with a similar function, since it can be used to store charges that are supplied to a circuit. The big advantage is that, as there is no chemical reactions involved, chargeing and unchargeing can be done very quickly and the capacitor is not unusable after several charges and discharges, which is what happens to a rechargeable battery. Imagine for example that instead of having to wait a few hours to recharge your battery, this stay was immediately recharged when plugged in, and that never to replace it with a new one. This is getting closer to being a reality, with the development of ultracapacitors.

The difficulty using a normal capacitor as a source is that as the capacitor discharges, the potential difference between their plate decreases rapidly. A further disadvantage is that the ability to store charge is not as high as in the batteries. Consider for example the rechargeable battery in issue 4 of Chapter 2. The value of the emf is 1.2 V and the stored maximum charge is 2300 mA · h = 8:28 kC. According to Equation 4.6 , 6.9 kF capacitor would be required to store this charge, with this potential difference.

Such a high capacity was unthinkable until the end of the last century. A traditional capacitor, the size of the stack, would have a capacity of around μF. Electrolytic capacitors reach higher capacities, but still short of kilo-farad. They have recently been produced ultracapacitors , with much higher capacities in the order of kilo-farad (Figure  4.8 ).

For example, the cylindrical ultracondensador located at the front in Figure  4.8 , has a capacity of 3000 volts to 2.7 farads. With these values, the chargeing that can store 8.1 is already close kC charge a rechargeable battery. The high capacity also means that it takes much longer to downcharge when connected to a circuit. There is still slightly reduce the size to be competitive with the current lithium-ion battery .

In ultracapacitors uses a porous medium to replace a reinforcement. The contact area between the electrolyte and electrodes is very high. The ultracapacitors are already used in combination with electric motors of cars that run on hydrogen with fuel cells (Figure  4.9 ), which are already being marketed in some countries.

The ultracondensador allows you to quickly accumulate the charges produced by the fuel cell or the electromagnetic brake, and this charge can be provided quickly in times when it is necessary to accelerate. The only chemical reactions produced in this type of vehicle is the combination of hydrogen with oxygen in the fuel cell that produces water vapor. No harmful gases are released to the atmosphere, and there are no batteries to produce corrosive chemicals.

The ultracapacitor can provide charge and be recharged far more rapidly than a battery without suffering wear and which causes the battery has a limited number of charge and discharge cycles.

## 4.4. electrical energy stored in a capacitor

To charge a capacitor, it is necessary to charge a cargo of plateed $Q$ and the other charge $-Q$. The process involves a charge transfer$Q$an armature to the other. This passage can be made by connecting two cables in the plate and the terminals of a battery (Figure  4.10 ).

To calculate the power supplied by the battery in the process, imagine that the total charge $Q$ It was transferred in small infinitesimal charges $\mathrm{d}q$from one of the plate to the other, as shown in Figure  4.10 . Each time a charge$\mathrm{d}q$ passes from negative to positive reinforcement, gains an electric potential energy

(4:15)

The total energy stored in the capacitor is obtained by integration, from $q$ = 0 until $q$ = $Q$ (Area under the line in the graph Δ$V$ in function of $q$In figure  4:11 ). The result is:

(4:16)

Using equation 4.6 , which relates the charge and the potential difference on any capacitor, the previous equation can be written in other two alternative forms:

(4.17)

The burden is not transferred to the plate instantly. When connecting a capacitor to a power source, the charge gradually increases to a final charge. The charge increasing process time is called response transient capacitor; if the resistance between the source and the armatures of the capacitor is not very high, the transient response is extremely fast and can be admitted that the charge on the capacitor already has its final stable value. In the chapter on signal processing it is shown how to determine the transient response.

## 4.5. Associations capacitors

A capacitor system may be replaced by a single equivalent capacitor. In cases where the capacitors are connected in series or in parallel, it is easy to calculate the equivalent capacitance of the capacitor.

Figure  4.12 shows two capacitors connected in series , between points A and B. If the capacitors are initially discharged by introducing a potential difference between points A and B, runs a charge$Q$entering by the higher potential point (A in the figure) and exits at point lower potential. In the central region, which connects the two armatures common to both capacitors, charges are induced$Q$ and $-Q$(Full charge in that region is zero). Thus, the charge stored in each of the capacitors is the same.

The potential difference between points A and B is the sum of potential differences at each of the capacitors

(4.18)

The system is then equivalent to a single capacitor whose capacity satisfies the equation:

(4.19)

The amount of charge stored in the capacitor equivalent is the same in each of the series capacitors.

Figure  4.13 shows one of two capacitors connected in system parallel between two points A and B. The potential difference is always the same for both capacitors, and equal to the potential difference between points A and B.

If the capacitor is initially discharged at the moment it is introduced a potential difference between points A and B, goes positive charge on the plate which are connected to the point with the greatest potential, and leaves the same amount of charge of armatures connected to the point with less potential. But the amount of charge entering each capacitor does not have to be the same; the total charge in and out between points A and B is:

(4.20)

That is, the system is equivalent to a single capacitor with a capacity equal to the sum of the capacities of the two capacitors

(4.21)

### example 4.2

Consider the circuit shown in the figure and calculate: ( a ) The equivalent capacity between A and B. ( b ) the charge stored in each capacitor when the voltage between A and B is Δ$V$= 200 V. ( c ) The total energy stored in the circuit when Δ$V$ = 200 V.

Resolution . The capacitors 4 μF and 15 μF are in series, and therefore may be replaced by a single capacitor capacity:

This capacitor is connected in parallel with the capacitor of 12 μF, so that the total capacity is 15:16 μF.

In both μF capacitor 12 is 3.16 μF voltage is the same and is equal to 200 V; Thus, charges such capacitors are:

The charges in the capacitors 4 μF and 15 μF are equal because they are connected in series:

The total stored energy can be calculated by summing the energies stored in each capacitor; the answer should be the same in any of the equivalent circuits. Using the simple circuit with one 15:16 μF capacitor, you get:

## questions

1. The electrical capacity of an insulated conductor:
1. Decreases if the driver has a dielectric around.
2. It does not depend on its size.
3. It is measured in J / C units.
4. It is equal to the work required to move a charge from infinity to the driver.
5. It is independent of the charge accumulated in the conductor.
2. What should be the capacity of a capacitor so that, when its voltage is 9.0 V, charging the equipment with negative charge is equivalent to 10 10 electrons?
1. 14 nF
2. 178 nF
3. 178 pF
4. 14 pF
5. 5.6 pF
3. What is the capacity of a capacitor of parallel circular plates, 5 cm radius, 1 cm apart?
1. 6.9 pF
2. 22.0 pF
3. 2.2 pF
4. 00:22 nF
5. 0.69 nF
4. Increasing the charge of a parallel plate capacitor 3 to 9 μC μC and increasing the separation between plates of 1 mm to 3 mm, the energy stored in the capacitor varies by a factor
1. 9
2. 3
3. 8
4. 27
5. 1/3
5. In one of two capacitors connected in parallel system, which of the following statements is true?
1. The equivalent capacity is less than the capacity of two capacitors.
2. The stored charge in both capacitors is the same.
3. The charge stored in capacitor will be larger with increased capacity.
4. The potential difference is higher in the capacitor with greater capacity.
5. The potential difference is higher in the capacitor with smaller capacity.

## Problems

1. A typical photographic flash provides 2 kW for approximately 2 ms. This energy is obtained by discharging a capacitor of 50 μF. ( A ) To what potential difference capacitor should be charged? ( B ) If the capacitor was replaced by a 250 μF until potential difference should be charged? ( C ) What would be the disadvantage of using the capacitor with higher capacity?
2. ( A ) Determine the ability of an isolated conducting sphere with radius of 4.0 cm and surrounded by air. ( B ) Above the preceding point of the sphere is placed a glass ball with a thickness of 1 mm and a dielectric constant of 5.6, and above it is placed a second metal sphere with radius of 4.1 cm, forming a capacitor spherical. Determine the capacity of this capacitor. ( C ) What is the relationship between the capacity of the capacitor and the sphere?
3. In three capacitors shown in Fig system, $C_1$ = 1.2 μF, $C_2$ = 4.3 μF and $C_3$= 2.5 μF. The voltage between points A and B is 9.0 V. ( a ) Calculate the charge stored in each capacitor. ( B ) Calculate the total energy stored in the system.
4. A capacitor flat and parallel plates of 12 cm 2 area and spaced 1 cm, it is completely filled by two dielectrics, each with a thickness of 0.5 cm and equal to the area of the plates. Calculate the capacity of the capacitor knowing that the constants of dielectrics are 4.9 and 5.6 (hint: admit that the capacitor is equivalent to two capacitors in series, each with a different dielectric).
5. Consider a capacitor flat and parallel plates, area 0.3 m 2 and 0.5 cm apart. Between the plates is an acrylic plate having the same area and thickness of 0.5 cm. The capacitor is charged until the potential difference being equal to 12 V and then shuts down the source used to charge it . ( A ) How much work is required to remove the acrylic plate between the capacitor plates? ( B ) Calculate the potential to break with dielectric and after it is removed.
6. Two capacitors 10 μF and 20 μF connect in series to a source of 1200 V ( a ) Calculate the charge on each capacitor. ( B ) The source is then switched off, binding together the two capacitors (positive reinforcement with positive and negative to negative). Compute and final charge voltage in each capacitor.
7. In the circuit of figure, calculate the equivalent capacity: ( a ) Between points B and D. ( b ) between points A and B.
8. The capacitors in the circuit of figure are initially discharged. Calculate the charge stored on the 2.4 pF capacitor when the voltage between points A and B is 5 V.

## replies

Questions: 1. E. 2. C. 3. A. 4. D. 5. C.

Problems

1. ( A ) 400 V ( b ) 179 V. ( c ) The increased capacity capacitor occupies a greater volume.
2. ( A ) 4:44 pF. ( B ) 01.02 nF. ( C ) 229.6
3. ( A )$Q_1$ = 3.38 μC, $Q_2$ = 12.1 μC and $Q_3$= 15.5 μC. ( B ) 69.6 mJ
4. 5:55 pF.
5. ( A ) 3.12 × 10 -7  J. ( B ) No dielectric, 15 kV; dielectric 200 kV.
6. ( A ) 8 mC. ( B ) Δ$V$ 1600/3 = V $Q_1$ 16/3 = mC, $Q_2$ 32/3 = mC.
7. ( A ) 12 pF. ( B ) 21.6 pF.
8. 3:15 pC.
Wrong

When placing a dielectric around the capacity increase, rather than decrease.

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Wrong

The larger the size of a conductor, the greater its ability to store charge.

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Wrong

J / C are units of energy per unit charge, while the charge capacity has units per unit potential.

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Wrong

This work is equal to the potential on the surface of the conductor times the charge; for the driver's capability is necessary to divide the charge on the surface potential on the surface.

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Right

The potential on the surface increases in proportion to the accumulated charge; capacity, obtained by dividing the voltage difference by the charge, then results independent of the charge and the potential on the surface.

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Wrong

The charge capacity is equal to that number of electrons divided by the potential difference of 9 V

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Wrong

The charge capacity is equal to that number of electrons divided by the potential difference of 9 V

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Right

The charge capacity is equal to that number of electrons divided by the potential difference of 9 V

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Wrong

The charge capacity is equal to that number of electrons divided by the potential difference of 9 V

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Wrong

The charge capacity is equal to that number of electrons divided by the potential difference of 9 V

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Right

The capacity is equal to the area of one of the armatures (circle area 5 cm radius) divided by 4 π  k d , where k is Coulomb's constant and d the distance between the two armatures.

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Wrong

The capacity is equal to the area of one of the armatures (circle area 5 cm radius) divided by 4 π  k d , where k is Coulomb's constant and d the distance between the two armatures.

(Click)

Wrong

The capacity is equal to the area of one of the armatures (circle area 5 cm radius) divided by 4 π  k d , where k is Coulomb's constant and d the distance between the two armatures.

(Click)

Wrong

The capacity is equal to the area of one of the armatures (circle area 5 cm radius) divided by 4 π  k d , where k is Coulomb's constant and d the distance between the two armatures.

(Click)

Wrong

The capacity is equal to the area of one of the armatures (circle area 5 cm radius) divided by 4 π  k d , where k is Coulomb's constant and d the distance between the two armatures.

(Click)

Wrong

The charge increases by a factor of 3. The ability decreases by a factor 1/3, as the distance increases 3 fold and therefore the potential difference increases a factor 9 (potential difference is equal to the charge on capacity). With these results calculate the increased energy U .

(Click)

Wrong

The charge increases by a factor of 3. The ability decreases by a factor 1/3, as the distance increases 3 fold and therefore the potential difference increases a factor 9 (potential difference is equal to the charge on capacity). With these results calculate the increased energy U .

(Click)

Wrong

The charge increases by a factor of 3. The ability decreases by a factor 1/3, as the distance increases 3 fold and therefore the potential difference increases a factor 9 (potential difference is equal to the charge on capacity). With these results calculate the increased energy U .

(Click)

Right

The charge increases by a factor of 3. The ability decreases by a factor 1/3, as the distance increases 3 fold and therefore the potential difference increases a factor 9 (potential difference is equal to the charge on capacity). With these results calculate the increased energy U .

(Click)

Wrong

The charge increases by a factor of 3. The ability decreases by a factor 1/3, as the distance increases 3 fold and therefore the potential difference increases a factor 9 (potential difference is equal to the charge on capacity). With these results calculate the increased energy U .

(Click)

Wrong

The equivalent capacity is equal to the sum of capacities, or larger than each of the two capacities.

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Wrong

The potential difference is the same in the two capacitors. As the charge is the product of the potential difference and the ability, the two charges would only be equal if the capabilities were equal.

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Right

As the potential differences in the two capacitors are equal, the charges will be directly proportional to the capacity and therefore the higher the capacitor with higher capacity.

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Wrong

In any two devices connected in parallel, the potential difference is always equal them.

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Wrong

In any two devices connected in parallel, the potential difference is always equal them.

(Click)