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Tutorial ID | Title | Tutorial | Video Tutorial | Revision Notes | Revision Questions | |
---|---|---|---|---|---|---|

14.7 | Capacitance and Capacitors |

In this Physics tutorial, you will learn:

- What is capacitance?
- How to find the capacitance of a charged conductor?
- What are capacitors?
- How many types of capacitors are there?
- How to find the capacitance of a specific capacitor based on its design?
- Why the series and parallel combinations of capacitors are used in electrical circuits?
- What are dielectrics and why do we insert them between the plates of a capacitor?
- How to find the energy stored in a capacitor or system of capacitors?

Do you think conductors can only act as a kind of route for electric charges or they are able to store charges and energy in them? Where do you base your opinion?

Give an idea on how to store electricity and to use it when needed.

In this tutorial, we will discuss about the ability of conductors to store energy in the form of electricity and the methods used to achieve this.

We have explained in the tutorial 14.1 "Electric Charges. Conductors and Insulators", that when some extra charge is given to a conductor, the charge spreads out on the outer surface of conductor. As a result, the conductor is charged by a single type of charge only. It is clear that greater the dimensions of conductor, more extra charges it can carry.

One of methods used for charging a conductor is by connecting it to an electric source (for example a battery). This is because a potential difference is applied on it. As a result, the conductor accumulates opposite charges on its sides. As discussed earlier, this method is called induction and the extra charges produced through this method are known as **induced charges**.

However, in order to produce long lasting extra charges, this method requires an insulating layer between the opposite groups of charges, otherwise they mix up immediately after the source stops providing free charges. The best (and cheapest) method is to use air as an insulating layer. This implies taking a two-pieces conductor which has a gap at middle and connecting each of the pieces with the poles of a battery. We have seen many examples in the previous tutorials where two parallel plates at a distance d between them are connected to the opposite poles of a battery. As a result, plates are charged by opposite signs.

The insulating layer (here the air) between the two charged plates is known as **dielectric**. Vacuum is the best dielectric but it is not always possible to use it. Therefore, air is often used for this purpose because it owns satisfactory dielectric properties (it is a good insulator).

By definition, **the amount of charge a conductor can store when a potential difference is applied is known as capacitance.**

From experiments, it is found that the amount of charge accumulated on the conductor increases with the increase in potential difference. Therefore, the capacitance C of a conductor is

C = *Q**/**∆V*

The unit of capacitance (here Coulomb per Volt) is known as **Farad** [F].

Capacitors are shown in electric circuits through the symbols

What is the capacitance of the parallel plate system shown in the figure if the plates attract each other by a 240 N force?

From the figure, we see that potential difference and electric field are given. We have ΔV = 6V and E = 120 ** V/m**. Thus, given that electric field E is calculated by

E = *F**/**Q*

we obtain for the electric charge Q on the plates

Q = *F**/**E*

=*240 N**/**120 **V**/**m*

= 2 C

=

= 2 C

Therefore, the capacitance of the system is

C = *Q**/**∆V*

=*2C**/**6V*

= 0.333F

=

= 0.333F

**Remark!**

Do not confuse the symbol of capacitance C used in formulae with the unit of electric charge [C], which stands for Coulomb.

Since Coulomb is a very large unit and the voltage operating values are a few volts, Farad is a large unit too. Typical values of capacitance vary from a few microfarads (10-6 F) to a few pico-farads (10-12 F).

A capacitor is a system composed by two conductors separated by a dielectric (usually air). The simplest and smallest capacitor possible would be a proton and an electron at a certain distance from each other. Two oppositely charged spheres not in contact are another example of capacitor.

Most capacitors however are of three designs:

- two parallel plates with some space between them,
- cylindrical capacitors, and
- concentric spheres

Cylindrical capacitors are the most widespread type of capacitors encountered in electronic devices. However, when explaining capacitor properties, it is more suitable to start with parallel plate capacitors, as they are easier to understand.

Once the capacitor is charged, the two conductors carry equal but opposite charges. Therefore, we consider the charge of one plate only when calculating the capacitance of a capacitor.

A parallel plate capacitor consists of two identical parallel plates, each of them having an area A and separated by a distance d between them.

When we connect the two plates with the poles of a battery, the capacitor is charged as discussed in the previous tutorials during the calculation of electric potential energy of a charge inside a uniform field. In such a system, the plate connected to the positive terminal of battery is charged positively (+Q) while the plate connected to the negative terminal of battery is charged negatively (-Q).

Let's see what are the factors affecting the capacitance of a parallel plate capacitor.

- When area A of plates increases, the capacitor is able to carry more charges than before. Thus, capacitance of parallel plate capacitors is proportional to the plates' area.
- If the distance d between plates increases, they are not able to induce as much charges as before because their attracting ability decreases. Therefore, capacitance is inversely proportional to the distance between plates.

Thus, we can write

C ~ *A**/**d*

As we have stated earlier, we must multiply a proportion by a constant to turn it into equation. The constant used here is the already discussed electric constant ϵ^{0}. Hence, we obtain for the capacity of a parallel plate capacitor

C = ϵ_{0} × *A**/**d*

The above formula is true in vacuum (and with a good approximation, in air). When the capacitor is placed inside a dielectric, we must include another factor in the formula. It is known as the dielectric constant ϵ, otherwise known as relative permittivity, which affects the value of capacitance, more precisely it increases the capacitance. Hence, we obtain

C = ϵ × ϵ_{0} × *A**/**d*

The relative permittivity of vacuum is 1, that of air very close to 1 (more precisely 1.00054) while for the other dielectrics, its value is greater than 1.

The unit of electric constant ϵ^{0} is **Farad / metre**. In the previous tutorial, we have mentioned another unit for ϵ^{0} that is C^{2}/N × m_{2}. Since these two units represent the same quantity, we obtain

1 [*C*^{2}*/**N × m*^{2}] = 1[*F**/**m*]

Or

1 [F] = 1[*C*^{2}*/**N × m*]

Two square shaped parallel plates of a capacitor are charged by +4nC and -4nC respectively when connected to the terminals of a powerful battery. The sides of each plate are 1 cm long.

- What is the capacitance if the plates are 8mm apart?
- What is the potential difference of battery?
- What will the new capacitance be if we place an 8mm thick class between the plates? The relative permittivity (dielectric constant) of glass is 10.

a) Let's convert the units into the standard form first. We have Q = 4 nC = 4 × 10^{-9} C, d = 8 mm = 8 × 10^{-3} m and L = 1 cm = 10^{-2} m.

Since the plates have a squared shape, their area is

A = L^{2}

= (10^{-2} m)^{2}

= 10^{-4} m^{2}

= (10

= 10

Thus, the capacitance is

C = ϵ_{0} × *A**/**d*

= 8.85 × 10^{-12} *F**/**m* × *10*^{-4} m^{2}*/**8 × 10*^{-3} m

= 1.1 × 10^{-13} F

= 0.11 pF

= 8.85 × 10

= 1.1 × 10

= 0.11 pF

b) The potential difference of battery is calculated by using the standard formula of capacitors. We have

C = *Q**/**∆V*

Rearranging, we obtain for the potential difference of battery:

∆V = *Q**/**C*

=*4 × 10*^{-9} C*/**1.1 × 10*^{-13} F

= 3.6 × 10^{4} V

= 36 kV

=

= 3.6 × 10

= 36 kV

c) If glass is placed between the plates, we must simply multiply the capacitance found at (a) by the relative permittivity of glass. Thus,

C^{1} = ϵ × ϵ_{0} × *A**/**d*

= ϵ × C

= 10 × 1.1 × 10^{-13} F

= 1.1 × 10^{-12} F

= 1.1 pF

= ϵ × C

= 10 × 1.1 × 10

= 1.1 × 10

= 1.1 pF

A cylindrical-shaped capacitor is a system composed by two coaxial cylinders. The parameters of a cylindrical capacitor are:

- Length, L
- Radius of the smaller cylinder a, and
- Radius of the smaller cylinder b.

Using integration methods (which are not worth to discuss here), we obtain for the capacitance of a cylindrical-shaped capacitor:

C = 2π × ϵ × ϵ_{0} × *L**/**ln **b**/**a*

From mathematics, it is known that if

a^{x} = b then log_{a} b = x

Also, it is known that

log *b**/**a* = log b - log a

When the base of logarithm is e (Euler number = 2.71828), the symbol of logarithm is written as **ln** (natural logarithm) instead of **log**. You can find the values of **log** and ln using a scientific calculator.

A cylindrical-shaped capacitor has the length of the smaller radius equal to 6 mm and that of the larger radius equal to 1 cm. The capacitor is 2 cm long and is placed inside water in 20°C, whose dielectric constant for this temperature is 80. Calculate the capacitance of this cylindrical capacitor. Refer to the figure shown above in theory for illustration.

Let's write the clues first. We have

a = 6 mm = 6 × 10^{-3} m

b = 1 cm = 1 × 10^{-2} m = 10 × 10^{-3} m

L = 2 cm = 2 × 10^{-2} m

ϵ = 80

ϵ^{0} = 8.85 × 10^{-12} *F**/**m*

C = ?

Applying the formula of cylindrical-shaped capacitor, we obtain

C = 2π × ϵ × ϵ_{0} × *L**/**In **b**/**a*

= 2 × 3.14 × 80 × 8.85 × 10^{-12} × *2 × 10*^{-2}*/**ln **10 × 10*^{-3}*/**6 × 10*^{-3}

=**8.89 × 10**^{-11}*/**ln **10**/**6*

=*8.89 × 10*^{-11}*/**ln 10 - ln 6*

=*8.89 × 10*^{-11}*/**2.3 - 1.8*

=*8.89 × 10*^{-11}*/**0.5*

= 1.78 × 10^{-10} F

= 178 pF

= 2 × 3.14 × 80 × 8.85 × 10

=

=

=

=

= 1.78 × 10

= 178 pF

A spherical capacitor is a system composed by two concentric spheres of radius a (the smallest) and b (the largest).

Again, using the integration methods, we obtain for the capacitance of such capacitor:

C = *4π × ϵ × ϵ*_{0}*/**1**/**a* - *1**/**b*

A spherical capacitor has the radii of the two concentric spheres 4 cm and 12 cm respectively. What is the capacitance if the capacitor

- Operates in dry environment
- Operates immersed inside a methanol container of dielectric constant equal to 30.

a) "Operates in dry environment" means "operates in air", whose dielectric constant is taken as 1. Given that a = 4 cm = 0.04 m and b = 12 cm = 0.12 m, we have

C = *4π × ϵ × ϵ*_{0}*/**1**/**a* - *1**/**b*

=*4 × 3.14 × 1 × 8.85 × 10*^{-12}*/**1**/**0.04* - *1**/**0.12*

=*111.156 × 10*^{-11}*/**3 - 1**/**0.12*

= 6.669 × 10^{-11} F

= 666.9 nF

=

=

= 6.669 × 10

= 666.9 nF

Capacitors have fixed values like most electric devices. This is because it is impossible to produce capacitors by demand. Certainly, we may try to insert dielectrics between the plates to change the capacitance but this process is too complicated. It is much easier to combine two or more capacitors in order to obtain the desired capacitance. Remember, we have used such a method when combining springs to obtain the desired spring constant. In that case, we used two main methods of spring combination: in **series** and in **parallel**. The same technique is also used to combine capacitors in an electric circuit.

Two or more capacitors are connected in series if they are placed one after another in the same conducting wire of an electric circuit, as shown in the figure below.

The system of two capacitors connected in series operates as follows:

Initially, both capacitors are uncharged. When the right plate of the second capacitor C^{2} is connected to the negative terminal of battery, it gains a negative charge -Q_{2} and repels the positive charges on the other plate of the same capacitor, producing a positive charge +Q_{2} on it. The charges -Q_{2} and +Q_{2} are opposite in sign but equal in magnitude. The positive charges on the left plate of C^{2} repel the negative charges on the conducting wire as far as possible, i.e. on the right plate of C^{1}, producing a negative charge -Q_{1} on it. As a result, a positive charge +Q_{1} is created on the other plate of the same capacitor. Since there is the same connecting wire between the two capacitors, the charge distribution between these two capacitors is even, as all negative charges that accumulate to the right plate of Q_{1} leave the same number of unpaired positive charges on the left plate of Q_{2}. Therefore, we can write for the two capacitors connected in series

Q_{1} = Q_{2} = Q_{s}

From the figure, it is clear that the total potential difference ΔVs produced by the source (the battery) is divided in two parts, ΔV_{1} and ΔV_{2}, i.e.

∆V_{s} = ∆V_{1} + ∆V_{2}

Substituting the quantities contained in the capacitor formula C = Q / ΔV in the above equation, we obtain after rearranging

Simplifying Q from both sides, we obtain for the capacitance of the series combination of a system composed by two capacitors

We can extend the above rule for more than two capacitors as well, that is

It is clear that when two capacitors are connected in series, the total capacitance of the system is smaller than that of a single capacitor. Therefore, series combination setups are used when in a circuit, a smaller capacitance than that of the individual capacitors available is needed.

A system composed by two capacitors connected in series is shown in the figure below.

What is the total charge stored in this system?

First, we must work out the capacitance of the system of capacitors. Since they are connected in series, we have

=

=

Thus,

C_{s} = *24**/**5* pF

= 4.8 pF

= 4.8 × 10^{-12} F

= 4.8 pF

= 4.8 × 10

Now, let's use the formula of capacitance to find the electric charge of the system. We have

C_{s} = *Q**/**∆V* ⇒ Q = C × ∆V

= 4.8 × 10^{-12} F × 6 V

= 28.8 × 10^{-12} C

= 28.8 pC

= 4.8 × 10

= 28.8 × 10

= 28.8 pC

If two or more capacitors are connected in two or more different branches of the same circuit, which have the same starting and ending point, we say they are connected in parallel, as shown in the figure below.

The system of two capacitors connected in series operates as follows:

Initially, both capacitors are uncharged. If a potential difference ΔV is applied across these capacitors, the plates are charged oppositely, depending on the terminal of battery they are connected with. Since these plates are connected to the same source of electricity, they have the same potential difference, which is equal to the potential difference of battery. We have

∆V_{1} = ∆V_{2} = ∆V

The battery charges the capacitors in accordance to their capacities. Therefore, we can write for the total electric charge stored in the two capacitors

Q_{p} = Q_{1} + Q_{2}

Applying the capacitance formula C = ** Q/ΔV** in the last equation, we obtain after rearranging:

C_{p} × ∆V = C_{1} × ∆V + C_{2} × ∆V

Simplifying ΔV from both sides, we obtain for the capacitance of a parallel system of two capacitors:

C_{p} = C_{1} + C_{2}

This formula is also true for a system composed more than two capacitors connected in parallel. Thus, we have

C_{p} = C_{1} + C_{2} + ... + C_{n}

As you see, the total capacitance of a parallel system is greater than the capacitance of each single capacitor. Therefore, such a combination is used when the capacitors available are smaller than needed.

A system composed by three capacitors is connected to a 12 V battery as shown in the figure.

Calculate:

- Capacitance of the system
- The total charge stored in the system
- The electric charge stored in each capacitor

First, let's appoint a symbol to each capacitor and write the corresponding capacitances. We can write

C^{1} = 3.2 pF = 3.2 × 10^{-12} F

C^{2} = 8 pF = 8 × 10^{-12} F

C^{3} = 12 pF = 12 × 10^{-12} F

C

C

Also, we have ΔV = 12 V.

a) First, we must find the capacitance of the series branch of capacitors that is the equivalent capacitance of C^{2} and C^{3}. Thus,

=

=

=

Hence,

C_{s} = *24**/**5* pF

= 4.8 pF

= 4.8 × 10^{-12} F

= 4.8 pF

= 4.8 × 10

Now, we can assume there are only two capacitors connected in the circuit: C^{1} and C^{p} which are in parallel, as shown below.

Therefore, we obtain for the total capacitance of the system:

C_{tot} = C_{1} + C_{s}

= 3.2 pF + 4.8 pF

= 8 pF

= 8 × 10^{-12} F

= 3.2 pF + 4.8 pF

= 8 pF

= 8 × 10

b) The total charge in the system is calculated by using the capacitance formula. We have

C_{tot} = *Q*_{tot}*/**∆V* ⇒ Q_{tot} = C_{tot} × ∆V

= 8 × 10^{-12} F × 12 V

= 96 × 10^{-12} C

= 96 nC

= 8 × 10

= 96 × 10

= 96 nC

c) The charge distribution is in accordance with the capacitance of each branch in the parallel section, in which the same potential difference (12 V) is applied. Also, we know that the electric charge is the same for each capacitor in a series combination of capacitors. Thus, we have

Q_{1} = C_{1} × ∆V

= 3.2 × 10^{-12} F × 12 V

= 38.4 × 10^{-12} C

= 38.4 pC

= 3.2 × 10

= 38.4 × 10

= 38.4 pC

Also,

Q_{2} = Q_{3} = Q_{s} = C_{s} × ∆V

= 4.8 × 10^{-12} F × 12 V

= 57.6 × 10^{-12} C

= 57.6 pC

= 4.8 × 10

= 57.6 × 10

= 57.6 pC

**Remark!**

It must be noted that when the capacitance of a system increases, its common potential difference ΔVcom decreases. This conclusion derives from the fact that capacitance and potential difference are inversely proportional for constant charge (C = Q / ΔV, so C × ΔV = C = constant). Let's explain this point through an example.

The capacitor C^{1} is charged by means of a 24 V battery as shown in the figure.

The capacitor C^{1} is then connected in parallel with an uncharged capacitor C^{2} of capacitance 24 nF as shown in the other figure shown below.

Calculate:

- The common capacitance after the capacitors are connected in parallel.
- The final charges in each capacitor.

a) The charge in the 8 nF capacitor when it is alone in the circuit is

Q_{1} = C_{1} × ∆V_{1}

= 8 nF × 12 V

= 96 nC

= 8 nF × 12 V

= 96 nC

Since the capacitors in the second figure are connected in parallel, the total charge is distributed in two branches. Thus, we have

Q_{tot} = Q_{1} + Q_{2}

Here the total charge Q_{tot} corresponds to the previous charge of Q_{1} since the capacitor Q_{2} was initially uncharged.

In addition, the total capacitance of this parallel system of capacitors is

C_{tot} = C_{1} + C_{2}

= 8 nF + 4 nF

= 12 nF

= 8 nF + 4 nF

= 12 nF

Therefore, the common potential difference after the capacitor C^{2} is added in the system is

V_{com} = *Q*_{tot}*/**C*_{tot}

=*96 nC**/**12 nF*

= 8 V

=

= 8 V

As you see, the new common difference is smaller than the initial potential difference when a single capacitor was present in the circuit.

b) We can apply the capacitance formula for each capacitor separately to find the charge stored in them. Thus, we obtain for the new charges on the two capacitors:

Q_{1}^{1} = C_{1} × ∆V_{com}

= 8 nF × 8 V

= 64 nC

= 8 nF × 8 V

= 64 nC

and

Q_{2}^{1} = C_{2} × ∆V_{com}

= 4 nF × 8 V

= 32 nC

= 4 nF × 8 V

= 32 nC

As you see, the charge is conserved as the total charge after is equal to the total charge before the second capacitor is connected in the circuit (both are 96 nC).

The source (here the battery) does some work to charge a capacitor from 0 to Q (i.e. increasing the charge of capacitor by ΔQ). As discussed in the previous tutorials of this section, the work W needed for such a process is

W = ∆Q × ∆V

This process brings a uniform increase of the charge as shown in the graph below.

It is clear that the area under the graph represents the work done by the source to charge the capacitor. Since the figure under the graph is a right triangle, whose area is A = a × b / 2, we obtain for the work done by the source to charge the capacitor

W = *Q × ∆V**/**2*

where Q is the charge when the capacitor is fully charged.

As stated earlier in this chapter, the work done on a system contributes in the increase of its potential energy. This is also true in the specific case, i.e. the work done by the source to charge the capacitor from 0 to Q brings an increase in the potential energy of capacitor from 0 to PE. Therefore, we obtain for the electric potential energy stored in a charged capacitor:

W = EPE_{C} = *Q × ∆V**/**2*

We can transform the above equation into other forms using the capacitor formula C = Q /ΔV. Thus, we can write

EPE_{C} = *Q × ∆V**/**2*

=*(C × ∆V) × ∆V**/**2*

=*C × ∆V*^{2}*/**2*

=

=

and

EPE_{C} = *Q × ∆V**/**2*

=*Q × **Q**/**C**/**2*

=*Q*^{2}*/**2C*

=

=

Obviously, all these formulae are equivalent as they are obtained through equivalent transformations.

Two capacitors C^{1} = 60 nF and 30 nF are connected to the terminals of a 12 V battery as shown in the figure below.

- What is the energy stored in the system?
- What is the energy stored in the capacitor C
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a) The total capacitance of this system of capacitors connected in series is

=

=

=

Thus,

C_{s} = *60**/**3* nF

= 20 nF

= 20 nF

Therefore, the energy stored in the system is

EPE_{C} == *C × ∆V*^{2}*/**2*

=*20 nF × (12 V)*^{2}*/**2*

= 1440 nJ

= 1.44 μJ

= 1.44 × 10^{-6} J

=

= 1440 nJ

= 1.44 μJ

= 1.44 × 10

b) The capacitor C^{1} carries the same charge as the entire system as it is connected in series with C^{2}. But first, we must find this charge using the capacitor formula. We have

Q_{1} = Q_{2} = Q

= C_{s} × ∆V

= 20 nF × 12 V

= 240 nC

= 2.4 × 10^{-7} C

= C

= 20 nF × 12 V

= 240 nC

= 2.4 × 10

Therefore, we obtain for the energy stored in the capacitor C^{1} (we have C^{1} = 60 nF = 6 × 10^{-8} F):

W_{1} = EPE_{C}1 = *Q*^{2}*/**2C*_{1}

=*(2.4 × 10*^{-7} C)^{2}*/**2 × 6 × 10*^{-8} F

= 0.48 × 10^{-6} J

= 4.8 × 10^{-7} J

=

= 0.48 × 10

= 4.8 × 10

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