Capacitor Notes

Capacitor

Capacitor is a device/conductor which is used to store the electrical energy in the form of charge. It is used in some electrical devices that requires very high current (25A - 50A) to start them due to static friction (like fan, motor, computer, camera flash). Capacitor stores the electrical energy in the form of charge and supply at once to such electrical in the form of large current. Battery/ cell also converts chemical energy to electrical energy in the form of current but that current is small.

Capacitance of a capacitor:

The capacity or ability of capacitor to store electrical energy in the form of charge is known as capacitance of capacitor. If Q be the charge given to a capacitor so that its potential increases to V then experimentally, it is found that the charge stored in the capacitor is directly proportional to the electrical potential (p.d). i.e.

$Q \propto V$ Or, Q = CV where C is proportionality constant and known as capacitance of a capacitor.

Or, $C = \frac{Q}{V}$

In other words, capacitance of a capacitor can also be defined as the ratio of charge to potential difference. Its unit is coulomb per volt $(CV^{-1})$ which also known as Farad (F).

If p.d. V = 1 V and Q = 1 C then C = 1 $CV^{-1} = 1$ Farad

Thus, capacitance of a capacitor is said to be of 1 Farad if 1Coulomb charge can raise its potential through 1 Volt. Capacitance of 1 Farad is extremely high value so generally capacitance is measures in $\mu$ F.

$1 \text{ mF} = 1 \times 10^{-3} \text{ F}$

$1 \mu F = 1 \times 10^{-6} F$

1 micro-micro Farad ( $\mu\mu$ F) = 1 pico-farad (pF) = 1×10^-12 F

Note: Although $C = \frac{Q}{V}$ , the capacitance of a capacitor does not depend upon charge and potential but depends upon nature of material and dimensions of capacitor.

Types of capacitor:

There different types of capacitor;

• (i) Isolated sphere as capacitor, (ii) Spherical capacitor (Two concentric spherical shell)
• (iii) Parallel plate capacitor

(i) Isolated sphere as capacitor:

Let us consider an isolated spherical conductor of radius ‘R’ and charge on its surface is ‘Q’.

The potential at any point on the surface is given by

$V = \frac{1}{4\pi\epsilon_o} \frac{Q}{R}$

, $4\pi\epsilon_o~R$ $\frac{Q}{V}=4\pi\epsilon_o R \qquad \text{where, } \epsilon_o~\text{is the permittivity of free space.}$

According to the definition of capacitance, $\frac{Q}{V} = C$ (capacitance), so we can write

Capacitance of an isolated sphere, $C = 4\pi\epsilon_0 R$

Thus capacitance of isolated sphere depends upon

• Radius as $C \propto R$ (i)
• Permittivity of medium as $C \propto \epsilon$ (ii)

In CGS- system, $4\pi\epsilon_0 = 1$

So in CGS-system, capacitance of an isolated charged sphere is numerically equal to its radius.

Note:

Capacitance of an isolated sphere, $C = 4\pi\epsilon_0 R$

For C = 1 F, R = $\frac{1}{4\pi\epsilon_0}$ = $9 \times 10^9$ m

This is greater than radius of earth. ( $R_E = 6.4 \times 10^6 \text{ m}$ )

For capacitance to be 1F, the size of charged sphere must be extremely greater than the earth.

Thus it can be concluded that the isolated charged sphere has very low capacity to store charge and hence can be regarded as not useful capacitor.

Parallel plate capacitor.

Let us consider two metallic plates P₁ and P₂ separated by a distance ‘d’ which is small as compared to the dimensions of plates. The plate P₁ is given a charge +Q and the plate P₂ is earthed as shown in figure above. When charge +Q is given to the plate $P_1$ , the charge -Q is induced on the inner face of plate $P_2$ and charge +Q on its outer face. When the plate $P_2$ is earthed, the charge +Q on its outer face is neutralized due to earthing. The charges +Q and -Q on plates $P_1$ and $P_2$ develop an electric field between them. If each plate has a uniform surface charge density $(\sigma)$ , then electric field E between the plates is uniform and is given by

$E = \frac{\sigma}{\varepsilon_0}$ …(i)

Here $\varepsilon_0$ is the permittivity of free space (as the plates are kept in vacuum or air).

Also the electric field developed between the plates can be written as

$E = \frac{V}{d} \dots (ii)$

Where V is the electric potential difference between the plates.

From equations (i) and (ii).

$\frac{V}{d} = \frac{\sigma}{\varepsilon_0}$

$V = \frac{d\sigma}{\varepsilon_0}$

Since surface charge density, $\sigma = \frac{Q}{A}$ then above equation becomes,

$\mathbf{V} = \frac{\mathbf{d} \mathbf{Q}}{\mathbf{\epsilon_0} \mathbf{A}}$ $\frac{\mathbf{Q}}{\mathbf{V}} = \frac{\mathbf{\epsilon_0} \mathbf{A}}{\mathbf{d}}$

Since $\frac{Q}{V}$ is capacitance (C), so above equation becomes

$C = \frac{\epsilon_0 A}{d}$

This is the required expression for capacitance of parallel plate capacitor.

But if the space between the plates is filled with a medium of absolute permittivity ’ $\epsilon$ ’ then capacitance of the capacitor is given by

$C = \frac{\varepsilon A}{d}$

Factors affecting the capacitance of parallel plate capacitor:

The expression for capacitance of parallel plate capacitor is given by

$C = \frac{\varepsilon_0 A}{d}$

Thus the capacitance of parallel plate capacitor is

• (i) directly proportional to permittivity of medium i.e. $C \propto \epsilon$
• (ii) directly proportional to area i.e. $C \propto A$
• (iii) inversely proportional to permittivity of medium i.e. $C \propto \frac{1}{d}$

Combination of Capacitors:

In many electric circuits, capacitors are combined suitably to obtain the desired capacitance. There are two basic ways of connecting the capacitors: series combination and parallel combination. When the capacitors are connected in series, the total capacitance decreases while the total capacitance increase when they are connected in parallel.

(i) Series Combinations of Capacitors:

In the combination, the second plate of the first capacitor is joined to the first plate of the second capacitor and so on. Below figure, shows three capacitors of capacitance $C_1$ , $C_2$ and $C_3$ connected in series.

When a battery is connected across the free ends of the series combination, a charge +Q is given to the plate A of the first capacitor it induces a charge -Q on the right face capacitor +Q on the left face of the third capacitor. Similarly two plates of the second capacitor are induced with charge of the same magnitude Q but opposite sign. Therefore, each plate capacitor in this combination contains the same amount of charge as given to the first capacitor.

Let V₁, V₂ and V₃ be the potential difference across C₁, C₂ and C₃ respectively. Then,

The resultant potential difference (V) across the combination is given by

$V = V_1 + V_2 + V_3$ [Since $Q = VC \rightarrow V = \frac{Q}{C}$ ]

$V = \frac{Q}{C_1} + \frac{Q}{C_1} + \frac{Q}{C_3}$

Or, $V = Q \left(\frac{1}{C_1} + \frac{1}{C_2} + \frac{1}{C_3}\right) \dots (i)$

If Ceq is the equivalent capacities of the series combination, then

$Q = VC_{eq}$

$V = \frac{Q}{C_{eq}} \dots (ii)$

From equations (i) and (ii)

$\frac{Q}{C_{eq}} = Q \left( \frac{1}{C_1} + \frac{1}{C_2} + \frac{1}{C_3} \right)$

$\frac{1}{C_{eq}} = \frac{1}{C_1} + \frac{1}{C_2} + \frac{1}{C_3}$

Similarly, if there are n capacitors connected in series, then the resultant capacitance, Ceq is given by

$\frac{1}{C_{eq}} = \frac{1}{C_1} + \frac{1}{C_2} + \frac{1}{C_3} + \dots + \frac{1}{C_n}$

Thus, when a number of capacitors are connected in series, the reciprocal of equivalent capacitance is equal to the sum of the reciprocals of the capacitances of each capacitor.

The resultant capacitance in the parallel combination is always less than that of any smallest individual, while the overall potential difference across the combination increases.

(ii) Parallel Combinations of Capacitors:

The capacitors are said to be connected in parallel if positive plates of all the capacitor are connected together at one point and negative plates are connected together at the other point. In the figure given below three capacitors of capacitance $C_1$ , $C_2$ and $C_3$ are connected in parallel.

Let the common terminals A and B be connected to a battery having a potential difference V. Let +Q be the charge supplied by the battery to the point A that is distributed among the three capacitors depending upon their capacities. The charge produces same potential across all capacitors. Let $Q_1$ , $Q_2$ and $Q_3$ be the charges distributed on the right side plates of capacitors $C_1$ , $C_2$ and $C_3$ respectively. Then, the total charge supplied

$Q = Q_1 + Q_2 + Q_3 \dots (i)$

By induction, $-Q_1$ , $-Q_2$ and $-Q_3$ charges will be developed in second plate of capacitors $C_1$ , $C_2$ and $C_3$ respectively. The potential difference across each capacitor is same, V but the charges stores are different as $Q_1$ , $Q_2$ and $Q_3$ . These are given as

$Q_1 = V C_1, Q_2 = V C_2 \text{ and } Q_3 = V C_3$

Then, Eq. (i) becomes

$O = V C_1 + V C_2 + V C_3$

$O = V (C_1 + C_2 + C_3) \dots (1)$

If C_eq be the resultant capacitance of the parallel combination, then

$Q = VC_{eq}$ …(2)

From equations (1) and (2)

$C_{eq}V = V (C_1 + C_2 + C_3)$

$\therefore C_{eq} = C_1 + C_2 + C_3$

Similarly, if there are n capacitors connected in parallel, then the resultant capacitance, C is given by

$C_{eq} = C_1 + C_2 + C_3 + ... + C_n$

Thus, when a number of capacitors are connected in parallel, the resultant capacitance is equal to the sum of capacitances of individual capacitor. The capacitance in parallel combination increases while the potential difference for the combination is same as that of individual capacitor. This combination is useful when high

value of capacitance is required at lower potential. In this combination at lower potential, large quantity of charge can be stored.

Energy stored in a charged capacitor:

When we connect a battery across the two plates of a capacitor a current flows in the circuit which charges the capacitor. If some charge is developed in the capacitor, an electric field is set up in the capacitor which opposes the charging current of the battery. Therefore, for the battery to charge the capacitor further, work will be done by the battery against the electric field set up in the capacitor. The work done in charging a capacitor is stored in the capacitor in the form of electric potential energy.

Let us consider a capacitor of capacitance C. Suppose that initially the capacitor does not carry charge but after some time it has charge Q and potential V, then

$O = V C$

Let the battery supplies a small amount of charge dQ at constant potential difference V, then a small amount of work done by the battery in charging the capacitor is

$dW = V dQ$

Or, $dW = \frac{Q}{C} dQ \dots (i)$

Then the total amount of work done in charging the capacitor from 0 to Q is obtained by integrating the above equation as

$$\begin{split} & W = \int_{0}^{Q} \frac{q}{C} \, dq \\ & \text{Or, } W = \frac{1}{C} \int_{0}^{Q} q \, dq \\ & \text{Or, } W = \frac{1}{C} \left[ \frac{q^2}{2} \right]_{0}^{Q} \end{split}$$

Or, W = $\frac{1}{C} \frac{Q^2}{2}$

This work done is stored inside the capacitor in the form of electric potential energy. Thus the energy stored in a capacitor is given by

$U = \frac{1}{2} \frac{Q^2}{C} \dots (i)$

Substituting Q = C V in equation (i), we get,

$U = \frac{1}{2} \frac{C^2 \, V^2}{C} =$

$U = \frac{1}{2} C V^2$ …(ii)

Substituting $C = \frac{Q}{V}$ in equation (ii), we get

$U = \frac{1}{2} \frac{Q}{V} V^2$

$U = \frac{1}{2} Q V .....(iii)$

If Q is in Coulomb, C is in Farad and V is in volt, the energy stored (U) in a capacitor is in Joule. As only half of the supplied energy QV is stored, the remaining part is lost as heat during charging the capacitor by the moving charge.

Uses of Capacitors

There are many applications of the capacitors; some of them are given below.

• 1. They can be used as device for storing charge.
• 2. Capacitors are used in increasing the efficiency of alternating current power transmission.
• 3. The ignition system of any automobile engine contains a capacitor to eliminate the sparking of the points when they open or close.
• 4. Capacitors are used in scientific investigation.
• 5. To give electric field of desired configuration.
• 6. They are also used in oscillators, in filter circuits, in time-delay devices and in many other electronic devices.

Dielectric

The material which cannot conduct electric current but can transmit electric field is known as dielectric. Paper, glass, air, pure water, mica etc. are some examples of dielectric.

Effect of Dielectric

Consider a parallel plate capacitor each plate of area A and separated by a distance ‘d’. If the medium between the plates is vacuum or air, the capacitance of capacitor is

$C_{air} = \frac{\varepsilon_o A}{d}$ …(i)

If the medium of dielectric constant K is placed between two plates of the capacitor the capacitance of capacitor will be

$C_m = \frac{\epsilon A}{d} = \frac{K \, \epsilon_o \, A}{d} \, \dots (ii)$

Dividing equation (ii) by (i), we get,

$\frac{C_m}{C_{air}} = K$

As the dielectric constant K for any medium is greater than that of the vacuum or air, so

$C_m > C_{air}$

Hence, the capacitance increases if a dielectric is placed between the two capacitor plates.

Types of dielectric molecules:

There are two types of dielectric; (i) Non polar dielectric molecules (ii) Polar dielectric molecules.

(i) Non polar dielectric molecules:

The molecules in which the centre of gravity of positive charge does not coincide with the centre of gravity of negative charge are known as non-polar molecules.

HCl, H₂O, CO, NH₃, N₂O are some examples of non-polar molecules.

(ii) Polar dielectric molecules:

The molecules in which the centre of gravity of positive charge exactly coincides with the centre of gravity of negative charge are known as non-polar molecules.

O₂, H₂, N₂, CH₄, are some examples of polar molecules.

Polarization:

When a non-polar dielectric is placed in an electric field, the centre of gravity of positive charge is pulled in the direction of applied electric field while the centre of gravity of negative charges is pulled in the direction opposite to electric field. The process of separation of centre of gravity of positive and negative charges by the application of an electric field is known as polarization.

Due to polarization, an internal electric field $(E_P)$ is setup but in the direction opposite to applied magnetic field $(E_o)$ . Due to which resultant electric field decreases and decreases the potential difference between the two plates of capacitor and this ultimately increases the capacitance of the capacitor.

Dielectric Constant (or Relative Permittivity)

The ratio of capacitance of a capacitor having a dielectric between the plates of the capacitance of the same capacitor without any dielectric (i.e. air or vacuum) is called of the material used. The relative permittivity of substance is denoted by $\epsilon_r$ or K and given by

$\epsilon_r = \frac{C_m}{C_{air}} = \frac{\epsilon \; A/d}{\epsilon_o A/d}$

$\varepsilon_r = \frac{\varepsilon}{\varepsilon_o}$

Therefore the relative permittivity or dielectric constant is the ratio of permittivity of medium to that of free space. It has no unit.

Capacitor Numericals

Warm up Numericals:

Q. Find the equivalent capacitance of the given figure.

(a)

Ans: 2µF

(b)

Ans: 3µF

(c)

Ans: 1µF

(d)

Given; $c2=c3=c5=2 \mu F$ ; $c1=c4=1 \mu F$ .

Ans: $2.667 \, \mu F$

b. In the given capacitors circuit applied potential between a b is 220 V. What is the equivalent capacitance of the network between a and b? Given: $C_1 = C_2 = C_3 = C_4 = C_5 = 2\mu F$ . (3)

Ans:?

Numerical Problems

Q.1 A capacitor is charged from a 20 V d.c. and is found to have carried a charge of 10 $\mu$ C. What was the capacitance of the capacitor and energy stored in it? Ans: $0.5\mu$ C and $1\times10^{-4}J$

Here,

Potential $(V) = 20 V$

Charge on capacitor, $Q = 10 \mu C = 10 \times 10^{-6} C$

Capacitance of capacitor, $C = ?$

Energy stored in first capacitor: $E = ?$

We have,

Charge, $Q = VC$

$C = \frac{Q}{V} = \frac{10 \times 10^{-6}}{20} = 0.5 \times 10^{-6} \ C = 0.5 \ \mu C$

And, energy stored, $E = \frac{1}{2} V^2C$

$E = \frac{1}{2} 20^2 \times 0.5 \times 10^{-6} = 1 \times 10^{-4} J$

Q.2 A 300V battery is connected across capacitors of $3\mu F$ and $6\mu F$ in parallel. Calculate the energy stored in each capacitors.

Here,

Potential $(V) = 300V$

Capacitance of 1^st capacitor (C₁) = $3\mu$ F = $3\times10^{-6}$ F

Capacitance of $2^{nd}$ capacitor (C₂) = $6\mu$ F = $6\times10^{-6}$ F

Energy stored in first capacitor: $E_1 = \frac{1}{2} V^2 C_1 = \frac{1}{2} \times 3 \times 10^{-6} \times 300^2 = 0.135 \text{ J}$

Energy stored in first capacitor: $E_1 = \frac{1}{2} V^2 C_2 = \frac{1}{2} \times 6 \times 10^{-6} \times 300^2 = 0.27 \text{ J}$

Q.3 A parallel plate air capacitor of capacitance 245 pF has a change of magnitude 0.148 $\mu$ C on each plate. Find the potential difference and electric field intensity between the plates if the distance between the plates is 5mm?

Here,

Capacitance of capacity (C) = $245 \text{ pF} = 245 \times 10^{-12} \text{ F}$

Charge on capacitor (q) = 0.418 $\mu$ C = 0.148 $\times 10^{-6}$ C

Distance between plate (d) = $5 \text{ mm} = 5 \times 10^{-3} \text{ m}$

P.d. between plate (V) = ?

Electric field intensity (E) = ?

We have

$q = CV$

Or, $V = \frac{q}{C} = \frac{0.148 \times 10^{-6}}{245 \times 10^{-12}} = 604 \text{ V}$

Again,

$E = \frac{V}{d} = \frac{604}{5 \times 10^{-3}} = 120816 \ Vm^{-1}$

Q. 3 A parallel plate air capacitor has a capacitance of $10^{-9}$ F. What is potential difference is required for a charge of $5 \times 10^{-5}$ C? What is the total energy stored in it?

Here,

Capacitance (C) = $10^{-9}$ F

Charge (q) = $5 \times 10^{-5}$ C

Potential Difference (V) = ?

Energy Stored (E) = ?

We know,

$q = CV$

$V = \frac{q}{C} = \frac{5 \times 10^{-5}}{10^{-9}} = 5 \times 10^4 \text{ V}$

Also,

Energy stored, $E = \frac{1}{2} CV^2 = \frac{1}{2} \times 10^{-9} \times (5 \times 10^4)^2 = 1.25 J$

Q4. Two capacitors of capacitance $4\mu F$ and $12\mu F$ respectively are connected in series with a battery of 220V. Find the charge and potential across each capacitor.

Here,

Capacitance of 1^st capacitor (C₁) = $4 \mu F = 4 \times 10^{-6} F$

Capacitance of $2^{nd}$ capacitor (C₂) = 12 $\mu$ F = 12×10^-6 F

Emf, V = 200 V

We know in series combination of capacitors, charge is same on each capacitor.

$q_1 = q_2 = q$

$\therefore V_1 = \frac{q}{C_1} \text{ and } V_2 = \frac{q}{C_2}$

Also in series combination of capacitors, potential is divided. i.e.

$V = V_1 + V_2$

Or, $200 = \frac{q}{C_1} + \frac{q}{C_2} = q \left( \frac{1}{C_1} + \frac{1}{C_2} \right)$

$\mathrm{Or},\, q = \frac{200}{1/C_1 + 1/C_2} = \frac{200}{1/4 \times 10^{-6} + 1/12 \times 10^{-6}} \, = 6 \times 10^{-4} \; \mathrm{C}$

And,

$V_1 = \frac{q}{C_1} = \frac{6 \times 10^{-4}}{4 \times 10^{-6}} = 150V$

$V_2 = \frac{q}{C_2} = \frac{6 \times 10^{-4}}{12 \times 10^{-6}} = 50V$

Q5. Two capacitors of capacitance $5\mu F$ and $15\mu F$ respectively are connected in series with a battery of 200V. Find the charge and potential across each capacitor.

Here,

Capacitance of 1^st capacitor (C₁) = $5\mu$ F = $5\times10^{-6}$ F

Capacitance of $2^{nd}$ capacitor $(C_2) = 15\mu F = 15 \times 10^{-6} F$

Emf, $V = 220 \text{ V}$

We know in series combination of capacitors, charge is same on each capacitor.

$q_1 = q_2 = q$

$\therefore V_1 = \frac{q}{C_1} \text{ and } V_2 = \frac{q}{C_2}$

Also in series combination of capacitors, potential is divided. i.e.

$V = V_1 + V_2$

Or, $220 = \frac{q}{C_1} + \frac{q}{C_2} = q(\frac{1}{C_1} + \frac{1}{C_2})$

Or, $q = \frac{220}{1/C_1 + 1/C_2} = \frac{220}{1/5 \times 10^{-6} + 1/15 \times 10^{-6}} = 8.26 \times 10^{-4} \text{ C}$

And, $V_1 = \frac{q}{C_1} = \frac{8.26 \times 10^{-4}}{5 \times 10^{-6}} = 165V$

$V_2 = \frac{q}{C_2} = \frac{8.26 \times 10^{-4}}{15 \times 10^{-6}} = 55V$

• Q6. A thundercloud and the earth can be regarded as a parallel plate capacitor. Taking the area of the thundercloud to be 50 km² and its height above the earth as 1 km and its potential 10 kV, calculate the energy stored. (Permittivity of free space, $\varepsilon_0 = 8.85 \times 10^{-12} \, \text{C}^2 \text{N}^{-1} \text{m}^{-2}$ ) Ans: 2212.5 J
• Q7. What distance should the two plates each of area $2 \times 10^2$ m² of an air capacitor be placed in order to have the same capacitance as a spherical conductor of radius 0.5m? Ans: 31.83m
• Q.8 A sheet of paper 40 mm wide and $1.5 \times 10^{-2}$ mm thick between metal foil of the same width is used to make a $2 \times 10^{-6}$ F capacitor. If the dielectric constant of the paper is 2.5, what length of the paper required? Ans: 33.9m