# CAPACITORS FOR CLASS XII

CAPACITANCE

When electric charge is given to a conductor, the potential of the conductor increases. The change in potential is found to be directly proportional to the charge given to the conductor. Thus, if a charge Q raises the potential of a conductor by V,

V α Q
Or Q α V
Q = CV

Where C is a constant known as the capacitance of the conductor.
From the above equation, if V = 1, C = Q

Therefore, the capacitance of a conductor may be defined, as the charge required raising its potential by unity.

The unit of capacitance is the Farad
From equation (1)
C = Q/V
Ie, Capacitance = Charge / Potential
Therefore 1 farad = 1coulomb/1 volt

Hence the capacitance of a conductor is one farad, if a charge of one coulomb raises its potential by one volt.

Farad is too large a unit for practical purposes. Hence capacitance is usually expresses in macro farad (mf or μF) or Pico farad (micro-micro farad, μμf, or pf )
1μf = 10-6F
1pf = 10-6μF=10-12F

Dimensions of Capacitance

Capacitance = Charge/Potential

Dimensions of capacitance = Dimensions of charge/ Dimensions of potential

IT/M1L2I-1T-3 = M-1L-2I2T4
Ie Capacitance has dimensions M-1L-2I2T4

Capacitor or condenser

Any arrangement by which the capacity of a conductor is increased it called a capacitor or electrical condenser. Thus a capacitor is an electrical device for storing electric charges. In its simplest practical form, a capacitor consists of two conducting plates arranged parallel and close to each other with a dielectric medium ( may be air) between them.

Principle of the Capacitor

Consider an insulated metal plate A. Let it be charged with a charge +Q so that its potential becomes V. Now the capacitance of the plate, C = Q/V

Let another insulated metal plate be placed near A. The plate B gets charged by induction. The side of B nearer to A gets negatively charged and the farther side gets positively charged. The negative charge on B tends to lower the potential of A, while the positive charge on B tends to raise the potential of A. But as the negative charge is nearer to A its effect will be more predominant on the positive charge on B. Hence the potential of A gets slightly lowered. Therefore the capacitance of A is slightly increased.

Now, let the plate B be earthed. The positive charge on the outer surface of B flows to the earth. The negative charge on B remains due to the electrostatic attraction with the positive charge on A. Therefore the potential of A is very much lowered. Hence the capacitance of A increases considerably. This is the principle of the capacitor.

A capacitor has a capacitance of one farad if a charge of one coulomb flows from one conductor to the other when the pd between the conductors is one volt.

Capacitance of a Parallel Plate capacitor

The parallel plate capacitor consists of two parallel metal plates kept separated by a small distance. The capacitance of a parallel plate capacitor is the ratio of the charge on any one plate to the potential difference between the plates.

Let A and B be two metal plates each of area A. Let them be kept separated by a small distance d. The plate B is earthed.

Let a charge +Q be given to the plate A. The surface charge density, or charge per unit area of the plate,

σ = Q/A
or Q = Aσ.
The charge +Q on A induces a charge –Q on the inner surface of B, and a charge +Q on the outer surface of B. The charge +Q on B leaks to the earth. The electric field between the plates is uniform.
The intensity of the electric field at any point between the plates is given by
E =σ /ε0 where ε0 is the permittivity of free space.

Therefore work done in moving unit positive charge from A to B = Ed = σ / ε0 d.

This is equal to the potential difference V between the plates.

V = σ d/ ε0

Therefore Capacitance of the parallel plate capacitor,
C = Charge on any one plate/Potential Difference = Q / σ d/ ε0
= A σ ε0/ σ d ( Q = A σ )

If the space between the plates is filled with a medium of relative permittivity εr,
C = A εr ε0/ d Farad.

From the above equation it is clear that the capacitance of a parallel plate capacitor may be increased by
1 increasing the area of the plates.
2.decreasing the separation between the plates, and
3. Using a medium having a high value for the dielectric constant.

If the parallel plate capacitor has n identical plates arranged at equal distance d apart and alternate plates connected together, its capacitance is given by C = εr ε0(n-1)A/d farad.

Here A represents the area of each plate.

Effect of Dielectric between the plates

A dielectric medium between the plates increases the capacitance. This is quite clear from the expression for the capacitance of the parallel plate capacitor. Also it is clear that thinner the dielectric, greater is the value fo the capacitance.

It keeps the plates very close together without actual contact.

It also prevents the electric discharge between the plates.

Different Types of Capacitors

a) Paper capacitor
In this type two long strips of thin tin foil or aluminum foil act as the conducting plates. Paraffin paper acts as the dielectric medium. It is placed between the metal foil and then rolled up into a cylindrical form. These capacitors are very cheap and convenient to handle. They are widely used in many electronic devices.

b) Electrolytic capacitor

It has been mentioned earlier that the capacitance of a capacitor can be increased by decreasing he thickness of the dielectric medium between the plates. In electrolytic capacitors, the thickness of the dielectric is extremely small (of the order of 10-6cm) Therefore the capacitance is very large. Hence these capacitors are used in electrical circuits where large capacitance is required.

There are two types of electrolytic capacitors – the wet type and the dry type. The wet type of capacitor consists of an aluminium cylinder A, kept dipped in a solution of ammonium borate taken in another cylindrical aluminium vessel C. A is the anode and C is the cathode. Electrolysis takes place when a D C potential of about 450 volts is applied between the anode and cathode. Then a very thin coating of aluminium oxide is formed on the anode. This film offers a very high resistance to the flow of current in one direction and a very low resistance to the current in the opposite direction. Thus the foil acts as a dielectric for currents in one direction. This behaves as a parallel plate capacitor. The thickness of the dielectric being very small, electrolytic capacitors have large capacitance.

Electrolytic capacitors are widely used tin power supply units for smoothening of varying currents. They can be used only with D C since a reversal of polarity causes the break down of the dielectric film.

Combination of Capacitors.

Capacitors may be connected either in series or in parallel.

a) Capacitors in series

Let three capacitors of capacities C1, C2 and C3 be connected in series as in figure between two points A and B. Let a charge +Q be given to a plate of capacitor C1. This induces a charge –Q on the other plate. Since the capacitors are connected in series, the charge on all the capacitors will be the same, but the potential across each capacitor will be different. Let V1, V2 and V3 be the potentials across the plates of C1, C2 and C3 respectively.

Then V1 = Q/C1, V2 = Q/C2, and V3 = Q/C3.

Let V be the pd between A and B.

Then V = V1 + V2 + V3

Ie V = Q/C1 + Q/C2 + Q/C3.

Let the three capacitors be replaced by a single capacitor or capacitance C such that when the pd between A and B is V, the charge on it is the same as before. Then C is called the effective of equivalent capacitance of the combination.

P.d between A and B, V = Q/C

From equn (4) and (5)

Q/C = Q/C1 + Q/C2 + Q/C3
1/C = 1/C1 + 1/C2 +1/C3

Thus, when a number of capacitors are connected in series, the reciprocal of the effective capacitance is the sum of the reciprocals of the individual capacitances.
It may be noted that when several capacitors are connected in series,

1. The charge on each capacitor is the same, and
2. The effective capacitance is less than the smallest individual capacitance.
The series combination is used to reduce the capacitance.

Capacitors in Parallel

Let three capacitors of capacitances C1,C2 and C3 be connected in parallel between A and B. Let V be the p.d applied between A and B. Therefore the Pd across the plate of each capacitor is V. Let Q1, Q2 and Q3 be the charges acquired by C1, C2 and C3 respectively.

Total Charge, Q = Q1 + Q2 + Q3
But Q1 = C1V, Q2 = C2V, Q3 = C3V

Total charge on the three capacitors
Q = Q1 + Q2 + Q3
= C1V + C2V + C3V
= (C1 + C2 + C3) V

Let the three capacitors be replaced by a single capacitor of capacitance C such that it acquires the same charge Q when the p.d between its plates is V. Then this capacitor is called the equivalent capacitor of the combination.

Then Q = CV
CV = (C1 + C2 + C3 ) V
Or C = C1 + C2 + C3

Thus, when a number of capacitors are connected in parallel, the effective capacitance is the sum of the capacities of the individual capacitors.

It may be noted that when several capacitors are connected in parallel,
The p.d across the plates of each capacitor is the same.
The effective capacity is larger than the largest individual capacitance.

Parallel combination of condensers is used to increase the capacity.

Energy of a charged capacitor.

Let a capacitor be charged with a charge Q. When it is charged, its potential increases uniformly from its initial value zero to the final potential V. Therefore the average potential of the capacitor in the processes of charging = 0 + V/2 = V/2. Let C be the capacitance of the capacitor.

The potential at a point is the work done in taking unit positive charge from zero potential to that point. Therefore the work done in giving a charge Q to the capacitor,
W = Charge x Average potential
= Q.V/2 = ½ QV

This work done will be stored in the capacitor as its electrical potential energy. Thus, energy of the charged capacitor, W = ½ QV
Or W = ½ CV2, since Q =CV
W = ½ Q2/C, since V = Q/C.

PROBLEMS

A parallel plate capacitor of area 1 square metre and relative permittivity 7 is charged to a potential of 300 V. If the distance between the plates is 10-4m, find the capacity and the energy stored in the capacitor.
A parallel plate capacitor of capacity 0.5 mf has its plates separated by 1mm. If the space between the plates is filled with a medium of relative permittivity 1.5, find the plate area of the capacitor.
The plates of a parallel plate capacitor of capacitance 0.01mF are 1m2 each in area. The dielectric constant of the medium between the plates is 2.5. Calculate the distance between the plates.
Find the area of the paper used in a capacitor of capacitance 0.5mF if the dielectric constant of paper is 2,5 and its thickness is 0.05 mm.
Find the equivalent capacitance of three capacitors of capacitance 2mF, 3mF and 5mF when connected (1) in series, and (2) in parallel.
Three capacitors of 2,3 and 4 mF are connected in series and then in parallel. Compare the effective capacitance in the two cases.
Calculate the capacitance of a capacitor which will acquire a charge of 0.05 micro coulomb at 150 volts
Calculate the current required to charge a capacitor of capacitance 25mF to a potential of 400V in 2 minutes.
Find the work done in charging a capacitor of capacitance 25mF to a potential of 300 volts.
Three capacitors fo 2mF,3mF and 4 mF are connected in series and charged by a potential of 260 volts. Calculate the p.d across each capacitor.
A capacitor of capacitance 1mF is charged with a potential 75 volts and another of 0.5 mF is charged with 80 volts. The two are then connected in parallel. Find the common potential and charges on each capacitor.

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