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Capacitance is defined as the ratio of the amount of the electric charge that is stored in the conductor to the difference in the electrical potential of that system. The unit of capacitance is Farad (F). Capacitance C is calculated as the ratio of the Q charge that is stored in the capacitor with a DC voltage U and F is denoted as:
C = \(\frac{Q}{U}\)
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Key Terms: Capacitance, Capacitor, Electric Charge, Electrical Potential, farad, self-capacitance, circuit, condenser, conductor
SI Unit of Capacitance
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Farad or F is the SI unit of the electrical capacitance of a capacitor. The ability of a body to store the electrical charge in itself is known as Farad. This can be expressed as
F = \(\frac{C}{V}\)
The capacitance of the conductor is also measured in microfarads and can be denoted as μF.
The video below explains this:
Capacitance Detailed Video Explanation:
What is Capacitance?
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Capacitance is defined as the ability of a capacitor to accumulate an electrical charge. It is an electrical property that is generally associated with a capacitor, condenser, or conductor. The capacitance can be of two forms: self-capacitance and mutual capacitance.
It can also be defined as the ability of a circuit or a component to collect and then store the energy in the form of an electrical charge.
The capacitance of a conductor can be calculated using the following capacitance formula:
Q = CV
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What is Capacitor?
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The capacitor is an electrical component that stores the electrical charge. The capacitor’s capacity to store this electrical charge is known as capacitance. The capacitor consists of 2 conducting material plates which are kept between an insulator, which is also known as a dielectric, made of film, glass, ceramic materials, etc.
Parallel Plate Capacitor
Types of Capacitors
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Capacitors are categorized into two types such as conductors and dielectrics used. The dielectric types of capacitors are made up of a solid or dielectric fluid. The plates of the dielectric capacitors are either made up of metals or in some cases, the metal is separated by an insulator.
The capacitors in which two parallel plates of metals are placed are called parallel-plate capacitors. In this type of capacitor, one of the parallel plates is insulated from the other plate and is also separated with the help of an insulator in between them. Here, the capacitance of the conductor is calculated by the sum of the charges which will be induced at the plates.
Capacitors are classified into the following main categories:
- Air-Insulated capacitors: The parallel plates of the capacitor are separated with the help of a dielectric medium.
- Oil-Immersed Capacitors: In these types of capacitors, the parallel plates are either immersed in an oil or any kind of dielectric medium. The plates are separated with the presence of a thin membrane in between them.
- Oil-Free Capacitors: The plates of the capacitors are separated by the vacuum, which is an electrical insulator.
The capacitors having the same charge, also have the same capacitance and the same voltage rating. To increase the voltage ratings of the capacitors, electrolytic capacitors can be used. The internal resistance of the electrolytic capacitor is lower than that of the dielectric medium of the capacitor.
Read More: Electrostatic Potential and Capacitance
Uses of a Capacitor
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Capacitor is useful in the following ways:
- It can be used to hold or store energy.
- The frequency can be increased or decreased with the help of a capacitor.
- Capacitors can hold a constant voltage for a specified time period.
- To smoothen the voltage spikes or to reduce them, capacitors can be used.
- It can also help in dampening out the AC noise by passing a high-frequency AC and low-frequency AC is blocked.
- For varying the pulse power or the high frequency can be increased or decreased by using a capacitor
- The impedance of the AC circuit can be increased or decreased with the use of a capacitor for the protection of the electronic circuits.
Things to Remember
- The capacitor is an energy storing device.
- The unit of capacitance is Farads.
- The plates of a parallel plate capacitor are separated by a dielectric medium.
- The charge of a capacitor is calculated by the formula, q = C V
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Sample Questions
Ques: Write in brief about 1 Farad capacitance of a capacitor. (2 marks)
Ans: The capacitance of a particle that has a 1 coulomb of charge and has a 1-volt potential difference is 1 Farad. Therefore, the unit of capacitance is Farad.
Ques: Is a capacitor an active or a passive device? (2 marks)
Ans: When the energy is supplied to a device, it is called an Active device. Since, the capacitor is a device that stores or holds the energy, hence it is called a passive device. The energy is stored by the capacitor for later use. Across the impedance of the source, the energy is consumed by the capacitor.
Ques: The area of each of the plates of a parallel plate capacitor is 4 cm2 and is separated by a distance of 2 mm. Calculate the capacitance H of the capacitor. (4 marks)
Ans: We have,
Area of the plates, A= 4 cm2 = 4 x 10-4 m2
Distance, d = 2 mm = 2 x 10-3 m
μo = 8.85 x 10-12 C2 N-1 m-2
H = μo x A / d = (8.85 x 10-12) x (4 x 10-4) / (2 x 10-3)
From this we will get,
Capacitance, H = 17.7 x 10-13 F
Ques. What will happen if the dielectric is placed in an external electric field? (3 marks)
Ans: When the dielectric is placed in an external electric field, the electric field present inside the electric field decreases. This is due to the polarization as an internal electric field is created which is of opposite polarity of the external electric field. Hence, the net electric field of the dielectric gets reduced.
Ques. There are 3 capacitors connected in series, each having a capacitance of 8 pF. Calculate the following:
a) Total capacitance of the combination
b) The potential difference across each capacitor if the combination is connected to a 120 Volt supply. (5 marks)
Ans: a) Capacitance of the capacitors, C′= 9 pF
Equivalent capacitance, C will be calculated using the following formula
\(\frac{1}{C} = \frac{1}{C'} + \frac{1}{C'} + \frac{1}{C'}\)
\(\frac{1}{C} = \frac{1}{8} + \frac{1}{8} + \frac{1}{8}\)
C = \(\frac{3}{8}\) μF
The total capacitance of the combination is, C = 0.375 μF
b) The supply voltage connected to the combination is, V = 120-volt
Potential difference can be calculated as,
V’= V/3= 120/3
⇒ V’= 40 volts
Hence, the potential difference across each capacitor of the combination is 40 volts.
Ques. In a combination, there are 3 capacitors placed parallelly with the capacitance of 3 pF, 4 pF, and 5 pF respectively. Find out the following:
a) Total capacitance of the combination
b) Charge on each capacitor if the combination is connected to a 120-volt supply. (5 marks)
Ans: a) Capacitance of the capacitors is of the combination is:
C1 = 3 pF
C2 = 4 pF
C3 = 5 pF
When the capacitors are placed in a parallel combination, the total capacitance C’ is calculated by
C’= C1 + C2 + C3
⇒ C’= 3 pF + 4 pF + 5 pF
⇒ C’= 12 pF
b) We have, supply voltage, V = 120 volts
As the capacitors are placed in a parallel combination, the voltage through each capacitor remains same as the supply voltage, which is 120 volts
The charge, q is calculated by:
q = C x V
i) For first capacitor placed in the combination,
Capacitance, C1 = 3 pF
charge is calculated as
q1= V x C1 = 120 x 3= 360 pC
q1= 3.6 x 10-10 C
ii) For second capacitor placed in the combination,
Capacitance, C22 = 4 pF
charge is calculated as
q2= V x C = 120 x 4= 480 pC
q2= 4.8 x 10-10 C
iii) For the third capacitor placed in the combination,
Capacitance, C3 = 5 pF
charge is calculated as
q3= V x C3 = 120 x 5= 600 pC
q3= 6 x 10-10 C
Ques. In a parallel plate capacitor, the parallel plates have air between them, and are separated by a distance of 4 mm. The area of each of the plates is 5 x 10-3 m2. Calculate the following:
a) Capacitance of the capacitor
b) Charge on each plate of the capacitor, if connected to a 120 volt supply (5 marks)
Ans: a) Area of each plate, A = 5 x 10-3 m2
Distance between the plates, d = 4 mm = 4 x 10-3 m
Capacitance, C = Aod ……… (i)
Εo= permittivity of the free space
⇒ Εo= 8.854 × 10-12 N−1 m−2 C−2
Substituting the values in the equation (i), we get
C = 8.854 x 10-12 x 5 x 10-34 x 10-3 / 4 x 10-3
⇒ C = 11.0675 x 10-12 F
C = 11.0675 pF
b) We have supply voltage, V = 120 volt
The calculated capacitance is, C = 11.0675 x 10-12 F
We know, q = CV
⇒ q = 11.0675 x 10-12 x 120
q = 1328.1 x 10-12
q = 1.3281 x 10-9 C
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