In an electrical circuit, the main components are current, voltage, and resistance. The electric current is the flow of charge. The voltage supply works as the energy source for these charges for movement. We can understand the concept of potential by taking the example of water flow. Water always flows from a higher level to a lower level. The potential energy in an electric field is a function of position, hence it has a different value for different points.
For a unit charge, the potential energy is called potential. If we place a unit of electric charge, it will move from a high potential to a lower one. Hence the electric current always flows from the high potential to the low potential. We can also store this electrostatic energy by some arrangement in the electric circuit. This new component that stores electrostatic energy is called a capacitor. The capacity of this component to store the energy is called capacitance.
The Capacitor consists of two conductors with opposite charges, placed very close to each other. This arrangement of conductors is known as capacitor. It can generate electrostatic energy between its plates. We can increase the capacitance of the capacitor by changing its shape or adding dielectric material between it.
There can be many types of capacitors. Some of them are mentioned below -
Parallel plate capacitors have two rectangular conducting plates placed at some distance.
In a spherical capacitor two concentric spherical conducting spheres are placed at some distance away.
In the cylindrical capacitor two coaxial conduction cylindrical shells are placed at a certain distance away.
The capacity to store the charge is called capacitance. In other words for a given voltage across the plates, the ability to store the charge is capacitance.
If the potential difference between the plates is V and let the charge in each plate is +Q and -Q, then we find that charge on plates is proportional to the voltage across the capacitor.
$$\mathrm{Q\:\propto\:V}$$
$$\mathrm{Q\:=\:C\:V}$$
Here C = Capacitance
From the above expression, we can see the units of the capacitance.
$$\mathrm{C=\frac{Q}{V}}$$
$$\mathrm{C=1\:Coulomb/ 1Volt}$$
It will be Coulomb per Volt.
In honor of Michael Faraday, the SI unit of capacitance is known as Farad. Farad is a very large unit of capacitance, we often use small units for capacitance.
$$\mathrm{1nF = 10^{−9} F}$$
$$\mathrm{1pF =10^{−12} F}$$
Its dimensional formula will be following
$$\mathrm{[C]=\frac{[Q]}{[V]}=\frac{[AT]}{[ML^{2} T^{−3} A^{−1}]}}$$
$$\mathrm{[C]=[M^{−1}L^{−2} T^{4} A^{2}]}$$
When we connect the capacitor to a voltage source, an electric field develops across the plates of the capacitor. One plate of capacitors is connected to the positive terminal of the DC source and another one to the negative terminal. Hence one plate collects the positive charges from the source and negative charges get collected in the other one. After a certain time, the plates of the capacitor get fully charged. The time taken by the capacitor to get fully charged is called the charging time of the capacitor.
Now if we switch off the battery, the plates of the capacitor still hold the charge in the circuit. And they can discharge it one some load connected to the circuit. The time taken in discharging the charge of the capacitor is called discharging time. Hence the capacitor can work as a temporary source of electrical energy.
We can calculate the capacitance of a capacitor using the definition of capacitance.
Which is
$$\mathrm{Q = CV}$$
Here we calculate the capacitance of a parallel plate capacitor as an example.
Let's say the plates are kept ‘d’ distance away, and the charge of the plates is Q. Area of plates is A. We know that the electric field between two charged plates is
$$\mathrm{\overrightarrow{E}=\frac{Q}{\epsilon_0A}}$$
And the potential difference is $\mathrm{V=Ed}$
We can write potential in terms of charge $\mathrm{V=d\frac{Q}{\epsilon_0 A}}$
Hence the capacitance of a parallel plate capacitor is
$$\mathrm{C=\frac{\epsilon_0d}{A}}$$
The capacitor can be used in the following ways
In a circuit, it can be used as a temporary battery.
A filter can be used to pass a certain range of frequencies. The capacitor is used in filter circuits.
Used in sensors and signal processing.
It is also used in Rectifiers.
It is used for smoothing the signals coming from transformers or rectifiers.
Capacitance is the ability of a capacitor to hold electrical energy. It is defined as the ratio of the total charge divided by the potential across it. It is measured in Farad. The Capacitor consists of two conductors with opposite charges, placed very close to each other. This arrangement of conductors is known as a capacitor. The capacitor is used in many circuits, for example, rectifiers, transformers, filters, etc. mainly there are three types of capacitors: parallel plate, cylindrical, and spherical capacitor.
Q1. What is the capacitance of the earth?
Ans. If we take the earth as a spherical capacitor then the capacitance will be
$$\mathrm{C=4\pi\epsilon_0 R}$$
where R = 6400 km
$$\mathrm{C=4\pi\epsilon_0 R=4\times 3.14\times 8.854\times 10^{−12}\times 6400\times 10^3}$$
$$\mathrm{C=711\:\mu F}$$
Q2. What is the capacitance of parallel plate capacitors, in the presence of dielectrics?
Ans. The capacitance in presence of dielectric is given by
$$\mathrm{C=\frac{k\epsilon_0A}{d}}$$
where k= dielectric constant.
Q3. What is the capacitance of spherical and cylindrical capacitors?
Ans.
For Spherical Capacitor
If the radius of the inner sphere is $\mathrm{r_1}$ and the outer sphere is $\mathrm{r_2}$, the charge on each sphere is Q.
Then the capacitance of a Spherical capacitor can be written as
$$\mathrm{C = 4\pi\epsilon_0\frac{r_1r_2}{r_2-r_1}}$$
For Cylindrical Capacitors
If the length of the cylinders is l and radius of the inner cylinder is $\mathrm{r_1}$ and the outer cylinder is $\mathrm{r_2}$. the charge on each shell is Q. Then their capacitance would be
$$\mathrm{C=2\pi\epsilon_0\frac{1}{ln(\frac{r_2}{r_1})}}$$
Q4. Write the expression of electrostatic energy of a capacitor.
Ans. Electrostatic energy of capacitor depends upon the voltage across it and capacitance. It is given by
$$\mathrm{U=\frac{1}{2}CV^2}$$
Where V = voltage across the capacitor
C = capacitance of the capacitor
Q5. In a parallel plate capacitor, the distance between plates is 6cm and the area of each plate is $\mathrm{30\:cm^{2}}$. If we apply a voltage of 5V across it, what is its capacitance?
Ans. We know that the capacitance of a parallel plate capacitor is
$$\mathrm{C=\frac{\epsilon_0A}{d}}$$
$$\mathrm{C=\frac{8.85\times 10^{−12}\times 30\times 10^{−4}}{6\times 10^{−2}}}$$
$$\mathrm{C=8.85\times 10^{−12}\times 5\times 10^{−2}=4.43\times 10^{−14} F}$$
$$\mathrm{C=4.43\times 10^{−14} F}$$
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