Parallel Plate Capacitor and Spherical Capacitor: Capacitors are considered to be a very important appliance in any electrical/electronic circuit. You can find a capacitor in almost every circuit of any appliance which you use in your daily life, like ceiling fans, the motherboard of a computer, remote controls, etc. But why is it so? Why do capacitors have found so much importance in such circuits? What is its use? How does it function?

You will find all the answers by reading this article. We will even look at two different types of capacitors – Parallel plate capacitor and spherical capacitor.

Capacitor and its Capacitance

A capacitor is an electrical device that draws energy from the battery and stores it inside itself temporarily in the form of an electric field. In our daily lives, we use three types of capacitors viz. \((1)\) parallel plate capacitor, \((2)\) spherical capacitor, and \((3)\) cylindrical capacitor. These capacitors are connected to the circuit according to their use. Some circuit needs to store more amount of energy while some circuit needs to store less amount of energy. Some use high voltage electricity while some low. Hence, different types of capacitors with different power ratings are used. Below are some of the pictures of each type of capacitor:

So, you can correctly guess from the pictures that the first picture is of a parallel plate capacitor, the second picture is of a cylindrical capacitor, and the third picture is of a spherical capacitor. 

The charge stored in a capacitor is directly proportional to the potential difference applied across its ends:

\(Q \propto V\)

Removing the proportionality sign, we get a constant:

\(Q=C V\)

Here, \(C\) is known as the capacitance of the capacitor. So, the capacitance will be the ratio of the charge stored inside the capacitor and the potential difference applied across its ends. Hence, capacitance can be defined as the ability of the capacitor to store charge by applying a potential difference across its ends. The SI unit of capacitance is \(\frac{\text { Coulomb }}{\text { Volt }}\) or \(\text {Faraday} (F)\). Generally, capacitors in the range of \(\mu F\) \((\text {microfaraday})\) and \(mF\) \((\text {millifaraday})\) are available in the market.

Parallel Plate Capacitor

A parallel plate capacitor is formed by placing two conducting plates parallel of equal cross-sectional area parallel to each other separated by some fixed distance. These plates can be circular or rectangular shaped. They are generally used in rechargeable systems. The picture given here shows a parallel plate capacitor.

When a parallel plate capacitor is connected across a potential difference, one of the plates becomes positively charged, and other plates become negatively charged due to high potential and low potential. Because of this potential difference, an electric field is induced between the plates from the plate with a positive charge to the plate with a negative charge, as shown in the diagram. 

As we know that electric field lines originate from a positive charge and terminate at a negative charge, the direction of the electric field from both the plates will be the same, and it will be from positive plate to negative plate. The electric field due to the individual plate can be formulated as;

\(E=\frac{\sigma}{2 \varepsilon_{0}}\)

Here, \(\sigma\) is the charge density of the plate, and \(\varepsilon_{0}\) is the permittivity in a vacuum as there is no medium between the plates of the capacitor. Since the electric field due to both the plates has the same magnitude and direction, the net electric field between the plates will be;

\(E_{n e t}=\frac{\sigma}{\varepsilon_{0}}\)

If the magnitude of the charge on each plate is \(Q\) and the cross-sectional area of each plate is \(A\), then the net electric field will be;

\(E_{n e t}=\frac{Q}{A \varepsilon_{0}}\)

As the potential difference connected across the plates is \(V\), and the distance between both the parallel plates is \(d\), we can write for the net electric field as;

\(E_{n e t}=\frac{V}{d}\)

Comparing both the equations for the electric field;

\(\frac{V}{d}=\frac{Q}{A \varepsilon_{0}}\)

\(\frac{Q}{V}=\frac{\varepsilon_{0} A}{d}\)

But the ratio of the charge stored per unit potential difference applied is known as capacitance. Hence the capacitance of a parallel plate capacitor can be written as;

\(C=\frac{\varepsilon_{0} A}{d}\)

From this, we can say that the capacitance of a parallel plate capacitor depends on \(– (1)\) cross-sectional area of the plates, \((2)\) distance between both the plates, and \((3)\) medium between both the plates (this part has been discussed later on in the same article). The circuit symbol for a parallel plate capacitor is

Spherical Capacitor

The aspherical capacitor is formed by using a hollow/solid spherical conductor surrounded by another concentric hollow spherical conductor. Both the conductors will be equally charged. Because of the same charge and different radius of each conductor, there arises a potential difference between both the conductors. The picture shows a cross-sectional view of a spherical capacitor. 

When it is connected to a potential difference, as shown in the picture, one of the spheres gets positively charged, and the other sphere gets negatively charged, as shown in the diagram

Electric field on the Gaussian surface, as shown above, will be given by:

\(E=\frac{Q}{4 \pi \varepsilon_{0} r^{2}}\)

Electric potential difference is also defined to be the line integral of the electric field. Therefore, the potential difference between both the spheres will be;

\(V=-\int E d r\)

\(V=-\int_{R_{2}}^{R_{1}} \frac{Q}{4 \pi \varepsilon_{0} r^{2}}\)

\(\therefore V=\frac{Q\left(R_{2}-R_{1}\right)}{4 \pi \varepsilon_{0} R_{1} R_{2}}\)

But, the capacitance of a capacitor is given by;

\(C=\frac{Q}{V}\)

\(\therefore C=\frac{4 \pi \varepsilon_{0} R_{1} R_{2}}{\left(R_{2}-R_{1}\right)}\)

As you can see from the above equation, the capacitance of a spherical capacitor depends on the radii of the inner and outer sphere and the medium inserted between both the spheres. If we take the limit for \(R_{1} \rightarrow R\) and \(R_{2} \rightarrow \infty\)

\(C=4 \pi \varepsilon_{0} R\)

This is the capacitance of a spherical capacitor if there is only one sphere in the capacitor. They are also known as isolated spherical capacitors.

Common Potential of Two Capacitors

Suppose you have two parallel plate capacitors with capacitance \(C_{1}\) and \(C_{2}\) and they are charged up to potential \(V_{1}\) and \(V_{2}\) and connected, then a charge will transfer from one capacitor to another capacitor till the potential on each capacitor becomes equal.

Since the final potential on both the capacitors is the same, we can write;

\(\frac{Q_{1}}{C_{1}}=\frac{Q_{2}}{C_{2}}\)

\(\therefore \frac{Q_{1}}{Q_{1}+Q_{2}}=\frac{C_{1}}{C_{1}+C_{2}}\)

Using \(Q=CV\) and the law of conservation of charge, we can write for the common potential on both the capacitors as;

\(V_{C}=\frac{C_{1} V_{1}+C_{2} V_{2}}{C_{1}+C_{2}}\)

And if the plates are connected such that their polarities oppose each other, then.

\(V_{C}=\frac{\left[C_{1} V_{1}-C_{2} V_{2}\right]}{C_{1}+C_{2}}\)

For the case of spherical capacitors, if the radii of both the spheres are \(R_{1}\) and \(R_{2}\) and electric potential \(V_{1}\) and \(V_{2}\) are connected via a thin conducting wire, then the common potential between both the spherical capacitors will be;

\(V_{C}=\frac{R_{1} V_{1}+R_{2} V_{2}}{R_{1}+R_{2}}\)

Here, radius appears instead of capacitance because the capacitance of a spherical capacitor depends on its radius.

Energy Stored in a Capacitor

A capacitor stores energy by creating dipoles between the space between its plates. Initially, the space between both the plates is non-polar. As soon as we apply a potential difference across it, an electric field is induced between both the plates. Due to this electric field, dipoles are created between the plates, and as we know, that dipoles store energy in them due to the presence of the electric field. Hence, in this way, a capacitor stores energy in the space. 

If the small amount of change in the potential difference between the plates is \(dV\), the change in its potential energy will be;

\(d U=Q d V\)

But, as we know that for a capacitor, \(Q=CV\), we can write;

\(dU=CVdV\)

Integrating on both the sides till complete potential difference,

\(\int d U=\int C V d V\)

\(\therefore U=\frac{1}{2} C V^{2}\)

Substituting \(C=\frac{Q}{V}\) in the above equation;

\(U=\frac{1}{2}\left(\frac{Q}{V}\right) V^{2}\)

\(\therefore U=\frac{1}{2} Q V\)

Substituting \(V=\frac{Q}{C}\) in the equation;

\(U=\frac{1}{2} Q\left(\frac{Q}{C}\right)\)

\(\therefore U=\frac{1}{2} \frac{Q^{2}}{C}\)

Hence, we have three equations for energy stored in a capacitor that is;

\(U=\frac{1}{2} C V^{2}=\frac{1}{2} Q V=\frac{Q^{2}}{2 C}\)

The energy density of a parallel plate capacitor

Energy density is the energy stored per unit volume inside the plates of a parallel plate capacitor. If the electric field between both plates is \(E\), the distance between both the plates is \(d\) and cross-sectional area of both the capacitors is \(A\); its energy density will be given by;

\(\rho_{E}=\frac{U}{\text { Volume }}\)

\(\rho_{E}=\frac{\frac{1}{2} C V^{2}}{A d}\)

Substituting \(C=\frac{\varepsilon_{0} A}{d}\) and \(V=E d\)

\(\rho_{E}=\frac{1}{2} \frac{\varepsilon_{0} A}{d} \frac{(E d)^{2}}{A d}\)

\(\rho_{E}=\frac{1}{2} \varepsilon_{0} E^{2}\)

Combination of Capacitors

When two or more capacitors are combined together, the net capacitance of the circuit may increase or decreases depending upon how they are combined. There are two types of combination which we are going to study \(– (1)\) series combination and \((2)\) parallel combination. Capacitors are combined with these kinds of different combinations based upon their use as both the combination will have their own importance.

(1) Series Combination:
Capacitors combined are said to be in series when the charge stored in them is the same and the potential supplied by the battery gets distributed among them based on their capacitance, as shown in the diagram:

Here, \(n\) number of capacitors are combined in series having a capacitance \(C_{1}, C_{2,}\ldots \ldots \ldots, C_{n}\). Let potential differences across plates of the capacitors be \(V_{1}, V_{2}, V_{3},\ldots \ldots \ldots \ldots ., V_{n}\)

Thus, we can write for the potential difference applied as;

\(V=V_{1}+V_{2}+\ldots \ldots \ldots \ldots+V_{n}\)

Since the charge stored on each capacitor is the same,

\(\frac{Q}{C_{e q}}=\frac{Q}{C_{1}}+\frac{Q}{C_{2}}+\ldots \ldots \ldots \ldots+\frac{Q}{C_{n}}\)

\(\therefore \frac{1}{C_{e q}}=\frac{1}{C_{1}}+\frac{1}{C_{2}}+\ldots \ldots \ldots \ldots \ldots \ldots+\frac{1}{C_{n}}\)

Here, \(C_{e q}\) is known as the equivalent capacitance of the combination.

(2) Parallel Combination:

Capacitors are said to be connected in parallel if the potential difference across their plates is the same and the charge supplied by the battery gets distributed among them based on their capacitance values as shown in the diagram

Here, \(n\) number of capacitors with capacitance \(C_{1}, C_{2}, \ldots \ldots \ldots ., C_{n}\) are connected in parallel If the charge stored in each capacitor is \(Q_{1}, Q_{2}, \ldots \ldots, Q_{n}\), then we can write for the total charge supplied by the battery as;

\(Q=Q_{1}+Q_{2}+\ldots \ldots \ldots \ldots+Q_{n}\)

Since the potential difference across each capacitor is the same

\(C_{e q} V=C_{1} V+C_{2} V+\ldots \ldots \ldots \ldots+C_{n} V\)

\(\therefore C_{e q}=C_{1}+C_{2}+\ldots \ldots \ldots \ldots .+C_{n}\)

Here, \(C_{e q}\) is the equivalent capacitance of the combination.

From the equations, we can see that for series combination, we add the reciprocals of capacitance, and for parallel combination, we directly add the values of capacitance.

Inserting Dielectrics

Dielectric is an insulating material in which is used to separate the plates of a parallel plate capacitor. By inserting dielectric between the plates of a parallel plate capacitor, its capacitance increases, and hence its ability to stored charge and energy also increases. The same concept goes for spherical capacitors too, if a dielectric material is inserted in the space between the inner sphere and outer sphere, its capacitance increases.

The diagram shows how we can insert a dielectric between plates of a capacitor. If the dielectric constant of this dielectric is \(K\) (which is also the relative permittivity of the material), then we can write for the capacitance of this capacitor as.

\(C=\frac{K \varepsilon_{0} A}{d}\)

\(\therefore C=K C_{0}\)

Here, \(C_{0}\) is the capacitance of the capacitor when

the dielectric is not inserted between the plates. From this equation, we can prove our previous argument that the capacitance of a parallel plate capacitor increases by inserting a dielectric between its plates.

Similarly, if we insert a dielectric material inside a spherical capacitor, we get the capacitance of the spherical capacitor to be;

\(C=\frac{4 \pi K \varepsilon_{0} R_{1} R_{2}}{R_{2}-R_{1}}\)

Let us say that we insert two dielectrics between the plates of a parallel plate capacitor. There can be two possibilities for this. In the first case, the distance between plates is divided, while in the second case, the area of plates is divided.

If we look closely, the dielectrics in both cases can be considered as an individual capacitor. Let us solve the first case;

Both the dielectrics seem to be connected in series. We can write for the capacitance of their individual capacitors as;

\(C_{1}=\frac{K_{1} \varepsilon_{0} A}{d_{1}} \quad C_{2}=\frac{K_{2} \varepsilon_{0} A}{d_{2}}\)

Hence, the equivalent capacitance will be;

\(\frac{1}{C_{e q}}=\frac{1}{C_{1}}+\frac{1}{C_{2}}\)

\(\frac{1}{C_{e q}}=\frac{1}{\varepsilon_{0} A}\left(\frac{d_{1}}{K_{1}}+\frac{d_{2}}{K_{2}}\right)\)

\(\therefore C_{e q}=\frac{\varepsilon_{0} A}{\frac{d_{1}}{K_{1}}+\frac{d_{2}}{K_{2}}}\)

Now, for the second case, both the dielectric seems to be connected in parallel. We can write for the capacitance of their individual capacitors as;

\(C_{1}=\frac{K_{1} \varepsilon_{0} A_{1}}{d} \quad C_{2}=\frac{K_{2} \varepsilon_{0} A_{2}}{d}\)

Hence, their equivalent capacitance will be;

\(C_{e q}=C_{1}+C_{2}\)

\(C_{e q}=\frac{K_{1} \varepsilon_{0} A_{1}}{d}+\frac{K_{2} \varepsilon_{0} A_{2}}{d}\)

\(\therefore C_{e q}=\left(K_{1} A_{1}+K_{2} A_{2}\right) \frac{\varepsilon_{0}}{d}\)

Similarly, we can add more and more dielectrics too and its capacitance will change accordingly.

Solved Examples

Q.1. What is the percentage change in capacitance of a parallel plate capacitor when a dielectric slab of dielectric constant \(3\) is inserted between its plates?
Ans: When a dielectric is inserted between the plates of a capacitor, its capacitance will be;
\(C=K C_{0}\)
\(\therefore C=3 C_{0}\)
Thus, a change in capacitance of the capacitor will be
\(\Delta C=C-C_{0}\)
\(\Delta C=3 C_{0}-C_{0}\)
\(\therefore \Delta C=2 C_{0}\)
The percentage change in capacitance will be given by;
\(\%\) change \(=\frac{\Delta C}{c_{0}} \times 100\)
\(\therefore \%\) change \(=200 \%\)
Hence, the percentage change in capacitance of this capacitor is \(200 \%\)

Q.2. Two spheres of radii \(R\) and \(2R\) are charged to an electric potential of \(2V\) and \(3V\) respectively. These spheres are connected using a conducting wire. Find the common potential of both the spheres once it attains equilibrium.
Ans: The common potential of two charged spherical capacitors is given by;
\(V_{C}=\frac{R_{1} V_{1}+R_{2} V_{2}}{R_{1}+R_{2}}\)
\(V_{C}=\frac{R(2 V)+2 R(3 V)}{R+2 R}\)
\(\therefore V_{c}=\frac{8 R}{3}\)
The common potential of both the spheres is \(8R/3\)

Q.3. A capacitor of capacitance \(100 \mu F\) is charged to a potential of \(20 V\). Find the amount of energy stored in it. What will be its energy after a dielectric slab of dielectric constant \(2\) is inserted between its plates?
Ans: The capacitance of the capacitor \((C)=100 \mu F\)
The potential difference across its plates \((V)=20 \mathrm{~V}\)
The dielectric constant of the dielectric slab \((K)=2\)
Energy stored in a parallel plate capacitor is given by the equation;
\(U=\frac{1}{2} C V^{2}\)
Substituting the values
\(U=\frac{1}{2}\left(100 \times 10^{-6}\right)(20)\)
\(\therefore U=1 \,mJ\)
Since the dielectric of constant \(2\) will make the capacitance double, its energy will also become double 
\(\therefore U^{\prime}=2 U\)
\(\therefore U^{\prime}=2 m J\)

Q.4. Find the equivalent capacitance across \(A\) and \(B\) in the given circuit:

Ans: As we can clearly see in the circuit, all the \(3 \,μF\) capacitors are connected in series. Thus, their equivalent capacitance will be given by;
\(\frac{1}{C_{1}}=\frac{1}{3}+\frac{1}{3}+\frac{1}{3}\)
\(\therefore C_{1}=1 \mu F\)
Similarly, \(2 \mu F\) and \(6 \mu F\) capacitors are also connected in series. Thus, their equivalent capacitance will be;
\(\frac{1}{C_{2}}=\frac{1}{2}+\frac{1}{6}\)
\(\therefore C_{2}=1.5 \mu F\)
Now, both these equivalent capacitors will be in parallel. Hence, the equivalent capacitance of the circuit between points \(A\) and \(B\) will be;
\(C=C_{1}+C_{2}\)
\(C=1+1.5\)
\(\therefore C=2.5 \,\mu F\)
The equivalent capacitance between \(A\) and \(B\) is \(2.5 \,\mu F\)

Summary

From this article, we learned about parallel plate capacitors and spherical capacitors in detail. The capacitor is a device that stores charge and energy in the form of an electric field.

The charge stored in a capacitor: \(Q=C V\)

The capacitance of a parallel plate capacitor: \(C=\frac{\varepsilon_{0} A}{d}\)

The capacitance of a spherical capacitor: \(C=\frac{4 \pi \varepsilon_{0} R_{1} R_{2}}{\left(R_{2}-R_{1}\right)}\)

Common potential for two capacitors: \(V_{C}=\frac{C_{1} V_{1}+C_{2} V_{2}}{C_{1}+C_{2}}\)

Series equivalent of capacitors: \(\frac{1}{c_{e q}}=\frac{1}{c_{1}}+\frac{1}{c_{2}}+\ldots \ldots \ldots \ldots \ldots .+\frac{1}{c_{n}}\)

The parallel equivalent of capacitors: \(C_{e q}=C_{1}+C_{2}+\ldots \ldots \ldots \ldots . . C_{n} \backslash\)

Energy stored in a capacitor: \(U=\frac{1}{2} C V^{2}=\frac{1}{2} Q V=\frac{Q^{2}}{2 C}\)

The energy density of a parallel plate capacitor: \(\rho_{E}=\frac{1}{2} \varepsilon_{0} E^{2}\)

Frequently Asked Questions

Q.1. How does a spherical capacitor work?
Ans: A spherical capacitor works by creating a potential difference between its inner and outer spheres. With the help of that potential difference, it can store charge and energy in the empty space between the spheres or when a dielectric material is inserted in it.

Q.2. What is the capacity of a spherical capacitor?
Ans: The capacity of a spherical capacitor depends upon its radius and it is given by the formula
\(C=\frac{4 \pi \varepsilon_{0} R_{1} R_{2}}{\left(R_{2}-R_{1}\right)}\)

Q.3. What is an isolated spherical capacitor?
Ans: For an isolated spherical capacitor, we can consider a solid charged sphere of a finite radius and other spheres of infinite radius with zero potential difference. The capacitance of such capacitor is given by \(C=4 \pi \varepsilon_{0} R\)

Q.4. What is a parallel plate capacitor and what are they used for?
Ans: A parallel plate capacitor is a device made up of two metallic plates with a fixed separation between them. Between the plates, there can be dielectric material or empty space based on its use and the required capacitance. They are mainly used in rechargeable batteries, LASER, radars, etc.

Q.5. How fast can a capacitor charge?
Ans: It usually takes a few seconds to charge. To charge a capacitor, it is connected to a resistor to limit the current supply. The charge stored at any time \(t\) is given by \(Q=Q_{0}\left(1-e^{\frac{-t}{R C}}\right)\)

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