What is a Capacitor?

A capacitor is a device that stores electrical energy in an electric field. It is a passive electronic component with two terminals.

A capacitor is a device that can store electrical energy. It is composed of two conductors, usually carrying equal and opposite charges, separated by an insulating medium. The insulating material can be an electric insulator, a vacuum, such as glass, paper, air, or a semiconductor, known as a dielectric.

Capacitors come in a variety of shapes and sizes and have many important applications in electronics.

Related Physics Concepts:

• Capacitor Types and Capacitance
• Combination of Capacitors
• Energy Stored in a Capacitor

What are Capacitors Used For?

• Storing electrical potential energy in batteries.
• Filtering out undesired frequency signals
• Delaying voltage changes when coupled with resistors.
• Used as a sensing device.
• Used in the audio system of the vehicle.
• Used to separate AC and DC.

One of the conductors has a positive charge of +Q and is at potential +V, whereas the other has an equal negative charge of -Q and is at potential -V.

Charge on Capacitor

Note: The capacitance of the capacitor is Q.

Total charge on the capacitor is 0: -Q + Q = 0

Circuit Symbols

Capacitance

The charge on the capacitor (Q) is directly proportional to the potential difference (V) between the plates, i.e. $Q\propto V$ or Q = CV.

The capacitance of the capacitor is referred to as the constant of proportionality (C).

Dimensional Formula and Unit of Capacitance

Unit of Capacitance: Farad (F)

The capacitor value can range from a fraction of a pico-farad to more than a micro-Farad, while voltage levels can range from a couple to a few hundred thousand volts.

Dimensional Formula: M^-1L^-2I²T⁴

Commonly Used Scales

• $\mu F$ =10^-6F
• nF = 9F
• pF = 12F

Factors That Influence Capacitance

Factors That Affect Capacitance

• 1. Plate area
• 2. Plate separation
• 3. Dielectric constant of material between the plates

Calculation of Capacitance

We will attempt to calculate the capacitance of variously shaped capacitors, by following these steps;

1. Assume the charge on the conductors (Q)

2. Calculate the Electric Field Between the Plates (E)

3. Calculate potential difference from electric field (V)

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

Types of Capacitor

• Parallel Plate Capacitor
• Spherical Capacitor
• Cylindrical Capacitor

Parallel Plate Capacitor

A parallel plate capacitor consists of two metal plates of Area, A and is separated by a distance d. The plate on the top is given a charge +Q and that at the bottom is given the charge -Q. A potential difference of V is developed between the plates.

The separation is so small in comparison to the plate’s dimensions that the **bending outward of electric field lines at the edges and the non-uniformity of surface charge density at the edges can be disregarded.

The magnitude of the charge density on each plate of a parallel plate capacitor is σ.

$\sigma =Q/A$

$E=\frac{Q}{{\epsilon }_{0}A}$

Also, E = V/d

Now taking into account the surface charges on the outside of the capacitor,

$\begin{array}{l}E=\frac{\sigma }{2{\epsilon }_0}-\frac{\sigma }{2{\epsilon }_0}=0\end{array}$

$\begin{array}{l}InsideE=\frac{\sigma }{2{\epsilon }_0}+\frac{\sigma }{2{\epsilon }_0}=\frac{\sigma }{{\epsilon }_0}=\frac{q}{A{\epsilon }_0}\end{array}$

$\begin{array}{l}\frac{v}{d}=\frac{q}{A{\epsilon }_0}\end{array}$

$\begin{array}{l}or,C=\frac{q}{v}=\frac{A{\epsilon }_0}{d}\end{array}$

For other medium, then capacitance will be $\begin{array}{l}C=\frac{kA{\epsilon }_0}{d}\end{array}$, where k is the dielectric constant of the medium.

$\begin{array}{l}{\epsilon }_0=Permittivityoffreespace=8.85\times {10}^{-12}{C}^2/N{m}^2\end{array}$

If there is a vacuum between the plates, k = 1.

Spherical Capacitor

Let’s consider a spherical capacitor that consists of two concentric spherical shells. Suppose the radius of the inner sphere, R_in = a and radius of the outer sphere, R_out = b. The inner shell is given a positive charge +Q and the outer shell is given –Q.

$\begin{array}{l}V=\frac{q}{4\pi {ϵ}_{0}ka}+\frac{-q}{4\pi {ϵ}_{0}kb}\end{array}$

$\begin{array}{l}V=\frac{q}{4\pi {ϵ}_{0}k}[\frac{1}{a}-\frac{1}{b}]\end{array}$

$\begin{array}{l}V=\frac{q}{4\pi {ϵ}_{0}k}\left[\frac{b-a}{ab}\right]\end{array}$

$\begin{array}{l}C=\frac{q}{V}=\frac{q}{\frac{q}{4\pi {ϵ}_{0}k}\left[\frac{b-a}{ab}\right]}\end{array}$

$\begin{array}{l}C=4\pi {ϵ}_{0}k\left[\frac{ba}{b-a}\right]\end{array}$

Cylindrical Capacitor

Consider a solid cylinder of radius a, surrounded by a cylindrical shell of radius b. The length of the cylinder, l, is much larger than a-b to avoid edge effects. The capacitor is charged so that the inner cylinder has a charge of +Q and the outer cylinder has a charge of -Q.

From Gauss’s Law:

$\begin{array}{l}E=\frac{Q}{2\pi {\epsilon }_{0}rl}=\frac{\lambda }{2\pi {\epsilon }_{0}r}\end{array}$

Where $\lambda =Q/l$, linear charge density

The potential difference of a Cylindrical Capacitor is given by:

$\begin{array}{l}\mathrm{\Delta }V=V_b–V_a=-{\int }_a^b{E}_{r}dr=-\frac{\lambda }{2\pi {\epsilon }_0}\ln \left(\frac{b}{a}\right)\end{array}$

As expected, the outer conductor with negative charge has a lower potential than where we have chosen the integration path to be along the direction of the electric field lines. That gives.

$\begin{array}{l}C=\frac{Q}{\left|\mathrm{\Delta }V\right|}=\frac{\lambda L}{\lambda \ln (b/a)/2\pi {\epsilon }_0}=\frac{2\pi {\epsilon }_{0}L}{\ln (b/a)}\end{array}$

Once again, we see that the capacitance C depends only on the geometrical L, a and b.

Cylindrical Capacitor

Problems on Capacitor and Capacitance

Problem 1: Find the capacitance of a conducting sphere with radius R.

Sol: The electric field outside a sphere with charge Q at a distance r is given by: $\begin{array}{l}E=\frac{kQ}{r^2}\end{array}$

$\begin{array}{l}Therefore-\frac{dV}{dr}=E\end{array}$

$\begin{array}{l}\therefore {\int }_0^{v}dV=-{\int }_{\mathrm{\infty }}^{R}Edr\end{array}$

$\begin{array}{l}\Rightarrow V=kQ{[-\frac{1}{r}]}_{\mathrm{\infty }}^R\end{array}$

$\begin{array}{l}\Rightarrow V=\frac{kQ}{R}\end{array}$

$\begin{array}{l}\therefore C=\frac{Q}{V}=\frac{R}{1/4\pi {\epsilon }_0}=4\pi {\epsilon }_{0}R\end{array}$

Problem 2: A parallel plate air capacitor is constructed using two plates that are 0.2m square and spaced 1cm apart. It is connected to a 50V battery.

• 1. What is capacitance?
• 2. How much is the charge per plate?
• 3. What is the electric field between two plates?
• 4. If the battery is disconnected and then the plates are pulled apart to a separation of 2cm, what are the answers to the questions posed above?

Sol:

$\begin{array}{l}{C}_0=\frac{{\epsilon }_{0}A}{d_0}=\frac{8.85\times {10}^{-12}\times 0.2\times 0.2}{0.01}\end{array}$

$\begin{array}{l}{Q}_0=C_0{V}_0=(35.4\times {10}^{-12}\times 50)C=1.77\times {10}^{-5}C=1770\times {10}^{-12}C\end{array}$

$\begin{array}{l}{E}_0=\frac{V_0}{d_0}=\frac{50}{0.01}=5000V/m\end{array}$

If the battery is disconnected, the charge on the capacitor plates remains constant while the potential difference between the plates can vary.

$\begin{array}{l}\Rightarrow C=\frac{A{\epsilon }_0}{2d}=1.77\times {10}^{-5}\mu f\end{array}$

$\begin{array}{l}\Rightarrow Q=Q_0=1.77\times {10}^{-3}\mu F\end{array}$

$\begin{array}{l}\therefore V=\frac{Q}{C}=\frac{Q_0}{C_b/2}=2V_0=100volts\end{array}$

$\begin{array}{l}\therefore E=\frac{V}{C}=\frac{2V_0}{2d_0}=E_0=5000V/m\end{array}$

Problem 3: What is the voltage of the battery if a parallel plate conductor connected to it has a plate area of 3.0 cm² and a plate separation of 3 mm, with a charge stored on the plate of 4.0 pc?

Sol:

Area A = 3.0 cm² = 3.0 × 10^-4 m²

$\begin{array}{l}{C}_a=\frac{{\epsilon }_{0}A}{d_0}\end{array}$

$\begin{array}{l}{C}_a=\frac{{\epsilon }_{0}A}{d_0}=\frac{8.85\times {10}^{-12}(3\times {10}^{-4})}{3\times {10}^{-3}}\end{array}$

C_a = 8.85 x 10^-13

$\begin{array}{l}C=\frac{Q}{V}\end{array}$

$\begin{array}{l}V=\frac{Q}{C}\end{array}$

$\begin{array}{l}V=\frac{4\times {10}^{-12}}{8.85\times \times {10}^{-13}}\end{array}$

V = 4.52 V

Dielectrics and Capacitance

What are Dielectrics?

Dielectrics are materials that are used to reduce or prevent the flow of electric current in an electric field. They are also used to insulate electrical components and to store electrical energy.

Dielectrics are an insulating material (non-conducting) with no free electrons, however a microscopic displacement of charges is observed in the presence of an electric field. It has been found that the capacitance increases when the gap between the conducting plates is filled with dielectrics.

Polar and Non-polar Dielectrics

Atoms are composed of a positively-charged nucleus surrounded by electrons. When the center of the negative charge does not align with the center of the nucleus, a permanent dipole moment is created. These molecules are known as polar molecules. When a polar dielectric is placed in an electric field, the individual dipoles experience a torque and attempt to align with the field.

In non-polar molecules, the centres of the positive and negative charge distributions coincide, meaning there is no permanent dipole moment created. However, when an electric field is present, the centres are slightly displaced, resulting in induced dipole moments.

Polarization of a Dielectric Slab

The dipole moment is created by inducing charges on the dielectric, and it appears in any volume of the dielectric. $\begin{array}{l}\text{The polarization vector}\overrightarrow{p}\text{is defined as the dipole moment per unit volume.}\end{array}$.

Dielectric Constant

The resultant field is: $\overrightarrow{E}=\overrightarrow{E_0}+\overrightarrow{E_p}$

The resultant field is in the direction of the applied field, but with a reduced magnitude. The induced electric field is in the opposite direction to the applied field.

$\begin{array}{l}\overrightarrow{E}=\frac{\overrightarrow{E_0}}{K}\end{array}$

Where ε is the dielectric constant or relative permittivity of the dielectric. For vacuum,

$$\begin{array}{l}\overrightarrow{E_p}=0,\epsilon =1\end{array}$$

Effect of Dielectric in Capacitance

Dielectric Slabs in Series

Two parallel plates of area A are separated by two dielectric slabs of thickness d₁ and d₂ with dielectric constants k₁ and k₂, respectively.

The equivalent capacitance C of a capacitor considered as a combination of two capacitors in series is given by:

$$\begin{array}{l}C=\frac{1}{\frac{1}{C_1}+\frac{1}{C_2}}\end{array}$$

$\frac{1}{C}=\frac{d_1}{k_1{\epsilon }_{0}A}+\frac{d_2}{k_2{\epsilon }_{0}A}$

$\begin{array}{l}C=\frac{{\epsilon }_{0}A}{\frac{d_1}{k_1}+\frac{d_2}{k_2}}\end{array}$

Dielectric Slabs in Parallel

Consider two capacitors in parallel, each with a dielectric slab of thickness $d$, area $A_1$ and $A_2$, and dielectric constants $K_1$ and $K_2$ respectively, as shown.

C1 + C2 = C

$$\begin{array}{l}C=\frac{{\epsilon }_0}{d}[k_1{A}_1+k_2{A}_2]\end{array}$$

Dielectric and Vacuum

The equivalent capacitance of a capacitor with a dielectric slab of thickness t between the plates, which are separated by a distance d, is given as:

$\begin{array}{l}C=\frac{{\epsilon }_{0}A}{\frac{t}{k}+\frac{d-t}{1}}(k=1forvacuum)\end{array}$

$$\begin{array}{l}C=\frac{{\epsilon }_{0}A}{\frac{t}{k}+d-t}\end{array}$$

The equivalent capacitance is not affected by changing the distance of slab from the parallel plates. If the slab is of metal, the equivalent capacitance is:

$$C=\frac{{ϵ}_{0}A}{d-t}$$

Problems Regarding Capacitance and Dielectrics

Problem 1: Find the equivalent capacitance of three 10μF capacitors connected as shown in the figure, where two of them are filled with dielectrics with K = 2 and K = 2.5.

After insertion of dielectrics, the electrical properties of a material change significantly.

C1 = 10μF;
C2 = KC0 = 2 x 10 = 20μF;
C3 = KC0 = 2.5 x 10 = 25μF

$\therefore Ceff=31\frac{2}{3}\mu F$

Problem 2: Find the equivalent capacitance of the system shown (assume square plates)?


Taking K1 = 2 to be series in K2 = 3

$\begin{array}{l}\Rightarrow \frac{1}{c_{left}}=\frac{1}{\frac{(2){\epsilon }_0\{(L)\left(\frac{L}{3}\right)\}}{\left(\frac{d}{3}\right)}}+\frac{1}{\frac{(3){\epsilon }_0\{(L)\left(\frac{L}{3}\right)\}}{\left(\frac{2d}{3}\right)}}\Rightarrow C_{left}=\frac{6{\epsilon }_0{L}^2}{7d}\end{array}$

$\begin{array}{l}\Rightarrow C_{right}=\frac{(4){\epsilon }_0\{(L)\left(\frac{2L}{3}\right)\}}{d}=\frac{8{\epsilon }_0{L}^2}{3d}\end{array}$

Now C_left and C_right are running parallel

$\begin{array}{l}\Rightarrow C_{eq}=C_{left}+C_{right}=\frac{6{\epsilon }_0{L}^2}{7d}+\frac{8{\epsilon }_0{L}^2}{3d}=\frac{74{\epsilon }_0{L}^2}{21d}\end{array}$

Problem 3: Calculate the effective capacitance when capacitors are connected in series and parallel to a 40 V battery? Also, calculate the voltage across the capacitors for each connection type.

Response:

C1 = 12°F

C2 = 6°F

When capacitors are connected in series, the total capacitance is less than the smallest capacitance of the individual capacitors.

$$\begin{array}{l}C=C_1+C_2\frac{1}{C}=\frac{1}{C_1}+\frac{1}{C_2}\end{array}$$

$$\begin{array}{l}\frac{1}{C}=\frac{1}{12}+\frac{1}{6}\end{array}$$

$\frac{1}{C}=0.25$

F = (C × 9/5) + 32

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

Q = CV Q = 160 Q = 160 C

For the 12 F capacitor:

$\begin{array}{l}V=\frac{160}{12}\end{array}$

V = 13.33V

For the 6F Capacitor:

$$\begin{array}{l}\frac{160}{6}=V\end{array}$$

V = 26.66 V

When capacitors are connected in parallel, the total capacitance of the circuit increases.

C = C1 + C2 C = 18 C = 18 × 9/5 + 32 F

The voltage is the same as 40V across each capacitor.

Frequently Asked Questions on Types of Capacitors and Capacitance

The capacitance of a capacitor is a measure of its ability to store an electrical charge. It is measured in Farads (F).

The capacitance of a capacitor is the ratio of the magnitude of the charge on either conductor to the potential difference between the conductors forming the capacitor.

What are the Ways to Increase the Capacitance of a Parallel Plate Capacitor?

Increasing the area of the plates or decreasing the distance between them.

Increasing the Area of the Plates

Decreasing the distance between the plates.

Using a medium with a higher dielectric constant.

Does doubling the charge on a capacitor change its capacitance?

Capacitance is independent of charge, so the capacitance stays constant.

How does the charge on the parallel plate capacitor change when the potential difference between the plates is doubled?

The charge is doubled.