What is Capacitor?

What is Capacitor? 

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 Capacitors

The capacitor is the passive component that is used in Electronics and Electrical Circuits. Which is made by placing two opposite plates with a dielectric material are in the form of plates which can accumulate charges. One plate is for a positive charge while the other is for a negative charge. 

Capacitance 

Capacitance is the effect of the capacitor. Capacitance is defined as the ratio of electric charge Q to the voltage V and it is expressed as

C = Q/V

Where.

• Q is the electric charge measured in coulombs
• C is the capacitance measured in farad
• V is the voltage across the plates 

 Unit of Capacitance and Dimensional Formula

Dimensional Formula 

M-1L-2I2T4

Commonly Used Scales

• $\mu F$ = 10-6F 
• nF = 10-9F
• pF = 10+2F

Unit of Capacitance Farad (F)

The capacitor value can vary from a fraction of pico-farad to more than a micro Farad. Voltage levels can range from a couple to a substantial couple of hundred thousand volts.

Capacitance depends on the following factor

1. Shape and size of the conductor
2. Medium between them
3. Presence of other conductors near it.

Types of Capacitor

Here a wide range of Capacitor are manufacturing according to their use,

like Shaps and Size, dielectric Mideam, trimmer, Variable, and SMDs as you can see in image.1 or 2, but technically we categorize in below.

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Type of Capacitors

• Parallel Plate Capacitor
• Spherical capacitor
• Cylindrical capacitor

Parallel Plate Capacitor

The 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 very small compared to the dimensions of the plate so that the effect of bending outward of electric field lines at the edges and the non-uniformity of surface charge density at the edges can be ignored.

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Parallel plate Capacitor

The charge density on each plate of parallel plate capacitor has a magnitude of σ

σ = Q/A

From Gauss law, E = Q/ε0A

Also, E = V/d

Now taking field due to the surface charges, outside of the capacitor,

$E = \frac{\sigma}{2\varepsilon _0}-\frac{\sigma}{2\varepsilon _0}=0$

$Inside\;E = \frac{\sigma}{2\varepsilon _0}+\frac{\sigma}{2\varepsilon _0}=\frac{\sigma}{\varepsilon _0}=\frac{q}{A\varepsilon _0}\;\;\;$

$\frac{v}{d} = \frac{q}{A\varepsilon _0}$

$or,C = \frac{q}{v} =\frac{A\varepsilon _0}{d}$

This result is valid for the vacuum between the capacitor plates. For another medium, then capacitance will be

 $C = \frac{kA\varepsilon _0}{d}$, where k is the dielectric constant of the medium,

$\varepsilon _0 = Permittivity\;of\;free\;space = 8.85\times 10^{-12}C^{2}/Nm^{2}$

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, Rin = a, and radius of the outer sphere, Rout = b. The inner shell is given a positive charge +Q and the outer shell is given –Q.

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Spherical Capacitor

The potential difference, $V = \frac{q}{4\pi \epsilon _{0}ka}+\frac{-q}{4\pi \epsilon _{0}kb}$q​+4πϵ0​kb−q​

$V = \frac{q}{4\pi \epsilon _{0}k}\left [ \frac{1}{a}-\frac{1}{b} \right ]$V=4πϵ0​kq​[a1​−b1​]

 

$V = \frac{q}{4\pi \epsilon _{0}k}\left [ \frac{b-a}{ab} \right ]$q​[abb−a​]

 

$C = \frac{q}{V}= \frac{q}{\frac{q}{4\pi \epsilon _{0}k}\left [ \frac{b-a}{ab} \right ]}$[abb−a​]q​

 

$C = 4\pi \epsilon _{0}k\left [ \frac{ba}{b-a} \right ]$ Farad

Cylindrical Capacitor

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

From gauss’s law,

Image.5

Cylindrical Capacitor

$E = \frac{Q}{2\pi \varepsilon_0 rl } = \frac{\lambda}{2\pi\varepsilon _0r}$

Where λ = Q/l, linear charge density

The potential difference of Cylindrical Capacitor is given by,

$\Delta V = V_b – V_a = -\int_{a}^{b}E_rdr = -\frac{\lambda}{2\pi\varepsilon _0}\ln \left ( \frac{b}{a} \right )$

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

$C = \frac{Q}{\left | \Delta V \right |} = \frac{\lambda L}{\lambda \ln (b/a)/2\pi\varepsilon _0} = \frac{2\pi\varepsilon _0L}{\ln(b/a)}$

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

Dielectrics and Capacitance

What are Dielectrics?

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

Polar and Non-polar Dielectrics

Each atom is made of a positively charged nucleus surrounded by electrons. If the center of the negatively charged electrons does not coincide with the center of the nucleus, then a permanent dipole (separation of charges over a distance) moment is formed. Such molecules are called polar molecules. If a polar dielectric is placed in an electric field, the individual dipoles experience a torque and try to align along the field.

In non-polar molecules, the centers of the positive and negative charge distributions coincide. There is no permanent dipole moment created. But in the presence of an electric field, the centers are slightly displaced. These are called induced dipole moments.

Image.7

Polar or Nonpolar Molecules

Polarization of a Dielectric Slab

It is the process of inducing charges on the dielectric and creating a dipole moment. Dipole moment appears in any volume of a dielectric. The polarization vector $\overrightarrow{p}$ is defined as the dipole moment per unit volume.

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Polarized

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Dielectric Constant

Let $\overrightarrow{{{E}_{0}}}$ be the electric field due to external sources and $\overrightarrow{{{E}_{p}}}$ be the field due to polarization (induced). The resultant field is

$\overrightarrow{E}=\overrightarrow{{{E}_{0}}}+\overrightarrow{{{E}_{p}}}$.

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

$\overrightarrow{E}=\frac{\overrightarrow{{{E}_{0}}}}{K}$​​ K is called the dielectric constant or relative permittivity of the dielectric. For vacuum, $\overrightarrow{{{E}_{p}}}$ = 0, K = 1. It is also denoted by ε

Effect of Dielectric in Capacitance

Dielectric Slabs in Series

A parallel plate capacitor contains two dielectric slabs of thickness d1, d2, and dielectric constant k1 and k2 respectively. The area of the capacitor plates and slabs is equal to A.

Considering the capacitor as the combination of two capacitors in series, the equivalent capacitance C is given by:

$\frac{1}{C}=\frac{1}{{{C}_{1}}}+\frac{1}{{{C}_{2}}}$1​+C2​1​

$\frac{1}{C}=\frac{{{d}_{1}}}{{{k}_{1}}\varepsilon _{0}A}+\frac{{{d}_{2}}}{{{k}_{2}}{{\varepsilon }_{0}}A}$ε0​Ad1​​+k2​ε0​Ad2​​

$C=\frac{{{\varepsilon }_{0}}A}{\frac{{{d}_{1}}}{{{k}_{1}}}+\frac{{{d}_{2}}}{{{k}_{2}}}}$d1​​+k2​d2​​ε0​A​

Summery

In this post, we learn about capacitors and their type and Working principles.

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