# Introduction to Capacitors

The capacitor is an electrical component which has charge storing capability and stores energy in the form of electric field and so a potential difference appears across it’s plates.

Capacitors come in different packages as well as they differ in size also but they all do the same thing i.e they store charge.

A very simple model of a capacitor consist of two parallel plates which are conducting in nature and separated by some predefined distance, these plates are filled with air or with some insulating materials such as mica, wax etc. This insulating material between the plates generally referred as dielectric.

this insulating material prevents flow of current between plates, these plates may have different shapes for example circular, rectangular or cylindrical. the shape of the plate may vary with the application of capacitor.

A capacitor blocks DC (i.e Direct Current) while it allows the flow of AC (i.e Alternating Current), so in some way we can understand that a capacitor offers infinite resistance to the DC whereas the resistance for AC is very small or negligible depending upon the type of the insulating material.

when a voltage or potential Vc is applied across the capacitor there comes the current and we know that the current is the flow of electric charge so when this charge flows onto the capacitor they reside on plates as they can’t pass through the insulating material (i.e Dielectric) .

Electrons – negatively charged particles – are sucked into one of the plates, and it becomes overall negatively charged. The large mass of negative charges on one plate pushes away like charges on the other plate, making it positively charged Or simply understand that the plate connected to positive terminal of battery becomes positive and the plate connected with negative terminal of battery become negatively charged.

Due to this stationary charge on plates i.e positive on one plate and negative on another plate there comes the electric field and the direction of the electric field is from positive charge to the negative charge as shown in figure below

which influence electric potential energy and voltage.

the charge stored by the capacitor is given by the product of capacitance and the potential applied across it. Mathematically Q= C x V . where Q is charge stored (unit : coulomb), C is capacitance (unit : Farad), V is the potential or voltage (unit : Volts), the relation can be used to find any of the unknown variable if two of them is given.

the strength of electric field depends on the amount of the charge stored on the plates of the capacitor i.e more the amount of charge stored on plates stronger the electric field between the plates.

### Capacitance of a capacitor

Capacitance is defined as the electrical property of a capacitor to store an electric charge or it can also be define as the ration of electrical charge to the electric potential. As we have already seen the capacitance is given by C = Q / V and so the S.I unit for capacitance is Farad or coulomb/volts.

### Example No. 1

Q. A capacitor of 500uF is charged by a power supply 4V through a 200 ohm resistor. Calculate (a) the initial charging current. (b) the final charge stored on the capacitor.

Solution:

a) initially the capacitor voltage is zero and 4V of power supply would be across resistor.

I = V / R  =  4V / 200ohm  =  0.02 A  =  2mA

b) at the end of the charging process, all 4V would be across the capacitor.

Q = C V  =  500uF x 4V = 2000uC

### Example No. 2

Q. A capacitor of 5000uF is charged by a 12V power supply and then discharged through a 150ohm resistor. Calculate (a) its initial charge. (b) the time constant. (c) the charge remaining after 1.5 seconds.

Solution:

a) since, Q = C V = 5000uF x 12V = 60000uC

b) time constant = R x C = 150ohm x 5000uF = 0.75 seconds

c) since, Q’ = Q e^(-t/RC) = 60000uC x e^(-1.5s/0.75s) = 8120uC

### Capacitance of a parallel plate capacitor

A parallel plate capacitor is a combination of two parallel plates separated by a small distance (say, d) and the region between the two plates can be air or some insulating material.

We define a parallel plate capacitor as a device which is capable of storing electrostatic energy in the form of charge in the dielectric medium between the plates, and thus it can be visualized as equivalent to a rechargeable DC battery.

The capacitance of a parallel plate capacitor is equal to A ε0 / d and mathematically can be derived as,

Here, εis permittivity of free space or vacuum and its value is equal to 1/(4π x 9×109) or  8.854 x 10-12 Farads per metre.

Generally, the conductive plates of a capacitor are separated by some kind of insulating material rather than a perfect vacuum, so when nothing is mentioned is any given problem then we can consider the permittivity of air.

### Example No. 3

Q : A parallel plate capacitor is placed in the air has an area of 0.44 m^2 and are separated from each other by distance 0.04m . Determine the capacitance parallel plate capacitor.

Solution:

Given: Area A = 0.44 m^2, distance d = 0.04 m, ϵ = 8.54  ×10^−12 F/m

The parallel plate capacitor formula is given by,  C = ϵoA/d

putting values in the formula,  C =  8.854×10^−12 × 0.44 / 0.04

C = 9.739 ×10 −11F.

Therefore, the capacitance of parallel plate capacitor is 9.739 ×10 −11 F.

### Example No. 4

Q : Calculate the area of parallel plate capacitor in the air if the capacitance is 22 nF and separation between the plates is 0.02m.

Solution:

Given : Capacitance = 22 nF, Distance d = 0.02 m, ϵo = 8.854 ×× 10^−12 F/m

The parallel plate capacitor formula is given by,   C = ϵoA/d
So that Area given as, A = dC/ϵo

A = 0.02 × 22×10^−9 / 8.854×10^−12 = 49.69 m^2.

Therefore, the area of parallel plate capacitor is 49.69 m^2.

### The Dielectric of a Capacitor

As we have seen the capacitance of a capacitor depends on the area of the plate and the spacing between the plates but there exist an another constant on which the capacitance depends and this constant is know as the Dielectric constant, k of the insulating material used between the plates.

we know the plate is conducting but the material filled between those plates are always insulating in nature and the value of dielectric constant depends on the type of the material used.

so that the permittivity of the overall system also get affected and given by the product of permittivity of free space (εo) and the relative permittivity (εr) of the material being used as the dielectric.

ε = εo . εr

Note : The dielectric constant – also called the relative permittivity indicates how easily a material can become polarized by imposition of an electric field on an insulator.

Now, the capacitance is given as

C = kϵoA / d   or   εrϵoA / d

lets’s see an example to understand the above illustration :

### Example No. 5

Q : Calculate the capacitance of a parallel plate air capacitor of plate area 30 m^2, the plates being separated by a dielectric 2 mm thick and of relative permittivity 6. If the electric field strength between the plates is 500 V/mm, calculate the charge on each plate.

Solution:

Capacitance,C = kϵoA / d = (6 x 8.854 x 10^-4 x 30) / 2 x 10^-3 = 0.797 × 10–6 F = 0.797 µF

P.D. across plates, V = E × d = 500 × 2 = 1000 volts

∴ Charge on each plate, q = CV = (0.797 × 10–6)1000 = 0.797 × 10–3 C = 0.797 mC

### Voltage rating of a capacitor

Like every electronic device capacitors also have voltage ratings and the voltage across these capacitors should never exceed the defined voltage ratings. if the value of voltage across capacitor exceed the rating, the dielectric will break down and the capacitor would be damaged.

the value of the voltage rating depends on the type of dielectric material used and the width of the dielectric, the voltage ratings are predefined by the manufacturer and clearly mentioned in the datasheet of the respective capacitors.