Physics‎ > ‎

### Capacitors.

posted Oct 26, 2017, 11:19 AM by Upali Salpadoru   [ updated Oct 29, 2017, 12:44 PM ]

A capacitor or a condenser is a passive component that can store energy in the form of static electrical charges. It is made up of two metal plates as close to each other yet separated by a suitable insulating material.  Mainly there are 3 types of capacitors. Non polar capacitors, Electrolytic capacitors and variable capacitors. When charged there is an electric field (E) between the two plates.

While the field is directly proportional to voltage, it is inversely proportional to the distance between the plates.

Electric field =  Volts / distance.  ……………..E  = V / d. Fig.1. Capacitors.

Capacity and Capacitance

The capacity to hold an electric charge by a capacitor depends on the following: Dielectric constant. - This is a ratio which gives the factor by which the capacity increases as compared with a vacuum in between.

 Substance Dielectric constant  orRelative permittivity. AirGlassMicaPaperPolytheneRubberVacuum 1.00064.75.03.92.37.01.000

2.     Area of the plates.  -   Greater the area greater the capacity.

3.    The gap between the plates..  Smaller the gap greater the capacity.

Electric field =  Volts / distance. E = V / d.

Capacitance....C.

The unit for measuring the capacitance is the Farad.  If a capacitor can hold a charge of 1 Coulomb on a voltage of 1Volt.   Capacitance =  Coulombs / Volts……… C=  Q / V.

This ratio will be a constant for a particular capacity will be marked on the cover.

As the farad  is a large unit normally  micro farads are used.

1.0 farad = 1000000 micro farad  ( 10 6 μ F )
Absolute Permittivity.

This is a measure of how an electric field is affected by, a dielectric medium.

A vacuum gives the lowest possible permittivity.

It is given as  …….. ε0 = 8.85×10−12 F/m.

When dielectric is air the formula to be used:-........C = ε0 xA / d    (A= area)

When the dielectric constant is not air, the above value has to be multiplied by the absolute permittivity of the material.  That is :......  C  =  ε0 x εr x A / d

Practice problem.1.

Relevant equation  E =  V/d

 Electric field  E Voltage   V Distance     d. Example:-  4000.V/m. 6.0  V 1.5 mm. (1.5x10-3) 2.      8000 V/m 12.0  V. 1.5 mm. 3.     16000 V/m 12.0 V 0.75 mm 3.        2000 V/m 3.0 V. 1.5 mm.

Practice problem 2.

Relevant equation.   C = Q/V.

Complete the blank spaces.

 Capacitance Charge  - Coulombs Volts Example:-    1.0 F 1 .0 C. 1 V 2 F 6 .0 C 3 V 0.5 F 0.5 C 1 V 1000 μ F   (1000x10-6) 0.012 C 12 V 100 μ F 100x10-6 x 6= 6x10-4 C. 6.0 V 100 μ F 100x10-6 x12 =  1,2 x 10-3C. 12.0 V
High lite to get the answers.

Practice problem 3.

Two metal foils are rolled together, with a polythene sheet in between to form a capacitor. Each sheet had a length of 50 cm and a width of 10 cm. Thickness of the polythene sheet =  0.1 mm ( 10-4 m)

Calculate the capacitance.

Area of a sheet =  0.5 x 0.1 = 0.05 m2

Relevant equation    C= ε0 x εr x A / d

C=  8.85×10−12 X 2.3 x 0.05 /  10-4

=1.02 x10-8

Capacitors in series and parallel.

 Series Parallel 1/C =  1/C1   +  1/C2 C =  C1  + C2

Charging a capacitor Vb = Battery voltage.
vr = Resistor voltage.
vc = Capacitor voltage.
I = Current.
R = Resistance.

Time constant.

This is the time for the voltage to change by 63%of the voltage. Time constant is shaded. Red for charging and green for discharging.  t = RC

Example.

A capacitor 0f 250μ F  is connected to a resistor of 2000Ω  to 10 V DC.source.

When the capacitor voltage falls by the time constant

1. Determine the current.

2. Find the time constant.

Capacitor Resistor circuits

 V = ir10 = i x 2000  I = 10/2000    =5x10-3 A. t = RCT =2000  x 5 x 10-3   =10 seconds.