CAPACITORS & CAPACITANCE OF PARALLEL PLATE CAPACITOR

CAPACITOR:-

               Capacitor is a charge storing device which consists of two parallel metal plates of same nature, surface area, magnitude of charge on each plate and surface charge density separated by small distance as
shown in fig
Magnitude of charge(Q) stored on each plate is directly proportional to potential difference(V) applied by the battery ie

CAPACITANCE OF CAPACITOR:-

  The ability of a capacitor to store charge is known as its capacitance. It is measured in farads (F) which is a higher unit other sub multiple units are

1µF=10^-6 F

1nF=10^-9 F

1pF=10^-12F

FARAD:-

     If a charge of 1 coulomb is stored on either plate having a potential difference of 1V then the capacitance of capacitor will be one farad.

CAPACITANCE OF PARALLEL PLATE CAPACITOR:-

If air or vacuum is used as a medium between the plates of capacitor as shown in the above fig then the capacitance can be written by using (ii)  

C_vacuumed= Q -------(iv)

                  V 

but surface charge density is  σ= Q  

                                                  A 

or Q=σA----(v)

also magnitude of electric field intensity as potential gradient is

E= V 

     d

or V=Ed----(vi)

by Gausses law application about parallel plate capacitor we know 

E= σ -----(vii)

     ε_o     

using (vii) in (vi) we get

V= σ  d

      ε_o  

using (v) and (viii) in (iv) we get

 C_vacuumed=σAε_o 

                  σd    

after cancellation we get

 C_vacuumed=A ε_o ----(ix)

                         d

Capacitance of a capacitor depends directly upon its area of each plate and inversely upon ε_{r is placed between the plates of capacitor as shown in fig then capacitance is given by}

separation between plates. However if any other medium of relative permittivity

COMBINATION OF CAPACITOR:-

Like any other form of electrical circuitry device, capacitors can be used in combination in circuits. These combinations can be in series (in which multiple capacitors can be found along the same path of wire) and in parallel (in which multiple capacitors can be found along different paths of wire).

CAPCITORS IN PARALLEL:-

    Capacitor connected parallel have following properties:

• Each capacitor connected to a battery of voltage V has same potential difference V across it ie

V₁=V₂=V₃ ------- = V_n

• The charge developed across the plate of each capacitor will be different due to different value of capacitances.
• The total charge Q supplied by the battery is divided among the various capacitors. Hence

Q= Q₁+Q₂+Q₃+--------+ Q_n

Q=C₁V+C₂V+C₃V+-------+C_nV

Or  Q= V( C₁+C₂+C₃+------+C_n)

-----(i)

        

                                                                                                                                                                              

 So (i) becomes C_eq= C₁+C₂+C₃+------+C_n-----(ii)

·          From (ii) we conclude that equivalent capacitance of a parallel combination of capacitors is greater than any of the individual capacitance.

CAPACITORS IN SERIES:-

Capacitors connected in series have following properties:

·         Capacitors have same charge across it. If the battery supplies +Q charge to the left plate of capacitor C₁ due to induction –Q charge is induced on its right plate and +Q on the left plate of the capacitor C₂. ie

Q₂= Q₂= Q₃=---------= Q_n  

·         The potential difference across each capacitor is different due to different values of capacitance.

·         The voltage of battery has been divided among the various capacitors. Hence

V=V₁+V₂+V₃+------+V_n

Some problems regarding equivalent are as follow

PROBLEM 1

What is the equivalent capacitance of the circuit shown below? 

The 1 μF and 3 μF capacitors are in parallel and are equivalent to a single 4 μF capacitor. The 6 μF and 2 μF capacitors are also in parallel and are equivalent to a single 8 μF capacitor.

The two 4 μF capacitors are in series, therefore 1/C_p = ¼ + ¼ = ½, or C_p = 2 μF. The two 8 F capacitors are in series, therefore 1/C_p = 1/8 + 1/8 = ¼, or C_p = 4 μF.

The 2 μF and 4 μF capacitors are in parallel and are equivalent to a single 6 μF capacitor. The equivalent capacitance is 6 μF.

PROBLEM 2

A system of four plates P, Q, R and S each having surface area A are adjust at a distance d from each other as shown in fig. What is the equivalent capacitance of arrangement between A and B.  

SOLUTION

Total capacitance can be find out by  finding out total path ways between two points for example in this question charge can move from A via P then Q to B, also charge can move from A via R then Q to B and also from A via R and then S to B. So we can say that there are three ways for charge to move from A to B so the capacitance of the arrangement between A and B is 3C (Capacitance) or  3ε_oA        

        d

similarly we can find out capacitance of this new arrangement between A and B in this example there are two ways for charge to move from A to B one is via P then Q to B and other is via S then R to B so total capacitance is 2C (capacitance) or 2ε_oA     

                                                                                                  d

PROBLEM 3

What is the equivalent capacitance between the two points A and B

SOLUTION

The circuit shown is parallel combination as we know that one of the property of parallel combination that if only one device is working properly then current will flow through the circuit. Here in this example we see that alternate paths are given so if only one device is working properly then current will flow through the circuit so  

C_equivalent =C₁ + C₂ + C₃

That is C_equivalent = 3µF + 3µF + 3µF

      C_equivalent = 9µF that is equivalent capacitance is equal to 9µF