Chapter 5 Electrostatic Potential and Capacitance (Electrostatics Part 5) – Physics free study material by TEACHING CARE online tuition and coaching classes

Chapter 5 Electrostatic Potential and Capacitance (Electrostatics Part 5) – Physics free study material by TEACHING CARE online tuition and coaching classes

File name : Chapter-5-Electrostatic-Potential-and-Capacitance-Electrostatics-Part-5.pdf

 

 

 Capacitance.

 

  • Definition : We know that charge given to a conductor increases it’s potential e.,

 

 

Q µ V

 

Þ Q = CV

 

Where C is a proportionality constant, called capacity or capacitance of conductor. Hence capacitance is the ability of conductor to hold the charge.

 

  • Unit and dimensional formula : I. unit is

Coulomb = Farad (F) Volt

 

Smaller   S.I.   units   are   mF,   mF,   nF     and   pF     ( 1mF = 10 -3 F ,

1pF = 1mmF = 10 -12 F )

1mF = 10 -6 F ,

1nF = 10 -9 F ,

 

 

C.G.S. unit is Stat Farad 1F = 9 ´ 1011 Stat Farad .   Dimension : [C] = [M -1 L-2T 4 A 2 ].

  • Capacity of an isolated spherical conductor : When charge Q is given to a spherical conductor of

 

radius R, then potential at the surface of sphere is

+ + +  Q

+               +

 

V = k . Q

ì   =    1  ü

+            R        +

 

î
0 þ

R               ík      4pe  ý

+                       +

+          O                +

+                    +

 

Hence it’s capacity C = Q = 4pe R

V               0

Þ  C = 4πε0 R =

1    .R

9 ´ 109

+ + + + +

 

in C.G.S.           C = R

 

Note : @            If earth is assumed to be spherical having radius

R = 6400 km.

It’s theortical capacitance

 

C =      1

9 ´ 109

  • 6400 ´ 103 = 711 mF . But for all practical purpose capacitance of earth is taken

 

infinity.

  • Energy of a charged conductor : When a conductor is charged it’s potential increases from 0 to V as shown in the graph; and work is done against repulsion, between charge stored in the conductor and charge coming from the source (battery). This work is stored as “electrostatic potential energy”

 

From graph :       Work done = Area of graph

= 1 QV

2

 

Hence potential energy U = 1 QV ; By using Q CV,

2

U = 1 QV = 1 CV 2 = Q2

 

we can write

 

2         2          2C

  • Combination of drops : Suppose we have n identical drops each having – Radius – r, Capacitance – c, Charge – q, Potential – v and Energy – u.

If these drops are combined to form a big drop of – Radius – R, Capacitance – C, Charge – Q, Potential – V

and Energy – U then –

  • Charge on big drop : Q = nq

 

 

 

 

  • Radius of big drop : Volume of big drop = n ´ volume of a single drop e.,

4 pR 3 = n ´ 4 pr 3 ,

 

R = n1/3r

 

 

  • Capacitance of big drop :

 

  • Potential of big drop :

C = n1/3c

V Q =  nq

3              3

 

 

V = n2/3v

 

C       n1 / 3 c

 

 

(v)  Energy of big drop :

U = 1 CV 2 = 1 (n1/ 3c)(n2 / 3v)2

 

U = n5/3u

 

2          2

  • Sharing of charge : When two conductors joined together through a conducting wire, charge begins to flow from one conductor to another till both have the same potential, due to flow of charge, loss of energy also takes place in the form of

 

Suppose there are two spherical conductors of radii r1 and r2 , having charge Q1

and

Q2 ,

potential V1

and

 

V2 , energies U1

and U 2

and capacitance C1

and C2

respectively, as shown in figure. If these two spheres are

 

connected through a conducting wire, then alteration of charge, potential and energy takes place.

 

 

  • New charge  :    According   to the  conservation of charge

Q1 + Q2  = Q

  • Q = Q

(say), also

 

1
2

Q     C V        4pe r        Q     r

Q          r

Q + Q     r + r

 

Q
Q
Q

    1 =    1      =      0 1 ,

    1 = 1

Þ   1 +    1 = 1 + 1 Þ

    1          2 = 1       2

 

Q
2
2
2

‘        C2 V        4per2

‘        r2

‘               r2

‘                  r2

 

Q
ê
2
Þ
2                          ú

Q é    r2     ù

r + r

and similarly

Q é    r1     ù

Q
ê
1                          ú

r + r

 

    1        2                                           

ë 1       2 û                                           ë 1       2 û

 

  • Common potential : Common potential (V ) = Total charge

Q1 + Q2

Q

  • Q

Þ V  C1 V1 + C2 V2

 

Total capacity

C1 + C2

C1 + C2

C1 + C2

 

(iii)   Energy loss : As electrical energy stored in the system before and after connecting the spheres is

 

1           1                        1

1             æ C V

  • C V ö 2

 

Ui =

C V 2 +

C V 2

and U   =

(C + C

). V 2 =

(C + C

  1   1         2   2 ÷

 

i               f

2   1 1

2   2   2

f           2   1         2

2   1         2 è

C1 + C2        ø

 

so energy loss

ΔU = UU =

C1C2

(V V )2

 

1
2

2(C1 + C2 )

 

 

+ +             + +

+        +       +

+         +                 +

++  +          +     +

Example: 95     Eight drops of mercury of same radius and having same charge coalesce to form a big drop. Capacitance of big drop relative to that of small drop will be                                              [MP PMT 2002, 1990; MNR 1999, 87; DCE 1998]

(a) 16 times                        (b) 8 times                       (c) 4 times                    (d) 2 times

Solution: (d)      By using relation C = n1/ 3 . c Þ    C = (8)1/ 3 . c = 2c

 

Example: 96 Two spheres A and B of radius 4 cm and 6 cm are given charges of

80mC

and 40mC respectively. If they are

 

connected by a fine wire, the amount of charge flowing from one to the other is                                      [MP PET 1991]

 

(a)

20 m C

from A to B           (b)

16 m C

from A to B (c) 32mC from B to A              (d)

32 m C  from A to B

 

 

Solution: (d)      Total charge

 

Q = 80 + 40 = 120 m C . By using the formula

Q  ‘ = Q é    r1     ù . New charge on sphere A is

 

1          ê r1 + r2 ú

ë             û

Q   = Q é    rA          ù = 120 é 4 ù = 48 m C . Initially it was 80 m Ci.e., 32 m C charge flows from A to B.

A               ê rA  + rB ú              êë 4 + 6 úû

ë              û

 

Example: 97     Two insulated metallic spheres of 3mF

and 5mF

capacitances are charged to 300V

and 500V

respectively.

 

The energy loss, when they are connected by a wire, is                        [Pb PMT 1999; CPMT 1999; KCET (Engg.) 2000]

 

(a)

0.012 J

(b)

0.0218 J

(c)

0.0375 J

(d)

3.75 J

 

 

Solution: (c)      By using ΔU =

C1C2

2(C1 + C2 )

(V1

  • V2

)2 ; ΔU = 0.375 J

 

Example: 98     64 small drops of mercury, each of radius r and charge q coalesce to form a big drop. The ratio of the surface density of charge of each small drop with that of the big drop is                                                [KCET 2002]

(a) 1 : 64                            (b) 64 : 1                         (c) 4 : 1                        (d) 1 : 4

 

s Small

q / 4pr 2

æ q öæ R ö 2

 

Solution: (d)                 =

= ç      ÷ç    ÷ ; since R = n1/3r and Q = nq

 

s Big

Q / 4pR 2

è Q øè r ø

 

So s Small      1    Þ s Small1

 

s Big

n1 / 3

s Big                4

 

 

 

 

Capacitor.

(1)   Definition : A capacitor is a device that stores electric energy. It is also named condenser.

or

A capacitor is a pair of two conductors of any shape, which are close to each other and have equal and opposite charge.

 

 

 

 

 

 

 

  • Symbol : The symbol of capacitor are shown below

or                                                    variable capacitor

 

  • Capacitance : The capacitance of a capacitor is defined as the magnitude of the charge Q on the positive

 

plate divided by the magnitude of the potential difference V between the plates i.e.,

C = Q

V

 

(4)   Charging : A capacitor get’s charged when a battery is connected across the plates. The plate attached to the positive terminal of the battery get’s positively charged and the one joined to the negative terminal get’s negatively charged. Once capacitor get’s fully charged, flow of charge carriers stops in the circuit and in this condition potential difference across the plates of capacitor is same as the potential difference across the terminals of battery (say V).

 

  • Charge on capacitor : Net charge on a capacitor is always zero, but when we speaks of the charge Q on a capacitor, we are referring to the magnitude of the charge on each plate. Charge distribution in making of parallel plate capacitor can easily be understand by reading carefully the following sequence of figures –

 

 

(6)   Energy stored : When a capacitor is charged by a voltage source (say battery) it stores the electric energy. If = Capacitance of capacitor; Q = Charge on capacitor and V = Potential difference across

 

capacitor then energy stored in capacitor

U = 1 CV 2 = 1 QV = Q2

 

2          2         2C

Note : @            In charging capacitor by battery half the energy supplied is stored in the capacitor and remaining half energy (1/2 QV) is lost in the form of heat.

  • Types of capacitors : Capacitors are of mainly three types as described in given table
Parallel Plate Capacitor                      Spherical Capacitor                       Cylindrical Capacitor

 

It consists of two parallel metallic plates (may be circular, rectangular, square) separated by a small distance

 

+Q                          Q

It consists of two concentric conducting spheres of radii a and b (a < b). Inner sphere is given charge

+Q, while outer sphere is earthed

It consists of two concentric cylinders of radii a and b (a < b), inner cylinder is given charge +Q while outer cylinder is earthed. Common length of the cylinders is l then

 

+              –                                                             –    –

+
+

–                  –

Q

–     +          +     –

b          a    q q

 

+      air –

+Q

+

A                                                                                                               –    +

a

+

–     +

–        +

+             –                                                          –  –

d

+ b                                                                        l

+      –

 

 

A = Effective overlaping area of

Capacitance C = 4πε . ab  

0

b a

Capacitance

 

each plate

d = Separation between the plates

Q = Magnitude of charge on the inner side of each plate

s = Surface density of charge of

in C.G.S. C =  ab  . In the presence of

b a

dielectric medium (dielectric constant K)

0

between the spheres C‘ = 4pe K ab  

b a

Special Case :

 

 

In the presence of dielectric medium (dielectric constant K) capacitance increases by K times and

 

each plate æ= Q ö

If outer sphere is given a charge +Q

C‘ =

2peKl

 

A
è         ø

ç         ÷                                         while inner sphere is earthed

a

log

æ b ö

 

V = Potential difference across the

Q¢ –         +Q

e ç a ÷

 

 

plates

E = Electric field between the plates

– – –    –                                                              è   ø

–             –

–               b

–       – –

 

æ      s ö

 

e

ç=      ÷

è       0 ø

Induced charge on the inner sphere

A

 

Capacitance :

 

in C.G.S. :

C =   0

d

C =     A

4πd

,

This arrangement is not a capacitor. But it’s capacitance is equivalent to the sum of capacitance of spherical

 

If a dielectric medium of dielectric constant           K             is          filled     completely

capacitor and spherical conductor

i.e.

 

between the plates then capacitance                b 2                      ab

 

increases by K times C’ = KC

4pe 0 . b – a = 4pe 0 b – a + 4pe 0 b

 

 

 

 

Concepts

It is a very common misconception that a capacitor stores charge but actually a capacitor stores electric energy in the electrostatic field between the plates.

Two plates of unequal area can also form a capacitor because effective overlapping area is considered.

 

 

A

 

 

 

If two plates are placed side by side then three capacitors are formed. One between distant earthed bodies and the first face of the first plate, the second between the two plates and the third between the second face of the second plate and distant earthed objects. However the capacitances of the first and third capacitors are negligibly small in comparision to that between the plates which is the main capacitance.

Capacitance of a parallel plate capacitor depends upon the effective overlapping area of plates (C µ A) , separation between the

plates æ C µ 1 ö and dielectric medium filled between the plates. While it is independent of charge given, potential raised or

d

ç              ÷

è              ø

nature of metals and thickness of plates.

The distance between the plates is kept small to avoid fringing or edge effect (non-uniformity of the field) at the bounderies of the plates.

+              –

 

 

 

 

 

+              –

 

Spherical conductor is equivalent to a spherical capacitor with it’s outer sphere of infinite radius.

A spherical capacitor behaves as a parallel plate capacitor if it’s spherical surfaces have large radii and are close to each other.

The intensity of electric field between the plates of a parallel plate capacitor (E = s/e0) does not depends upon the distance between them.

The plates of a parallel plate capacitor are being moved away with some velocity. If the plate separation at any instant of time is

‘d’ then the rate of change of capacitance with time is proportional to  1 .

d 2

Radial and non-uniform electric field exists between the spherical surfaces of spherical capacitor.

Two large conducting plates X and Y kept close to each other. The plate X is given a charge Q1 while plate Y is given a charge

Q2 (Q1 > Q2 ) , the distribution of charge on the four faces a, b, c, d will be as shown in the following figure.

 

 

X

Q1                                   Q2

– æ Q1  Q2 ö

 

 

b                                     d              Þ

ç                ÷

Y
X

è       2       ø

 

æ Q1 + Q2 ö

æ Q1  Q2 ö

æ Q1  Q2 ö

 

ç
ø

a                                      c                                          è       2       ÷

ç                ÷

è       2       ø

ç               ÷

è       2       ø

 

 

+  –

Example: 99     The capacity of pure capacitor is 1 farad. In D.C. circuit, its effective resistance will be                            [MP PMT 2000]

 

 

  • Zero (b) Infinite                        (c) 1 ohm                            (d)

1 ohm

2

 

Solution: (b)     Capacitor does not work in D.C. for D.C. it’s effective resistance is infinite i.e. it blocks the current to flow in the circuit.

Example: 100 A light bulb, a capacitor and a battery are connected together as shown here, with switch S initially open. When the switch S is closed, which one of the following is true                                                                 [MP PMT 1995]

  • The bulb will light up for an instant when the capacitor starts charging
  • The bulb will light up when the capacitor is fully charged
  • The bulb will not light up at all
  • The bulb will light up and go off at regular intervals

Solution: (a)     Current through the circuit can flow only for the small time of charging, once capacitor get’s charged it blocks the current through the circuit and bulb will go off.

 

Example: 101 Capacity of a parallel plate condenser is 10mF when the distance between its plates is between the plates is reduced to 4cm, its capacity will be

8 cm . If the distance

 

[CBSE 2001; Similar to CPMT 1997; AFMC 2000]

(a) 10mF                                      (b) 15mF                                   (c) 20mF                              (d) 40mF

 

 

Solution: (c)

C e 0 A µ 1

 

\     C1  = d2 

or    C

= d1 ´ C

 

= 8 ´ 10 = 20m F

 

d         d                                      C2       d1

2     d2        1      4

 

Example: 102 What is the area of the plates of a 3F

parallel plate capacitor, if the separation between the plates is 5 mm

[BHU 2002; AIIMS 1998]

 

(a)

1.694 ´ 109 m2

(b)

4.529 ´ 109 m2

(c)

9.281 ´ 109 m2

(d)

12.981 ´ 109 m2

 

Solution: (a)      By using the relation C = e 0 A

d

Þ  A = Cd =

e 0

3 ´ 5 ´ 10 -3

8.85 ´ 10 -12

= 1.694 ´ 109 m2 .

 

Example: 103 If potential difference of a condenser (6 m F) is changed from 10 V to 20 V then increase in energy is

[CPMT 1997, 87]

 

(a)

2 ´ 10 -4 J

(b)

4 ´ 10-4 J

(c)

3 ´ 10 -4 J

(d)

9 ´ 10-4 J

 

Solution: (d)      Initial energy Ui = 1 CV 2 ;               Final energy U    = 1 CV 2

2     1                                              f          2     2

\   Increase in energy DU = U f   Ui   = 1 C (V 2V 2 ) = 1 ´ 6 ´ 10 6 (202 – 102 ) = 9 ´ 104 J.

2       2         1         2

 

 

 

Example: 104 A spherical capacitor consists of two concentric spherical conductors. The inner one of radius

R1 maintained

 

at potential V1 and the outer conductor of radius R2

at potential

V2 . The potential at a point P at a distance

 

x from the centre (where R2 > x > R1 ) is                                                                                     [MP PMT 1997]

 

 

(a)

V1 – V2 (x R )

 

(b)

V1 R1 (R2 – x) + V2 R2 (x R1 )

 

 

R2 – R1             1

(R2 – R1 ) x

 

(c)

V1 +

V2 x

(R2 – R1 )

(d)

(V1 + V2 ) x

(R1 + R2 )

 

Solution: (b)      Let Q1

and Q2

be the charges on the inner and the outer sphere respectively. Now

V1 is the total potential

 

on the sphere of radius R1,

 

So, V

Q1 + Q2 

…….. (i)

 

  • R1 R2

and V2 is the total potential on the surface of sphere of radius R2 ,

 

So, V

Q2 + Q1

 

…….. (ii)

 

  • R2 R2

If V be the potential at point P which lies at a distance x from the common centre then

 

 

Q1       Q2 Q1 + V  – Q1  = æ 1 –  1 ö + V   Q1(R1 – x)

1                     1 ç               ÷       1

+         =

x       R           x

Rx RøxR1

 

V =

2                              1             è            1

  • V1

……..(iii)

 

Substracting (ii) from (i)

 

V V

= Q1 – Q2  Þ (V

  • V )R R

= R Q

  • R Q

Þ Q    = (V1 – V2 )R1 R2

 

 

1        2      R1      R2

1        2     1   2

2   1        1   1            1

R2 – R1

 

Now substituting it in equation (iii), we have

V = (R1 – x)(V1 – V2 )R1 R2 + V        Þ V V1 R1(R2x) + V2 R2 (x R1)

 

xR1 (R2 – R1 )             1

x(R2 – R1 )

 

Example: 105 The diameter of each plate of an air capacitor is 4 cm. To make the capacity of this plate capacitor equal to that of 20 cm diameter sphere, the distance between the plates will be                                                      [MP PET 1996]

 

(a)

4 ´ 10 -3 m

(b)

1 ´ 10-3 m

  • 1 cm (d)

1 ´ 10 3 cm

 

Solution: (b)     According to the question

e A = 4pe R   Þ

d                0

d =   A   =

4pR

p (2 ´ 10-2 )2 4p ´ 10 ´ 10-2

= 1 ´ 10-3 m.

 

Example: 106 A spherical condenser has inner and outer spheres of radii a and b respectively. The space between the two is filled with air. The difference between the capacities of two condensers formed when outer sphere is earthed and when inner sphere is earthed will be                                                                                                         [MP PET 1996]

 

  • Zero (b)

4pe 0 a

(c)

4pe 0b

(d)

4pe 0

æ   b    ö

a

ç b a ÷

 

è           ø

 

Solution: (c)      Capacitance when outer sphere is earthed

C1 = 4pe 0 . ab

b   a

and capacitance when inner sphere is earthed

 

C2 = 4pe 0

b 2

ba

. Hence C2

  • C1

= 4pe 0 .b

 

Example: 107 After charging a capacitor of capacitance

4 m F

upto a potential 400 V, its plates are connected with a

 

resistance of 1kW . The heat produced in the resistance will be                                                     [CBSE PMT 1994]

(a) 0.16 J                                     (b) 1.28 J                                 (c) 0.64 J                             (d) 0.32 J

Solution: (d)     This is the discharging condition of capacitor and in this condition energy released

 

 

 

U = 1 CV 2

2

= 1 ´ 4 ´ 10 -6 ´ (400)2 = 0.32J 2

= 0.32 J.

 

Example: 108 The amount of work done in increasing the voltage across the plates of a capacitor from 5V to 10V is W. The work done in increasing it from 10V to 15V will be

(a) 0.6 W                                     (b) W                                        (c) 1.25 W                           (d) 1.67 W

Solution: (d)     As we know that work done = U final Uinitial = 1 C(V 2V 2 )

2       2        1

When potential difference increases from 5V to 10V then

 

W = 1 C(102 – 52 )

2

……..(i)

 

When potential difference increases from 10V to 15V then

 

W‘ = 1 C(152 – 102 )

2

On solving equation (i) and (ii) we get

W‘ = 1.67 W.

……..(ii)

 

 

 

Dielectric.

Dielectrics are insulating (non-conducting) materials which transmits electric effect without conducting we know that in every atom, there is a positively charged nucleus and a negatively charged electron cloud surrounding it. The two oppositely charged regions have their own centres of charge. The centre of positive charge is the centre of mass of positively charged protons in the nucleus. The centre of negative charge is the centre of mass of negatively charged electrons in the atoms/molecules.

 

 

 

  • Type of Dielectrics : Dielectrics are of two types –
    • Polar dielectrics : Like water, Alcohol, CO2 , NH3 ,

 

(ii)     Non polar dielectric : Like

 N 2 ,

 O2 , Benzene,

 

HCl etc. are made of polar atoms/molecules.

In polar molecules when no electric field is applied centre of positive charges does not coincide with the centre

Methane etc. are made of non-polar atoms/molecules. In non-polar molecules, when no electric field is applied the centre of positive charge coincides with the centre of

 

of negative charges.          O– –

 

105o

negative charge in the molecule. Each molecule has zero dipole moment in its normal state.

 

 

H+                               r                    H+

P

–             + +

r

P = 0

 

A polar molecule has permanent electric dipole

 

moment (p) in the absence of electric field also. But a

When electric field is applied, positive charge

 

polar dielectric has net dipole moment is zero in the absence of electric field because polar molecules are randomly oriented as shown in figure.

experiences a force in the direction of electric field and negative charge experiences a force in the direction opposite to the field i.e., molecules becomes induced

 

 

 

 

In the presence of electric field polar molecules tends to line up in  the direction of electric field, and the

electric dipole.                                        r

–                                      E

 

+   +

 

 

 

substance has finite dipole moment.

  • In general, any non-conducting, material can be called

as a dielectric but broadly non conducting material having

 

–  +         –  +         – +

–                                              +

–  +         –  +         – +

 

P

non polar molecules referred to as dielectric because induced dipole moment is created in the non polar molecule.

 

 

  • Polarization of a dielectric slab : It is the process of inducing equal and opposite charges on the two faces of the dielectric on the application of electric

Suppose a dielectric slab is inserted between the plates of a capacitor.

As shown in the figure.

 

Induced electric field inside the dielectric is

Ei , hence this induced

 

electric field decreases the main field E to

E Ei

i.e., New electric field

 

between the plates will be

E‘ = E Ei .

 

(3)   Dielectric constant : After placing a dielectric slab in an electric field. The net field is decreased in that region hence

 

If E = Original electric field and

E‘ = Reduced electric field. Then

 EK E

where K is called dielectric constant

 

K is also known as relative permittivity (e r ) of the material or SIC (Specific Inductive Capacitance)

 

The value of K is always greater than one. For vacuum there is no polarization and hence

E = E

and

K = 1

 

 

  • Dielectric breakdown and dielectric strength : If a very high electric field is created in a dielectric, the outer electrons may get detached from their parent The dielectric then behaves like a conductor. This phenomenon is known as dielectric breakdown.

The maximum value of electric field (or potential gradient) that a dielectric material can tolerate without it’s electric breakdown is called it’s dielectric strength.

 

S.I. unit of dielectric strength of a material is

V but practical unit is

m

kV .

mm

 

Variation of Different Variables (Q, C, V, E and U) of Parallel Plate Capacitor.

Suppose we have an air filled charged parallel plate capacitor having variables are as follows :

 

Charge – Q,    Surface charge density – s =

Q ,  Capacitance – C =

A

e 0 A d

 

Potential difference across the plates – V = E . d

 

 

Electric field between the plates –

E = s

e 0

=  Q Ae 0

 

 

 

Energy stored – U 1 CV 2  = Q 2

= 1 QV

 

2             2C       2

  • When dielectric is completely filled between plates : If a dielectric slab is fills completely the gap

 

between the plates, capacitance increases by K times i.e., C‘ = Ke 0 A Þ

d

C‘ = KC

 

The effect of dielectric on other variables such as charge. Potential difference field and energy associated with a capacitor depends on the fact that whether the charged capacitor is disconnected from the battery or battery is still connected.

Note : @ If nothing is said it is to be assumed that battery is disconnected.

 

 

  • When dielectric is partially filled between the plates : If a dielectric slab of thickness t (t < d) is

 

inserted between the plates as shown below, then

E = Main electric field between the plates,

Ei = Induced electric

 

+
+

Ei

+
 
+
+
+

 

field in dielectric.

E‘ = (E Ei ) = The reduced value of electric field in the dielectric. Potential difference between the

 

two plates of capacitor is given by

 

V ‘ = E (d t) + Et = E (d t) +

 

E . t K

 

Þ            æ                 t  ö       s  æ

 

t ö        Q   æ                 t ö

 

V ‘ = E ç d – t K ÷ = e   ç d – t K ÷ = Ae   ç d – t K ÷

è                    ø        0 è                     ø            0 è                    ø

Now capacitance of the capacitor

 

C‘ = Q

V

Þ       C‘ =

e 0 A d t + t

K

 

 

 

 

 

C‘ =                                             e0 A                                          

C‘ =

e 0 A

 

 

æ t1      t2       t3                 ö

æ t1       t 2         t 3         t 4 ö

 

d – (t1 + t2 + t3 + ……..) + ç K   + K   + K   +…….. ÷

ç        +         +         +         ÷

K        K         K         K

 

è   1         2         3               ø                                    è   1          2          3          4 ø

  • When a metallic slab is inserted between the plates :

 

 

 

Capacitance

C‘ =   e 0 A

(d t)

C‘ = ¥

(In this case capacitor is said to be short circuited)

 

  • When separation between the plates is changing : If separation between the plates changes then it’s

 

capacitance also changes according to

C µ 1 . The effect on other variables depends on the fact that whether the

d

 

charged capacitor is disconnected from the battery or battery is still connected.

 

(i)  Separation is increasing

 

QuantityBattery is removed

 

 

 A

 

 

 d¢

Battery remains connected

 

 

A  

 

 

 d¢

 

V

CapacityDecreases because C µ 1 i.e., C‘ < C

d

Decreases i.e.,   C‘ < C
ChargeRemains constant because a battery is not present

i.e.,                   Q‘ = Q

Decreases because battery is present i.e., Q‘ < Q

Remaining charge (Q Q‘ ) goes back to the battery.

Potential differenceIncreases because V Q Þ V µ 1 i.e., V ‘ > V

C                   C

 V¢ = V (Since Battery maintains the potential

difference)

Electric fieldRemains constant because E s      Q 

e 0      Ae 0

i.e.,      E‘ = E

Decrease because E    Q       Þ E µ Q

Ae 0

i.e.,                     E‘ < E

 

Energy

Increases because U = Q 2  Þ  U µ 1

2C                    C

i.e.,                   U‘ > U

Decreases because U = 1 CV 2 Þ U µ C

2

i.e.,                      U‘ < U

 

  • Separation is decreasing

 

QuantityBattery is removedBattery remains connected
CapacityIncreases because C µ 1 i.e., C‘ > C

d

Increases i.e.,    C‘ > C
ChargeRemains constant because battery is not present

i.e.,             Q‘ = Q

Increases because battery is present i.e., Q‘ > Q

Remaining charge (Q‘ – Q) supplied from the battery.

Potential differenceDecreases because V Q Þ V µ 1 i.e., V ‘ < V

C                 C

 V¢ = V     (Since

difference)

Batterymaintainsthepotential
Electric fieldRemains constant because E s      Q 

e 0      Ae 0

i.e.,      E‘ = E

Increases because E =      Þ  E µ Q

Ae 0

i.e.,                    E‘ > E

 

Energy

Decreases because U = Q 2  Þ  U µ 1

2C                     C

i.e.,       U‘ < U

Increases because U = 1 CV 2 Þ U µ C

2

i.e.,                       U‘ > U

 

 

 Force Between the Plates of a Parallel Plate Capacitor.

 

Field due to charge on one plate on the other is

E = s , hence the force

2e 0

F = QE

 

 

æ s ö       s 2

 

F = –sA ´ ç

è 2e

Þ                        | F |= s 2 A =

÷ = –        A

0 ø       2e 0

Q2

 

2e 0     2e 0 A

Energy Density Between the Plates of a Parallel Plate Capacitor.

The energy stored in a capacitor is not localised on the charges or the plates but is distributed in the field. And as in case of a parallel plate capacitor field is only between the plates i.e. in a volume (A× d), the so called energy density.

 

 

 Energy

CV 2

1 ée

A ù V 2

1    æ V ö 2    1

 

Hence Energy density =

= 2       =      ê       ú

=    e 0 ç     ÷   =

e 0 E 2 .

 

   0

Volume

Ad           2 ë d   û Ad

2    è d ø       2

 

 

Example: 109 The mean electric energy density between the plates of a charged capacitor is (here Q = Charge on the capacitor and A = Area of the capacitor plate)                                                                               [MP PET 2002]

 

 

(a)

Q 2

2e 0 A 2

 

     Q    

2e 0 A 2

 

Q2

 

2e 0 A

  • None of these

 

 

Solution: (a)     Energy density =

e 0 E 2 =

æ          ö            Q2

1
Q
2
=

e 0 ç           ÷                       .

 

 

2             2     è Ae 0 ø

2e 0 A2

 

Example: 110 Plate separation of a 15m F capacitor is 2 mm. A dielectric slab (K = 2) of thickness 1 mm is inserted between

the plates. Then new capacitance is given by                                                       [BHU 1994, Similar to BHU 2000]

 

(a)

15m F

(b)

20m F

(c)

30m F

(d)

25m F

 

Solution: (b)      Given C = e 0 A = 15m F

d

……..(i)

 

 

 

Then by using C‘ =      e0 A      =                   e0 A                

= 2 ´ e0 A ´ 103 ; From equation (i) C‘ = 20m F.

 

d t + t  

K

2 ´ 103 – 103 + 10-3       3

2

 

Example: 111 There is an air filled 1 pF parallel plate capacitor. When the plate separation is doubled and the space is filled

with wax, the capacitance increases to 2 pF. The dielectric constant of wax is                                            [MNR 1998]

(a) 2                                  (b) 4                                (c) 6                            (d) 8

Solution: (b)      Given that capacitance C = 1 pF

After doubling the separation between the plates C‘ = C

2

and when dielectric medium of dielectric constant k filled between the plates then C‘ = K C

2

 

According to the question, C‘ = K C = 2

2

Þ     K = 4.

 

Example: 112 If a slab of insulating material 4 ´ 10 -5 m thick is introduced between the plate of a parallel plate capacitor, the distance between the plates has to be increased by 3.5 ´ 10-5 m to restore the capacity to original value. Then the dielectric constant of the material of slab is                                                                                               [AMU 1999] (a) 10            (b) 12                                       (c)  6                                (d) 8

 

Solution: (d)      By using

K =    t   ; here t = 4 ´ 105 m ; d‘ = 3.5 ´ 105 m Þ

t d

K =               4 ´ 10 -5            = 8

4 ´ 10 -5 – 3.5 ´ 10 -5

 

Example: 113 The force between the plates of a parallel plate capacitor of capacitance C and distance of separation of the plates d with a potential difference V between the plates, is                                                              [MP PMT 1999]

 

 

(a)

CV 2                                       (b)

2d

C 2 V 2

2d 2

(c)

C 2 V 2

d 2

(d)

V 2d C

 

 

Solution: (a)      Since F =

Q2

 

2e 0 A

Þ F = C V 2

2e 0 A

CV 2 .

2d

 

Example: 114 A capacitor when filled with a dielectric

K = 3

has charge

Q0 , voltage V0

and field

E0 . If the dielectric is

 

replaced with another one having K = 9, the new values of charge, voltage and field will be respectively

 

 

(a)

3Q , 3V , 3E

(b)

Q  , 3V  , 3E

(c)

Q  , V0 , 3E

 

(d)

Q  , V0 , E0

 

 

 

0        0         0

0        0         0

0     3        0

0    3     3

 

Solution: (d)     Suppose, charge, potential difference and electric field for capacitor without dielectric medium are Q, V and E

respectively

 

With dielectric medium of K = 3

Charge Q0 = Q

With dielectric medium of K = 9

Charge Q‘ = Q = Q0

 

Potential difference V0 = V

3

Potential difference V ‘ = V

9

V0 

3

 

 

Electric field

E   = E

Electric field E‘ = E = E0 .

 

 

0     3                                                  9      3

 

 

Example: 115 A slab of copper of thickness b is inserted in between the plates of parallel plate capacitor as shown in the

 

figure. The separation between the plates is d. If

b = d

2

then the ratio of capacities of the capacitor after and

 

before inserting the slab will be                       [IIT-JEE 1976; Similar to Orissa JEE 2002, KCET 2001, MP PMT 1994]

 

(a)        : 1                     (b) 2 : 1                           (c) 1 : 1                        (d) 1 :

 

Solution: (b)      Capacitance before inserting the slab C = e 0 A

d

and capacitance after inserting the slab C‘ = e 0 A .

d t

 

Where t = b = d

so C‘ = 2e 0 A

hence,

C‘ = 2 .

 

2                d                      C       1

 

Example: 116 The capacity of a parallel plate condenser is C0 . If a dielectric of relative permitivity e r

and thickness equal to

 

one fourth the plate separation is placed between the plates, then its capacity becomes C. The value of C

C0

will be

 

 

(a)

5e r

 

4e r + 1

(b)

   4e r

3e r + 1

(c)

   3e r

2e r  + 1

(d)

  2e r

e r + 1

 

 

Solution: (b)      Initially capacitance C0

e 0 A d

……..(i)    Finally capacitance C =           e 0 A       

d – dd / 4

……..(ii)

 

 

 

By dividing equation (ii) by equation (i)

 C  =   4e r 

4      e r

 

C0        3e r + 1

 

 

 

Grouping of Capacitors.

Series grouping                                                           Parallel grouping

 

(1) Charge on each capacitor remains same and equals to the main charge supplied by the battery

Q
1

C1                   C2                 C3

  • otential difference across each capacitor remains same and equal to the applied

 

+Q1

+

+  –

+

 

Q1

 

+Q         Q              +Q         Q               +Q         Q

                                                             +  –               +  –                + –

potential difference                                +

 

+  –               +  –                 + –

+Q2        –Q2

 

V = V

+  –               +  –                +  –                                                                                            +  –

Q

+ V  + V                       +  –               +  –                 +  –                                                                                            +  –

 

1         2         3

V1                  V2                   V3

+  –

V

Q = Q1

+ Q2

+ Q3

Q2

 

Q                   Q3

+

+

+Q3

+

+

Q3

 

  • Equivalent capacitance + –

+  –

 

1

Ceq

= 1 + 1 + 1 or

C1      C2      C3                                                                                                                                                                         V

 

  • In series combination potential difference and energy distribution in the reverse ratio of capacitance e.,
  • Ceq = C1 + C2 + C3

 

V µ 1

C

and U µ 1 .

C

  • In parallel combination charge and energy distributes in

the ratio of capacitance i.e. Q µ C and U µ C

 

  • If two capacitors having capacitances C1 and C2 are

connected in series then Ceq = C1C2  = Multiplication

  • If two capacitors having capacitance C1 and C2

respectively    are     connected     in     parallel     then

 

C1 + C2         Addition

Ceq = C1 + C2

 

æ     C1       ö

and

æ     C2        ö

 

V1 = ç C   C   ÷. V               V2  = ç C   C   ÷. V

and

 

è   1         2 ø                              è   1         2 ø

 

  • If n identical capacitors each having capacitances C are connected in series with supply voltage V then

(5) If n identical capacitors are connected in parallel

 

n

Equivalent    capacitance    Ceq   = C         and    Potential

Equivalent capacitance Ceq   = nC and Charge on each

Q

 

difference across each capacitor V ‘ = V .

n

capacitor Q‘ = n

 

 

Redistribution of Charge Between Two Capacitors.

When a charged capacitor is connected across an uncharged capacitor, then redistribution of charge occur to equalize the potential difference across each capacitor. Some energy is also

wasted in the form of heat.

 

Suppose we have two charged capacitors C1

and C2

after disconnecting

 

these two from their respective batteries. These two capacitors are connected to each other as shown below (positive plate of one capacitor is connected to positive plate of other while negative plate of one is connected to negative plate of other)

       æ     C1        ö             æ     C2        ö

Charge on capacitors redistributed and new charge on them will be Q1 = Q ç C   C   ÷ , Q2 = Q ç C   C   ÷

 

The common potential V = Q1 + Q2

C1 V1 + C2 V2

è   1

and loss of energy DU =

2 ø

C1C2

è   1         2 ø

1
2

(V V )2

 

C1 + C2

C1 + C2

2(C1

+ C2 )

 

Note : @Two capacitors of capacitances C1 and C2 are charged to potential of V1 and V2 respectively. After disconnecting from batteries they are again connected to each other with reverse polarity i.e., positive plate

 

of a capacitor connected to negative plate of other. So common potential V = Q1 – Q2

C1 + C2

 

C1 V1 – C2 V2  .

C1 + C2

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