CAPACITORS

SESSION – 1, 2 AND 3 AIM To introduce the Concept of Battery. To introduce Concept of Capacitor and term capacitance along

with properties of capacitor To introduce finding the capacitance of all type of capacitors THEORY: 1) BATTERY: Battery is a combination of cells (will be discussed

in “current Electricity”) What do you think a battery can do? a) Is it a source of charge? b) Is it a source of Electrical Energy? c) Is it the source of both of them? Out of these three questions only ‘b’ is correct i.e. battery is the

source of Electrical Energy which can flow the charges across a closed circuit because battery maintains a fixed potential difference across its terminal (‘–ve’ and ‘+ve’). The charge always moves from higher (‘+ve’) to lower (‘–ve’) potential. The electrical energy actually moves the charge from lower potential to higher potential (‘-ve’ to ‘+ve’) which is exactly opposite to the direction of flow outside the cell, but it makes a closed circuit.

Noteworthy point here is closed circuit. You can see the figure

above; when switch of the cell is open, the circuit cannot produce current to glow the bulb, but current flows only when switch is closed and the bulb starts glowing.

WORK DONE BY CELL: Since cell supplies all the charge at constant voltage (EMF or

Electromotive force −휖), the net work done by the cell is

푊 = ∫ 푉푑푞 = ∫ 휖푑푞 = 푄휖

푊 = 푄휖 Note: Whenever cell is connected at higher potential, it can absorb

Electrical Energy. Hence in this case. 푊 = −푄퐸 Charge flows from ‘+ve’ to ‘-ve’ in the cell, opposite to EMF,

hence -ve.

O

V

q

(0, )

Capacitor: Do you know our body can also store charge. This is a very

common phenomenon but might have missed being observed by you. If you work for a while sitting on a plastic chair wearing synthetic or woolen cloths your body gets slightly charged. Now, if you touch a conductor you get a tickling effect. This effect is produced due to transfer of charge, between your body and conductor.

What did you learn here. You can also store some charge. Is there a limit of charging? With increase of charge, potential and Electric field will also increase. When the maximum electric field near any surface exceeds dielectric strength, break down of dielectric starts and ionization of medium occurs which keeps the charge at the highest possible limit, to sustain the dielectric medium.

Relation between charge and potential: since potential is directly proportional to charge

푄 ∝ 푉 푄 = 퐶푉 퐶 = 퐶 푉⁄ is called the capacitance of any object. We have already seen

relation for all types of Electric field (s)/ and potential. So by

QPote

ntia

l

V

Charge

finding potential due to a charge on a body we can determine the capacitance.

Capacitance: Capacitance of any body is its ability to store the charge. We can determine it as the ratio of charge to the potential of the body (charge per unit voltage).

We can use this concept for a shell. CAPACITANCE OF SHELL Let charge ‘Q’ be given to a spherical shell of radius R.

Its potential 푉 =

hence its capacitance 퐶 = = 4휋휖 푅

Unit is Farad, Practical unit µF, and pF. Capacitor: Capacitors are special devices designed to store the

charge. Usually a capacitor is designed by two sheets, separated by some (distance) dielectric medium. These devices when are in a closed circuit or have one of their terminals at Earth work as capacitor else treated as isolated capacitor.

TYPES OF CAPACITORS: 1. Parallel plate capacitor (PPC): Parallel plate capacitor is

formed by using two sheets of area ‘A’ separated by a separation ‘d’ with or without dielectric.

R

Q

Whatever charge given to plate1 charge will be induced on inner side of plate 2 only due to earthing of plate 2.

So charge density of plates is and hence field near the

surfaces

퐸 = = 퐸 = =

Since both of them are in same direction hence 퐸 = 퐸 + 퐸

= + =

Since potential at (2) is zero due to earth so. On plate (1) let it

be 푉 ∫ 푑푣 = −∫ 퐸푑푟

푂 − 푉 = −퐸 ∫ 푑푟 E = constant 푉 = . [푟]

푉 =

+++ +

+++++++++ +

++++++++

Q

1

-Q

2

E+E+ E-

hence 퐶 = = for Air 퐶 = for dielectric

Note: For isolated PPC

From figure within the plates

퐸 = 퐸 − 퐸 = − 퐸 = − 휎 = 푄 퐴⁄휎 = 푄 퐴⁄

퐸 = hence 푉 = 퐸푑 = 푑 푉 =

2] Spherical capacitor: It is an assembly of two concentric shells (radius a and b) of dissimilar radii (a<b) separated by air or dielectric.

As we have seen earlier that if we give charge ‘+Q’ to the inner

shell and earthing the outer, total induction on inner surface of outer shell is ‘-Q’.

Now field within shell can be given as per Gauss theorem as

퐸 =

Q1 Q2

11 12

E1

E1

E2

E2

+Q

++++

+++++

+ ++++++

++ + +++

b

a

-Q

Now inner shell is at higher potential due to +ve charge while outer is at zero potential due to earthing.

hence

∫ 푉 = ∫ 퐸⃗.푑푟⃗ 퐸⃗ is parallel to 푑푟⃗

푂 − 푉 = − ∫ 푉 = ⟹ 푉 = −

hence capacitance is 퐶 = = =

Case-1: 푏 = ∞ 퐶 = = 4휋휖 푎

Same as spherical shell (when isolated)

Case-2: 푎 ≈ 푏 푏 − 푎 = 푑 퐶 = = =

nearly same as parallel plate capacitor Case-3: If earthing is done at inner surface with outer surface

having charge Q. The charge induced on inner shell is

푞 = as seen in charge distribution in electrostatics, So

퐶 = 퐶 + 퐶

bQa

bQa

a

b

Where 퐶 = like spherical capacitor 퐶 = 4휋휖 푏 like

isolated sphere hence 푐 =

Note: Distribution of charge on outer shell leads to parallel combination (discussed in Electrostatics)

Cylindrical Capacitors: It is assembly of linear conductor surrounded by cylindrical shell, having dielectric separating the two. Example Dish TV cable.

Radius of inner conductor is ‘a’, outer shell radius is ‘b’. If

charge ‘Q’ is distributed over inner conductor (length l) and outer is earthed, charge on outer shell will be -Q (same concept as spherical shells)

Using Gauss law again we can write the field within the shells as

퐸 =

++++++++++++++

+ +q

a

++++++ + ++

++

+ + b

Again potential on inner shell is ‘V’ and outer is zero due to earthing

∫ 푑푉 = −∫ 퐸푑푟 푂 − 푉 = ∫ 푉 =

ln(푏/푎)

hence 퐶 = = ( / ) ≈ ( / ) 푝퐹

4. Two linear conductor separated by a fixed distance: It is an assembly of two thin conductors separated by a fixed

distance by some means (like dielectric). Example T-V Antenna wire.

In this example we consider two wires of infinite length and radius ‘a’ separated by a distance ‘d’ (d>>>a).

Using application of Gauss theorem. Electric field at a distance from charged wire

퐸 = 퐸 + 퐸 = + ( )

Now +ve charge wire is at potential V and ‘-ve’ charge wire is at potential zero due to earthing.

hence ∫ 푑푉 = ∫ 퐸.푑푟 푂 + 푉 = + ∫ + 푑푟

++++++++++++

1 1

x

E+

a

Ed

d

l

+Q-Q

푉 = [ln 푟 − ln(푑 − 푟) ] = ln − ln

퐶 = = ≈

SESSION – 4, 5, 6 AND 7 AIM To introduce kirchoff’s Law for capacitors To introduce concept of series and parallel combination To introduce concept of filling of dielectrics Examples of special combination of capacitors THEORY: Kirchoff’s Law: 1] Kirchoff’s Junction Rule: The first law is based on the

principle of conservation of charge. The net incoming charge (current) at a junction must always equal to net outgoing charge (current) at the junction.

Q1+ Q2 = Q3 + Q4

[Independent of other part of circuit] 2] Kirchoff’s Loop Law: The second law is based on conservation

of Energy 푝표푡푒푛푡푖푎푙푑푖푓푓푒푟푒푛푐푒 = . The sum of potential

differences along a loop is always zero. To consider the potential difference following rules are to the

followed a) If we pass from ‘-ve’ to ‘+ve’ terminal of a cell, potential

difference is +휖(퐸푀퐹) and vice verse.

Q1Q2

Q3 Q4

b) Through a capacitor, charge passes from ‘+ve’ to ‘-ve’ terminal. So in directions of charge the potential difference is ‘-ve’ and vice verse.

c) Similarly for a resistance, current passes from ‘+ve’ to ‘-ve’

direction. So potential difference in direction of current is ‘-ve’ and vice verse.

for ‘capacitor’ only circuit, use the following example

Consider the above shown loop. Take charge through all the

capacitors. Define polarity of capacitors (charge flows from ‘+ve’ to ‘-ve’). Now use the law starting from A.

− + 휖 + − 휖 − + 휖 + 휖 − = 0

CQ+

+Q/C

-Q/C

I+

R

-

-IR

+IR

A +Q1

C1 -- +

1 B

C4+C2

Q4+

+ +Q2

+Q3

+CD

4

1

2

3

Note: Once you assign polarity of capacitors (Polarity of cell(battery) is fixed; bigger arm ‘+ve’ smaller arm ‘-ve’), you can see that the potential difference takes the sign of the cell’s terminal through which we come out; same for capacitor (or resistance).

Using these two rules we can discuss combinations of capacitors.

2] Combination of Capacitors: a) Series Combination: Let three capacitors be connected

through a cell of EMF 휖 and same charge passes through all the capacitors. So for the complete loop: =

휖 − − − = 0 ⟹ 휖 = + +

but net capacitance is or 휖 푄⁄ = + +

but 퐶 =

So = + + for two capacitors 퐶 = for series

combination Note: Series means same charge through single path.

+

Q +

+

C1 C2 C3

b) Parallel Combination: Let three capacitors be connected in parallel combination through a cell. You can notice that all the three capacitors here are connected across AB i.e. same potential difference 휀.

for loop ABCD 휀 − = 0 ⟹ 푄 = 퐶 휀

for loop ABEF 휀 − = 0 푄 = 퐶 휀

for loop ABGH 휀 − = 0 푄 = 퐶 휀

But you can also notice the charge Q is being divided in 푄 ,푄 and 푄 .

hence 푄 = 푄 + 푄 + 푄 푄 = 휀퐶 + 휀퐶 +휀퐶

so 퐶 = 퐶 + 퐶 + 퐶 Note: Parallel combination means same potential difference across

all capacitors.

3] Filling the dielectrics: - As discussed in electrostatics, dielectrics tend to reduce the electric field. Dielectric reduces the electric field due to polarisation, which causes induction of charge. Since induced charge is opposite in nature, electric field within dielectric is lower than that in air.

QA + B

Q1+ C1

CD

Q2+ C2

EFQ2

+ C2EF

Q3+ C3

GH

In the above figure you can see that electric field in air (E0) is

higher than that in dielectric (퐸 ) due to polarisation of charge 푄 . Following the derivation of capacitance Electric field in

air 퐸 =

Electric field in dielectric 퐸 = (charge on the dielectric

edges −푄 )

But field in dielectric 퐸 =

So 퐸 = ⟹ =

푄 − 푄 = 푄 = 1 − 푄 is the charge induced due to

polarisation. Note: For all type of dielectrics, the method of polarisation is same.

The difference in method of polarisation of polar and non polar dielectrics do not change the above concept.

Case- 1: a) Parallel to plates filling of dielectrics As you can see in the figure, all the dielectrics have same

charge passing through them due to single path, hence we have capacitors in series. If plate area is A.

+Q -Q+++++++

E0

-

------

Ed

+

+

+-

- QP

= + + = + + 퐶 =

Case – 1 b) A dielectric of thickness t(<d) is placed within plates of capacitor.

This can be considered as a combination of air capacitor of thickness (d - t); and dielectric capacitor of thickness ‘t’ and dielectric constant ‘k’.

So = + = +

. 퐶 =

Case 1 (c) If we replace dielectric with conductor. 퐾 = ∞ 퐶 =

Case 1 (d) If t = d 퐶 =

Case 1 (e) If plates are connected by a conductor 퐶 = ∞

++++++++

K

td

Case 2: Filling of dielectrics parallel to separation.

As you can see in figure the dielectrics will behave as

combination in parallel. If area associated with the dielectrics is 퐴 ,퐴 ,퐴 respectively.

퐶 = 퐶 + 퐶 + 퐶 = + + = (퐾 퐴 +

퐾 퐴 + 퐾 퐴 ) 4] Some special combinations:- Combination – 1.

Case A: If 푆 is earthed all charges will get induced towards plate 1

and 푄 will go to earth.

++++++

------

K1

K2

K3

d

Q1 Q2Q3

1 2 3

d1 d2

SS1 S2

So 푉 = 0

푉 = 퐸 푑 = ( ) ( ) 푑 = 푑

푉 = 푉 + ( )푑 푉 = 푑 + 푑

Case B: Similarly closing 푆 , 푄 will go to earth and net charge will induce towards plates 3.

Sol: Again 푉 = 0

푉 = 퐸 푑 = 푑

푉 = 푉 + 퐸 푑

푉 = 푑 + 푑

Case - C: If both the a switches 푆 , 푆 are closed, 푄 and 푄 will go to earth and 푄 will be divided

Q-q2-q

S1

S2

d1

d2

1 2 3

-(Q-q)2

q

Sol:

푉 = 0,푉 = 0 and 푉 = 푉 ⇒ 푉 = 푑 = 푑

⟹ 푞(푑 + 푑 ) = 푄 푑

푞 = with plate 1 and 푄 − 푞 = 푑 with plate 3

Case: D If Plate 2 is earthed, 푄 will go to earth and induction will be towards plate 2

푉 = 0; 푉 = 푑 ; 푉 = 푑

Note: In all the cases net charge transferred to earth is

(푄 + 푄 + 푄 ). Combination – 2: This is a very interesting case. If you have four

plates (or more), you can have four types of different combinations

Q1 Q1Q3

Q3

d1 d2

A] It is the simplest of its kind. Consecutive plates are connected

to different terminals. Try to rearrange them after numbering You see three capacitance in parallel. Hence

so for C = 3. 푛 plates, (n–1) capacitors in parallel

⟹ 퐶 = (푛 − 1)

B]

Here you see 2 capacitors in series hence 퐶 =

C] Here you see division of charge at plate 2 and 3 with charge at

plate-1 is pushed to plate -4.

+

1 2 3 4

(A)

21 3 4

12

2 3

4 3

Rearranging will make idea more clear. Now two capacitors in series

퐶 = with one in parallel hence

퐶 = .

D) Here you can see again charge dividing at plate-2 and 3 but transferred from 1 to 3.

Rearrangement will make it clear.

Now you have two capacitors parallel 퐶 = and one in

series, 퐶 =

Note: Here K is assumed as dielectric constant of the medium filled between the plates.

Combination - 3: Combinations of spherical capacitors (concentric

shells) Case–A: Charge is given to inner most shell and outer most shell is

earthed.

1 2 43

12

2 3

3 4

If you follow lines of electric field, you will see induction is

outwards along the direction of electric field. Charge ‘q’ is going to be induced on each surface as ‘-q’ and ‘+q’ in turn, but on outer most surface net charge is zero due to earthing.

Due to same charge, both the capacitors formed here are in series.

퐶 = =.

퐶 = ( ) ( )

Case: B Now with charge q at inner most shell, middle shell will not

have any charge on it’s outer surface or you can say induction will stop at middle shell only. So outermost shell becomes useless. Hence

++++++

++ + + +

+++

++qK1

b

K2

C

a

A++++

++ + + +

+++

++q K1

bK2

Ca

K1

qq

+ +q + + + ++

+

++++ + + +

++

+

+ ++

++

퐶 =

Note: Even if outer most is connected to earth, no charge will take

place. Case C: Now, let charge ‘q’ be given to the middle shell and outermost

shell be earthed. In this case induction will come to inner surface of outermost shell only

hence 퐶 =

Case D: The charge ‘q’ be again given to middle shell but inner most

shell be earthed. For this case, division of charge will take place. Inner surface of middle shell will take q’ = 푞. which will

induce −푞 on inner most shell at its outer surface. Remaining

+++++ + + ++

++

+q

K2

K1

q

q

K1

bC

a

+ +

q

+ + + ++

+

+++

+ + + +

++

+ ++

+++

K2

푞 will lie on outer surface of middle shell causing induction

to outer most shell.

Now outermost shell will also behave as isolated sphere,

having its capacitance = 4휋휀 푐. This will remain in series with outer

spherical capacitor of capacitance . Combination of

these two is in parallel with inner spherical capacitor of

capacitance

Now you can easily combine these three capacitors Case E: Again same charge ‘q’ is given to middle shell and both

inner most and outermost shells are earthed. Like case - D induction will take place but on outermost surface charge will be zero. Hence

퐶 = +

Cba

++++

++

+ + + ++ ++

+ ++

+K1

qa+b

qab

++++

K2

+++++ + + q+ + +

++

+ + +qb-ab+++

b-ab

++

++

++

++ + +

++

+

++

++ +

q b-ab

C4 0 bcbck4 20

ababk4 10

Because charge distributes at middle shell and remaining both the shells are at (zero) same potential.

Case E: Isolated Parallel Plate Capacitor Electric field between the plates

퐸 = 퐸 − 퐸 ⇒ 푄 > 푄 = −

have 푉 = 퐸푑 = 푑 =

Cba

+++

++

+ + + ++ ++

+ ++

+K1

+

qab

++++

++++ + + + + +

++

+ + +

++

q b-ab

q b-ab

K2

q b-ab

AE1 E1

E2

A

E2 E1E2

d2 1FluxQ Q

SESSION – 8, 9 AND 10 AIM To determine energy stored in capacitor To determine work done by battery and difference between

charging and discharging of capacitor. To study redistribution of charge and loss of energy. To determine heat generated in the circuit by charging

potential difference To determine new charges developed on capacitor when

circuits are changed. Energy density, Electric field and force between the plates of

capacitor. Force on dielectric slabs 1] Energy Stored in capacitor: Since we know, (V) potential

difference across a capacitor is directly proportional to charge, V-Q graph is linear.

Hence Energy stored in electric field within a capacitor is U = Area under graph

= 푄푉 = 퐶푉

V

Q

= (since Q = CV)

푈 = ∫ 푉푑푞

= ∫ 푑푞 = =

2] Work done by Battery: - As we discussed in session-1 of

Capacitors. Battery is a device that maintains fixed potential difference. Hence during transfer of charge potential difference of battery will remain constant. Therefore,

푊 = 푄푉 = 퐶푉 = were C is the capacitor being charged

and Q is the charge given and V is the E.M.F of the battery. The point to be noted here is that half of the work done by

battery is used in process of charging the capacitor and the remaining half is lost in the form of heat. So, capacitor can store only half of the work done by the battery.

Charging of Capacitor: When charge is given by a cell or battery like in the above case work is done by battery, this is called charging.

Discharging of Capacitor: Now consider a capacitor charged to a potential higher then the cell or battery. Now if we connect the capacitor to the battery some charge will flow from positive to negative terminals of battery. This direction is just opposite to working of battery. So here work is done on the

battery. If EMF of the battery is 휀 and charge flown is Q, work done on battery is

푊 = −푄휀 (negative sign because work is done on the battery ) We will see its application in further sessions. 3] Redistribution of charge & loss of Energy as heat. Case - A Two capacitors only.

We can join two capacitors at voltages 푉 &푉 of capacitance

퐶 &퐶 like shown here in figures. Charge will start to flow from higher potential to lower potential till both the capacitors attain same potential V. Therefore before the connections being made

푄 = 퐶 푉 and 푄 = 퐶 푉 ⇒ 푈 = 퐶 푉 and 푈 = 퐶 푉

After redistribution of charge (Let the common potential reached be V)

푄 = 퐶 푉 ⇒ 푄 = 퐶 푉

푈 = 퐶 푉 ⇒ 푈 = 퐶 푉

Since two capacitors will always make parallel combination only.

V1

C1 C2

V2

+V1 C1-

+ -C2 V

Hence 퐶 = 퐶 + 퐶 ⇒ 푄 + 푄 = 푄 + 푄

Therefore 푉 = =

Loss of energy = Δ푈 = 푈 − 푈 = 푈 + 푈 − 푈 − 푈

= = 퐶 푉 + 퐶 푉 − (퐶 +

퐶 )

= = . [ ]

Redistribution of charge on two capacitors will always cause loss of Energy in form of heat, light and sound (sparking).

Note:

a] If polarity is opposite 푉 = | |

Loss of Energy = [ ]

b. For uncharged capacitor voltage will be zero. Case A: Reconnecting the capacitor to a cell: Let a capacitor of capacitance C be charged by cell of EMF 휀

and latter connected to another cell of EMF 휀 Hence charge on capacitor 푄 = 퐶휀

V1

C1 C2V2+

+V1C1 -

+-C2 V2

Energy stored in electric 푈 = 퐶휀

Now connection to second cell can be of two types. a) Same polarity

New charge is 푄 = 퐶휀

New Energy stored 푈 = 퐶휀

Work done by cell 푊 = Charge flown × 휀 = 푄 − 푄 휀 Now Heat generated is = 푊 + 푈 − 푈 b) Opposite Polarity

푄 = 퐶휀 ⇒ 푈 = 퐶휀

Charge flown = 푄 + 푄 to charge polarity hence

푊 = 푄 + 푄 휀 Heat generated = 푊 + 푈 − 푈

Case – B Inserting dielectric (By filling the total space between the

plates) a) Cell is connected Before inserting 푄 = 퐶휀

푈 = 퐶휀

After inserting 푄 = 퐾퐶휀 ⇒ 푈 = 퐾퐶휀

Heat produced = 푊 + 푈 − 푈

= 푄 − 푄 휀 + 퐶휀 − 퐶휀 푘

Note: Potential difference across the plates remains same. b) Cell disconnected after charging

Before inserting 푄 = 휀퐶;푈 = 퐶휀

After inserting 푉 = = =

푈 = 퐾퐶푉 =

heat Produced = 푈 − 푈 Note: The amount of charge before and after inserting dielectric is

same. Case–A: Charge Q is given to one plate of Isolated capacitor (PPC)

퐶 and it is connected to another capacitor 퐶 . Find charges on each plate.

Hence after connections are made we can apply Kirchoffs law

( ) − ( ) = 0 ⇒ = find ‘q’ by solving

For 퐶 = 퐶 = 퐶 ⇒ 푄 − 2푞 = 2푞 ⟹ 푄 = 4푞 ⇒ 푞 =

Case -B:

i) If 푆 , 푆 both are open(as above) 퐶 &퐶 are in series Apply

kirchoffs Law and find q

ii) If 푆 is closed (as above), 퐶 and 퐶 are in parallel and

combination in series with 퐶 . Apply kirchoff’s law iii) Similarly is if 푆 is closed 퐶 and 퐶 are in parallel and

combination is in series with 퐶 . Apply Kirchoff’s Law.

++++

++++

Q O

C1

+++

C1+

+++

C2

C3

S1 S2+

S1

+++

C1q1

+

q1

q2q1

S2

-q -q1 2

++

+++

C2q1

+

+q2

++q2

q2C3

+-

+q2

q1

Case C:

If any of the switch is open, no change in charge distribution

will take place. If all the switches are closed assume a charge q flowing in

arbitrary direction and apply Kirchoff’s law Here we have assumed 푞 is maximum. Now apply Kirchoff’s Law

− − + = 0

Now you can find charges on each capacitor and their voltages. 6] Energy density, Electric field and force between the plates. Consider a Parallel Plate Capacitor (PPC) of plate area A and

separation d. When the capacitor is having charge Q.

Capacitance 퐶 =

Energy stored 푈 = =

Hence energy density

푢 = = × = =

휎 is charge density ⟹ 퐸

퐸 = for capacitor

So Energy density (u) = 휀 퐸

This we use to find Energy distribution of sphere and shell in Electrostatics.

Note: The relation 푢 = 휀 퐸 is of course derived for uniform PPC

but it is true for all distributions. GENERAL FORM

= 휀 퐸

⟹ where dV is elementary volume not potential difference. ⟹ Electric field and force between the plates.

-Q+Q

A A

d

⟹ Here we have to consider electric field of one of the plate to get force on other plate.

Field of positive plate in the vicinity of the plate is

퐸 = =

Hence force on −푣푒 charge plate due to ‘ + 푣푒’ charge plate is

F = QE 퐹 = 퐹 =

(attraction due to unlike charges) Note: Pulling the plates of capacitor apart with isolated capacitor

carrying charge Q

= (푑 − 푑 ) (퐶 < 퐶 )푎푠

Work done by external force

= − (푑 > 푑 )

= Increase in PE of capacitor Note: Pulling the plates of capacitor apart when the capacitor is

connected to the battery (V= EMF of battery)

퐹 = = = . . =

Q -Q

E+

Work done by external force = 푤 = ∫ 퐹.푑푥

=12 휀 퐴푉

1푥 푑푥 =

12 휀 퐴푉

1푑 −

1푑

= 퐶 푉 − 퐶 푉

(퐶 < 퐶 푎푠푑 > 푑 ) Decrease in P.E. of capacitor = 푤

Work done on the battery = 푤 = 푉 푄 − 푄 = 푉(퐶 푉 −

퐶 푉) = 퐶 푉 − 퐶 푉

Conservation of energy in proved Work done by external force + decrease in PE of capacitor =

Work done on the battery 푤 + 푤 = 푤

7] Force on dielectric slab. Case A: Slab being pulled in by capacitor connected to cell.

Electric field at the ends of boundaries of parallel plate

capacitor is not uniform. We only assume is uniform within the plates, away from boundaries. If can be understand to following figure

+

-K

A

F d+ + +

Axl

You can see here, field at boundaries is not uniform. Hence it

induces the charge on a dielectric slab which is even not inside the capacitor. Hence it is pulled in as shown in previous figure.

Capacitance of capacitor here is

퐶 = = [(푘 − 1)푥 + 푙]

If Slab is pulled by a further distance dx, Change in Capacitance

푑푐 = (푘 − 1)푑푥

Now using work energy theorem = ∫ 퐹.푑푥

푑푊 + 푑푊 = 푑푈 ↓↓ work done by Battery ↓ Change in PE Work done by force applied to hold the slab

−퐹.푑푥 + 휀푑푄 = 휀푑푄

⟹−퐹.푑푥 = − = − . (푘 − 1)푑푥

- ve sign means 퐹 is opposite to force on slab

⟹ 퐹 = 휖 ( ) independent of ‘x’ hence constant

++++++++

+Q -Q

Case B: If capacitor is not connected to cell.

Again dC = (푘 − 1)푑푥 and (k-1)dx and

퐶 = [ ] [(푘 − 1)푥 + 푙]

푑푊 = 푑푈

−퐹푑푥 = 푑 [( ) ] = ( )[( ) ]

퐹 = ( )[( ) ] not a constant