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 − ), the net work done by the cell is =∫
=∫
=
= Note: Whenever cell is connected at higher potential, it can absorb Electrical Energy. Hence in this case. =− Charge flows from ‘+ve’ to ‘-ve’ in the cell, opposite to EMF, hence -ve. V (0, )
O
q
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 Potential
V
Charge
∝
=
Q
= ⁄ is
called the capacitance of any object. We have already seen relation for all types of Electric field (s)/ and potential. So by
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. Q R
Its potential
=
hence its capacitance
=
=4
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.
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 + ++ + -Q + Q ++ + + + 2 + 1 ++ + + + + + + ++ + EE+
=
=
=
=
Since both of them are in same direction hence =
+
=
+
=
Since potential at (2) is zero due to earth so. On plate (1) let it be −
∫
= −∫ =− ∫
E = constant =
=
.[ ]
hence
=
=
for Air
=
for dielectric
Note: For isolated PPC Q1
Q2
E1
E1
E2
11
21
=
−
= 2]
E2
From figure within the plates =
−
hence
= =
−
=
= =
⁄ ⁄
=
Spherical capacitor: It is an assembly of two concentric shells (radius a and b) of dissimilar radii (a
+Q
+ + + + +
+ + +
+b + + + a ++ + ++ + + +
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 =
Now inner shell is at higher potential due to +ve charge while outer is at zero potential due to earthing. hence = ∫ ⃗ . ⃗
∫ −
=−
⃗ is parallel to ⃗
∫
=
hence capacitance is Case-1:
= ∞
=
=
⟹
=
=
−
=
= 4
Same as spherical shell (when isolated) Case-2:
≈
−
=
=
=
=
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 Qa b a
Qa b
b
= =
as seen in charge distribution in electrostatics, So +
Where
=
like spherical capacitor
=4
like
isolated sphere hence = 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. + + + + + + + + + + + + + + +
q +
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 + ++ + +b + + + + a ++ + +
=
Again potential on inner shell is ‘V’ and outer is zero due to earthing = −∫
∫
−
=
=
∫
ln( / ) hence 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 1 +Q
l
=
+
=
+
1 a -Q + + + + E+ + +x Ed + + + + + + d
(
)
Now +ve charge wire is at potential V and ‘-ve’ charge wire is at potential zero due to earthing. hence ∫
=∫
.
+
=+
∫
+
[ln − ln( − )
= =
=
≈
]=
ln
− ln
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. Q2
Q1 Q3
Q4
Q 1+ Q 2 = Q 3 + Q 4 [Independent of other part of circuit] 2]
Kirchoff’s Loop Law: The second law is based on conservation of Energy
=
. 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 + (
) and vice verse.
b)
Through a capacitor, charge passes from ‘+ve’ to ‘-ve’ terminal. So in directions of charge the potential difference is ‘-ve’ and vice verse. -Q/C + Q
C
+Q/C
c)
Similarly for a resistance, current passes from ‘+ve’ to ‘-ve’ direction. So potential difference in direction of current is ‘-ve’ and vice verse. -IR R
I
-
+ +IR
for ‘capacitor’ only circuit, use the following example + C1 -
A Q1 + +
4
1 -
B +
C4
+
Q4 +
D
Q3
+
3
C2 +Q2
2
C
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. −
+
+
−
−
+
+
−
=0
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 and same charge passes through all the capacitors. So for the complete loop: = +
Q
+
+ C1
−
−
−
=0⟹
=
C2
+
C3
+
but net capacitance is or ⁄ but So
=
+
+
=
+
+
= for two capacitors
=
combination Note: Series means same charge through single path.
for series
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 . A D F H
Q
B
+ Q 1+ C 1
C
Q 2+ C 2
E
Q 3+ C 3
G
for loop ABCD −
=0
⟹
=
for loop ABEF −
=0
=
for loop ABGH −
=0
=
But you can also notice the charge Q is being divided in and
.
hence so
,
= =
+
+
+
=
+
+
+
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.
+Q + + + + + + + E
-Q - QP + -
+ Ed
+
-
0
In the above figure you can see that electric field in air (E0) is higher than that in dielectric (
) due to polarisation of charge
. Following the derivation of capacitance Electric field in air
=
Electric field in dielectric edges −
−
(charge on the dielectric
)
But field in dielectric So
=
= =
⟹
= =
= 1−
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.
=
+
+
=
+
+
=
Case – 1 b) A dielectric of thickness t(
K
t d
So =
+
=
+
.
=
Case 1 (c) If we replace dielectric with conductor. Case 1 (d) If t = d
=
=∞
Case 1 (e) If plates are connected by a conductor
=∞
=
Case 2: Filling of dielectrics parallel to separation. + K 1 + + K2 + + K3 + d As you can see in figure the dielectrics will behave as combination in parallel. If area associated with the dielectrics is
,
,
=
+ +
4]
respectively. +
=
+
+
=
(
+
)
Some special combinations:-
Combination – 1. Q1
Q2
3
2
1 S1
S2 d1
Case A: If and
Q3
d2
is earthed all charges will get induced towards plate 1 will go to earth.
So
=0 = =
(
= (
+
) (
)
)
=
=
Case B: Similarly closing
,
+
will go to earth and net charge will
induce towards plates 3.
Sol: Again
=0
= =
= +
=
+
Case - C: If both the a switches earth and
,
are closed,
will be divided -q
q
Q-q 2 2
1
3 S2
S1 d1
-(Q-q) 2 d2
and
will go to
Sol: = 0, ⟹ ( =
= 0 and +
=
⇒
=
=
)=
with plate 1 and
−
Case: D If Plate 2 is earthed,
=
with plate 3
will go to earth and induction will
be towards plate 2 = 0;
=
;
=
Q1
Q1
d1
Q3 Q3
d2
Note: In all the cases net charge transferred to earth is ( + + ). Combination – 2: This is a very interesting case. If you have four plates (or more), you can have four types of different combinations
+
1
(A)
3
2
4
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. ⟹
= ( − 1)
plates, (n–1) capacitors in parallel
B] Here you see 2 capacitors in series hence 2
1
2
3
4
2
1
= 4
3
3
C] Here you see division of charge at plate 2 and 3 with charge at plate-1 is pushed to plate -4.
Rearranging will make idea more clear. Now two capacitors in series =
with one in parallel hence
= . D) Here you can see again charge dividing at plate-2 and 3 but transferred from 1 to 3. Rearrangement will make it clear. 2
1
2
3
4
2
1
3 3
Now you have two capacitors parallel series,
4
=
and one in
=
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.
K2
q K1 + + +
+ + a + + + + + b+ C
+ + + + +
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. q + + + ++ + q K1 + + + + A q K1 + + + + q+ + C + + K2 + + + b a + + + + + + ++ + + ++ + + + +
=
=
.
=
(
)
(
)
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
= K2
q K1 q + + + + + + + ++ +++
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
= +
+
+ + +q + + +
+ + + +
C
+
+q
+ + K1 K2 + ++ + + + + b
a
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’ = induce −
. which will
on inner most shell at its outer surface. Remaining
will lie on outer surface of middle shell causing induction to outer most shell. + + + + + + + + + a + q b-a + ++ + + + + qb- b +qa + b + + K + b+ + ++K ++ + C b + + ++ + a + ++ + + + qa + + b+ + + + ++b-a+q+ + + + b + + + +
2
1
Now outermost shell will also behave as isolated sphere, having
its
capacitance = 4
. This will remain in series with outer
spherical capacitor of capacitance
. Combination of
these two is in parallel with inner spherical capacitor of capacitance 40C
4 0 bck 2 cb
4 0 k1ab ba
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 =
+
Because charge distributes at middle shell and remaining both the shells are at (zero) same potential. q b-a
+ + +b + ++ + +q b-a+ + q b-a + b + b + + ++ ++ + + b C K+ K + + a + + + q a + + + + + +b + + + + + 1
2
Case E: Isolated Parallel Plate Capacitor Electric field between the plates E1 A
E2
E1
A E2
E2 FluxQ2
=
−
⇒
have
=
=
>
= =
d
E1 Q1
−
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. V
Q
Hence Energy stored in electric field within a capacitor is U = Area under graph =
=
= (since Q = CV) =∫ =∫ 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, =
=
=
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 and charge flown is Q, work done on battery is =−
(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. C1
C2 V2
V1 +
V1 C1
-
-
+ C2 V
We can join two capacitors at voltages &
&
of capacitance
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 =
and
=
⇒
=
and
=
After redistribution of charge (Let the common potential reached be V) = =
⇒
= ⇒
=
Since two capacitors will always make parallel combination only.
Hence
=
Therefore
+ =
⇒
+
=
=
Loss of energy = Δ =
−
=
+
=
=
+ −(
+
−
− +
) =
=
.
[
]
Redistribution of charge on two capacitors will always cause loss of Energy in form of heat, light and sound (sparking). Note: C1
C2 V2
+ V1 +
C1V1
-
-
+ C2 V2
a]
If polarity is opposite 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 and latter connected to another cell of EMF Hence charge on capacitor
=
Energy stored in electric
=
Now connection to second cell can be of two types. a) Same polarity
New charge is
=
New Energy stored Work done by cell
= = Charge flown ×
Now Heat generated is =
+
=
−
−
b) Opposite Polarity
=
⇒
Charge =
=
flown
=
+
to
+
Heat generated =
+
−
charge
polarity
hence
Case – B Inserting dielectric (By filling the total space between the plates) a)
Cell is connected Before inserting
=
= After inserting
=
⇒
Heat produced = =
−
+
+
= −
−
Note: Potential difference across the plates remains same. b)
Cell disconnected after charging Before inserting After inserting =
= =
;
=
=
=
=
heat Produced =
−
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) and it is connected to another capacitor each plate.
. Find charges on
QO + + + +
+ + + +
C1
Hence after connections are made we can apply Kirchoffs law (
)
For
− =
(
)
=
=0⇒ ⇒
=
find ‘q’ by solving
−2 =2 ⟹
=4 ⇒
=
Case -B: + + +
C2
C1
+ + + +
C3
S1
S2 +
i)
If
,
both are open(as above)
&
are in series Apply
kirchoffs Law and find q q 1C1 q 1 + + + + + +
q1
q2 q2
q1+q2C2
q2 C3
+ + + + + +
-q1-q2 S2
S1
q1 +q2
+ -
ii)
If
is closed (as above),
combination in series with iii) Similarly is if
is closed
combination is in series with
and
are in parallel and
. Apply kirchoff’s law and
are in parallel and
. Apply Kirchoff’s Law.
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
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.
+Q
-Q
A
A
d
Capacitance Energy stored
=
=
=
Hence energy density =
=
×
=
=
is charge density ⟹
=
for capacitor
So Energy density (u) = This we use to find Energy distribution of sphere and shell in Electrostatics. Note: The relation
=
is of course derived for uniform PPC
but it is true for all distributions. GENERAL FORM = ⟹ where dV is elementary volume not potential difference. ⟹ Electric field and force between the plates.
⟹ 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 Q
-Q
E
=
+
=
Hence force on − F = QE
charge plate due to ‘ +
=
=
’ charge plate is
(attraction due to unlike charges) Note: Pulling the plates of capacitor apart with isolated capacitor carrying charge Q (
=
)
−
(
<
)
Work done by external force =
(
−
)
>
= Increase in PE of capacitor Note: Pulling the plates of capacitor apart when the capacitor is connected
to
the
(V= EMF of battery) =
=
=
.
.
=
battery
Work done by external force = =
1 2
1
= (
=
=∫
1 2
1
−
. 1
− <
>
)
Decrease in P.E. of capacitor =
Work done on the battery = ) =
=
−
= (
−
−
Conservation of energy in proved Work done by external force + decrease in PE of capacitor = Work done on the battery + 7]
=
Force on dielectric slab.
Case A: Slab being pulled in by capacitor connected to cell. A
+
K
F
+++
d
-
A
x
l
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
+Q
+ + + + + + + +
-Q
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 =
=
[( − 1) + ]
If Slab is pulled by a further distance dx, Change in Capacitance ( − 1)
=
Now using work energy theorem = ∫ +
=
.
↓ ↓ work done by Battery ↓ Change in PE Work done by force applied to hold the slab − .
+
⟹− .
= =−
=−
- ve sign means ⟹
=
(
.
( − 1)
is opposite to force on slab )
independent of ‘x’ hence constant
Case B: If capacitor is not connected to cell. ( − 1)
Again dC = =
[ ]
and (k-1)dx and
[( − 1) + ]
= −
= =
[(
(
)
) [(
)
]
]
=
( [(
not a constant
) )
]
