Chapter
stringclasses 18
values | sentence_range
stringlengths 3
9
| Text
stringlengths 7
7.34k
|
|---|---|---|
1 | 2090-2093 | (2 35) ] However, for d2 << A, these effects can be ignored in the regions
sufficiently far from the edges, and the field there is given by Eq (2 |
1 | 2091-2094 | 35) ] However, for d2 << A, these effects can be ignored in the regions
sufficiently far from the edges, and the field there is given by Eq (2 41) |
1 | 2092-2095 | ] However, for d2 << A, these effects can be ignored in the regions
sufficiently far from the edges, and the field there is given by Eq (2 41) Now for uniform electric field, potential difference is simply the electric
field times the distance between the plates, that is,
0
1 Qd
V
E d
A
ε
=
=
(2 |
1 | 2093-2096 | (2 41) Now for uniform electric field, potential difference is simply the electric
field times the distance between the plates, that is,
0
1 Qd
V
E d
A
ε
=
=
(2 42)
The capacitance C of the parallel plate capacitor is then
Q
C
=V
=
=εd0A
(2 |
1 | 2094-2097 | 41) Now for uniform electric field, potential difference is simply the electric
field times the distance between the plates, that is,
0
1 Qd
V
E d
A
ε
=
=
(2 42)
The capacitance C of the parallel plate capacitor is then
Q
C
=V
=
=εd0A
(2 43)
which, as expected, depends only on the geometry of the system |
1 | 2095-2098 | Now for uniform electric field, potential difference is simply the electric
field times the distance between the plates, that is,
0
1 Qd
V
E d
A
ε
=
=
(2 42)
The capacitance C of the parallel plate capacitor is then
Q
C
=V
=
=εd0A
(2 43)
which, as expected, depends only on the geometry of the system For
typical values like A = 1 m2, d = 1 mm, we get
12
2
–1
–2
2
9
3
8 |
1 | 2096-2099 | 42)
The capacitance C of the parallel plate capacitor is then
Q
C
=V
=
=εd0A
(2 43)
which, as expected, depends only on the geometry of the system For
typical values like A = 1 m2, d = 1 mm, we get
12
2
–1
–2
2
9
3
8 85
10
C N m
1m
8 |
1 | 2097-2100 | 43)
which, as expected, depends only on the geometry of the system For
typical values like A = 1 m2, d = 1 mm, we get
12
2
–1
–2
2
9
3
8 85
10
C N m
1m
8 85
10
F
10
m
C
−
−
−
×
×
=
=
×
(2 |
1 | 2098-2101 | For
typical values like A = 1 m2, d = 1 mm, we get
12
2
–1
–2
2
9
3
8 85
10
C N m
1m
8 85
10
F
10
m
C
−
−
−
×
×
=
=
×
(2 44)
(You can check that if 1F= 1C V–1 = 1C (NC–1m)–1 = 1 C2 N–1m–1 |
1 | 2099-2102 | 85
10
C N m
1m
8 85
10
F
10
m
C
−
−
−
×
×
=
=
×
(2 44)
(You can check that if 1F= 1C V–1 = 1C (NC–1m)–1 = 1 C2 N–1m–1 )
This shows that 1F is too big a unit in practice, as remarked earlier |
1 | 2100-2103 | 85
10
F
10
m
C
−
−
−
×
×
=
=
×
(2 44)
(You can check that if 1F= 1C V–1 = 1C (NC–1m)–1 = 1 C2 N–1m–1 )
This shows that 1F is too big a unit in practice, as remarked earlier Another way of seeing the ‘bigness’ of 1F is to calculate the area of the
plates needed to have C = 1F for a separation of, say 1 cm:
0
Cd
A
=ε
=
2
9
2
12
2
–1
–2
1F
10
m
10 m
8 |
1 | 2101-2104 | 44)
(You can check that if 1F= 1C V–1 = 1C (NC–1m)–1 = 1 C2 N–1m–1 )
This shows that 1F is too big a unit in practice, as remarked earlier Another way of seeing the ‘bigness’ of 1F is to calculate the area of the
plates needed to have C = 1F for a separation of, say 1 cm:
0
Cd
A
=ε
=
2
9
2
12
2
–1
–2
1F
10
m
10 m
8 85
10
C N m
−
×−
=
×
(2 |
1 | 2102-2105 | )
This shows that 1F is too big a unit in practice, as remarked earlier Another way of seeing the ‘bigness’ of 1F is to calculate the area of the
plates needed to have C = 1F for a separation of, say 1 cm:
0
Cd
A
=ε
=
2
9
2
12
2
–1
–2
1F
10
m
10 m
8 85
10
C N m
−
×−
=
×
(2 45)
which is a plate about 30 km in length and breadth |
1 | 2103-2106 | Another way of seeing the ‘bigness’ of 1F is to calculate the area of the
plates needed to have C = 1F for a separation of, say 1 cm:
0
Cd
A
=ε
=
2
9
2
12
2
–1
–2
1F
10
m
10 m
8 85
10
C N m
−
×−
=
×
(2 45)
which is a plate about 30 km in length and breadth 2 |
1 | 2104-2107 | 85
10
C N m
−
×−
=
×
(2 45)
which is a plate about 30 km in length and breadth 2 13 EFFECT OF DIELECTRIC ON CAPACITANCE
With the understanding of the behaviour of dielectrics in an external
field developed in Section 2 |
1 | 2105-2108 | 45)
which is a plate about 30 km in length and breadth 2 13 EFFECT OF DIELECTRIC ON CAPACITANCE
With the understanding of the behaviour of dielectrics in an external
field developed in Section 2 10, let us see how the capacitance of a parallel
plate capacitor is modified when a dielectric is present |
1 | 2106-2109 | 2 13 EFFECT OF DIELECTRIC ON CAPACITANCE
With the understanding of the behaviour of dielectrics in an external
field developed in Section 2 10, let us see how the capacitance of a parallel
plate capacitor is modified when a dielectric is present As before, we
have two large plates, each of area A, separated by a distance d |
1 | 2107-2110 | 13 EFFECT OF DIELECTRIC ON CAPACITANCE
With the understanding of the behaviour of dielectrics in an external
field developed in Section 2 10, let us see how the capacitance of a parallel
plate capacitor is modified when a dielectric is present As before, we
have two large plates, each of area A, separated by a distance d The
charge on the plates is ±Q, corresponding to the charge density ±s (with
s = Q/A) |
1 | 2108-2111 | 10, let us see how the capacitance of a parallel
plate capacitor is modified when a dielectric is present As before, we
have two large plates, each of area A, separated by a distance d The
charge on the plates is ±Q, corresponding to the charge density ±s (with
s = Q/A) When there is vacuum between the plates,
0
0
E
εσ
=
Factors affecting capacitance, capacitors in action
Interactive Java tutorial
http://micro |
1 | 2109-2112 | As before, we
have two large plates, each of area A, separated by a distance d The
charge on the plates is ±Q, corresponding to the charge density ±s (with
s = Q/A) When there is vacuum between the plates,
0
0
E
εσ
=
Factors affecting capacitance, capacitors in action
Interactive Java tutorial
http://micro magnet |
1 | 2110-2113 | The
charge on the plates is ±Q, corresponding to the charge density ±s (with
s = Q/A) When there is vacuum between the plates,
0
0
E
εσ
=
Factors affecting capacitance, capacitors in action
Interactive Java tutorial
http://micro magnet fsu |
1 | 2111-2114 | When there is vacuum between the plates,
0
0
E
εσ
=
Factors affecting capacitance, capacitors in action
Interactive Java tutorial
http://micro magnet fsu edu/electromag/java/capacitance/
Rationalised 2023-24
Physics
70
and the potential difference V0 is
V0 = E0d
The capacitance C0 in this case is
0
0
0
Q
A
C
V
εd
=
=
(2 |
1 | 2112-2115 | magnet fsu edu/electromag/java/capacitance/
Rationalised 2023-24
Physics
70
and the potential difference V0 is
V0 = E0d
The capacitance C0 in this case is
0
0
0
Q
A
C
V
εd
=
=
(2 46)
Consider next a dielectric inserted between the plates fully occupying
the intervening region |
1 | 2113-2116 | fsu edu/electromag/java/capacitance/
Rationalised 2023-24
Physics
70
and the potential difference V0 is
V0 = E0d
The capacitance C0 in this case is
0
0
0
Q
A
C
V
εd
=
=
(2 46)
Consider next a dielectric inserted between the plates fully occupying
the intervening region The dielectric is polarised by the field and, as
explained in Section 2 |
1 | 2114-2117 | edu/electromag/java/capacitance/
Rationalised 2023-24
Physics
70
and the potential difference V0 is
V0 = E0d
The capacitance C0 in this case is
0
0
0
Q
A
C
V
εd
=
=
(2 46)
Consider next a dielectric inserted between the plates fully occupying
the intervening region The dielectric is polarised by the field and, as
explained in Section 2 10, the effect is equivalent to two charged sheets
(at the surfaces of the dielectric normal to the field) with surface charge
densities sp and –sp |
1 | 2115-2118 | 46)
Consider next a dielectric inserted between the plates fully occupying
the intervening region The dielectric is polarised by the field and, as
explained in Section 2 10, the effect is equivalent to two charged sheets
(at the surfaces of the dielectric normal to the field) with surface charge
densities sp and –sp The electric field in the dielectric then corresponds
to the case when the net surface charge density on the plates is ±(s – sp) |
1 | 2116-2119 | The dielectric is polarised by the field and, as
explained in Section 2 10, the effect is equivalent to two charged sheets
(at the surfaces of the dielectric normal to the field) with surface charge
densities sp and –sp The electric field in the dielectric then corresponds
to the case when the net surface charge density on the plates is ±(s – sp) That is,
0
P
E
σ
σ
−ε
=
(2 |
1 | 2117-2120 | 10, the effect is equivalent to two charged sheets
(at the surfaces of the dielectric normal to the field) with surface charge
densities sp and –sp The electric field in the dielectric then corresponds
to the case when the net surface charge density on the plates is ±(s – sp) That is,
0
P
E
σ
σ
−ε
=
(2 47)
so that the potential difference across the plates is
0
P
V
E d
d
σ
σ
−ε
=
=
(2 |
1 | 2118-2121 | The electric field in the dielectric then corresponds
to the case when the net surface charge density on the plates is ±(s – sp) That is,
0
P
E
σ
σ
−ε
=
(2 47)
so that the potential difference across the plates is
0
P
V
E d
d
σ
σ
−ε
=
=
(2 48)
For linear dielectrics, we expect sp to be proportional to E0, i |
1 | 2119-2122 | That is,
0
P
E
σ
σ
−ε
=
(2 47)
so that the potential difference across the plates is
0
P
V
E d
d
σ
σ
−ε
=
=
(2 48)
For linear dielectrics, we expect sp to be proportional to E0, i e |
1 | 2120-2123 | 47)
so that the potential difference across the plates is
0
P
V
E d
d
σ
σ
−ε
=
=
(2 48)
For linear dielectrics, we expect sp to be proportional to E0, i e , to s |
1 | 2121-2124 | 48)
For linear dielectrics, we expect sp to be proportional to E0, i e , to s Thus, (s – sp) is proportional to s and we can write
P
σK
σ
−σ
=
(2 |
1 | 2122-2125 | e , to s Thus, (s – sp) is proportional to s and we can write
P
σK
σ
−σ
=
(2 49)
where K is a constant characteristic of the dielectric |
1 | 2123-2126 | , to s Thus, (s – sp) is proportional to s and we can write
P
σK
σ
−σ
=
(2 49)
where K is a constant characteristic of the dielectric Clearly, K > 1 |
1 | 2124-2127 | Thus, (s – sp) is proportional to s and we can write
P
σK
σ
−σ
=
(2 49)
where K is a constant characteristic of the dielectric Clearly, K > 1 We
then have
0
0
d
Qd
V
K
A
K
εσ
ε
=
=
(2 |
1 | 2125-2128 | 49)
where K is a constant characteristic of the dielectric Clearly, K > 1 We
then have
0
0
d
Qd
V
K
A
K
εσ
ε
=
=
(2 50)
The capacitance C, with dielectric between the plates, is then
0KA
Q
C
V
d
ε
=
=
(2 |
1 | 2126-2129 | Clearly, K > 1 We
then have
0
0
d
Qd
V
K
A
K
εσ
ε
=
=
(2 50)
The capacitance C, with dielectric between the plates, is then
0KA
Q
C
V
d
ε
=
=
(2 51)
The product e0K is called the permittivity of the medium and is
denoted by e
e = e0 K
(2 |
1 | 2127-2130 | We
then have
0
0
d
Qd
V
K
A
K
εσ
ε
=
=
(2 50)
The capacitance C, with dielectric between the plates, is then
0KA
Q
C
V
d
ε
=
=
(2 51)
The product e0K is called the permittivity of the medium and is
denoted by e
e = e0 K
(2 52)
For vacuum K = 1 and e = e0; e0 is called the permittivity of the vacuum |
1 | 2128-2131 | 50)
The capacitance C, with dielectric between the plates, is then
0KA
Q
C
V
d
ε
=
=
(2 51)
The product e0K is called the permittivity of the medium and is
denoted by e
e = e0 K
(2 52)
For vacuum K = 1 and e = e0; e0 is called the permittivity of the vacuum The dimensionless ratio
0
K
=εε
(2 |
1 | 2129-2132 | 51)
The product e0K is called the permittivity of the medium and is
denoted by e
e = e0 K
(2 52)
For vacuum K = 1 and e = e0; e0 is called the permittivity of the vacuum The dimensionless ratio
0
K
=εε
(2 53)
is called the dielectric constant of the substance |
1 | 2130-2133 | 52)
For vacuum K = 1 and e = e0; e0 is called the permittivity of the vacuum The dimensionless ratio
0
K
=εε
(2 53)
is called the dielectric constant of the substance As remarked before,
from Eq |
1 | 2131-2134 | The dimensionless ratio
0
K
=εε
(2 53)
is called the dielectric constant of the substance As remarked before,
from Eq (2 |
1 | 2132-2135 | 53)
is called the dielectric constant of the substance As remarked before,
from Eq (2 49), it is clear that K is greater than 1 |
1 | 2133-2136 | As remarked before,
from Eq (2 49), it is clear that K is greater than 1 From Eqs |
1 | 2134-2137 | (2 49), it is clear that K is greater than 1 From Eqs (2 |
1 | 2135-2138 | 49), it is clear that K is greater than 1 From Eqs (2 46) and
(2 |
1 | 2136-2139 | From Eqs (2 46) and
(2 51)
0
C
K
=C
(2 |
1 | 2137-2140 | (2 46) and
(2 51)
0
C
K
=C
(2 54)
Thus, the dielectric constant of a substance is the factor (>1) by which
the capacitance increases from its vacuum value, when the dielectric is
inserted fully between the plates of a capacitor |
1 | 2138-2141 | 46) and
(2 51)
0
C
K
=C
(2 54)
Thus, the dielectric constant of a substance is the factor (>1) by which
the capacitance increases from its vacuum value, when the dielectric is
inserted fully between the plates of a capacitor Though we arrived at
Rationalised 2023-24
Electrostatic Potential
and Capacitance
71
EXAMPLE 2 |
1 | 2139-2142 | 51)
0
C
K
=C
(2 54)
Thus, the dielectric constant of a substance is the factor (>1) by which
the capacitance increases from its vacuum value, when the dielectric is
inserted fully between the plates of a capacitor Though we arrived at
Rationalised 2023-24
Electrostatic Potential
and Capacitance
71
EXAMPLE 2 8
Eq |
1 | 2140-2143 | 54)
Thus, the dielectric constant of a substance is the factor (>1) by which
the capacitance increases from its vacuum value, when the dielectric is
inserted fully between the plates of a capacitor Though we arrived at
Rationalised 2023-24
Electrostatic Potential
and Capacitance
71
EXAMPLE 2 8
Eq (2 |
1 | 2141-2144 | Though we arrived at
Rationalised 2023-24
Electrostatic Potential
and Capacitance
71
EXAMPLE 2 8
Eq (2 54) for the case of a parallel plate capacitor, it holds good for any
type of capacitor and can, in fact, be viewed in general as a definition of
the dielectric constant of a substance |
1 | 2142-2145 | 8
Eq (2 54) for the case of a parallel plate capacitor, it holds good for any
type of capacitor and can, in fact, be viewed in general as a definition of
the dielectric constant of a substance Example 2 |
1 | 2143-2146 | (2 54) for the case of a parallel plate capacitor, it holds good for any
type of capacitor and can, in fact, be viewed in general as a definition of
the dielectric constant of a substance Example 2 8 A slab of material of dielectric constant K has the same
area as the plates of a parallel-plate capacitor but has a thickness
(3/4)d, where d is the separation of the plates |
1 | 2144-2147 | 54) for the case of a parallel plate capacitor, it holds good for any
type of capacitor and can, in fact, be viewed in general as a definition of
the dielectric constant of a substance Example 2 8 A slab of material of dielectric constant K has the same
area as the plates of a parallel-plate capacitor but has a thickness
(3/4)d, where d is the separation of the plates How is the capacitance
changed when the slab is inserted between the plates |
1 | 2145-2148 | Example 2 8 A slab of material of dielectric constant K has the same
area as the plates of a parallel-plate capacitor but has a thickness
(3/4)d, where d is the separation of the plates How is the capacitance
changed when the slab is inserted between the plates Solution Let E0 = V0/d be the electric field between the plates when
there is no dielectric and the potential difference is V0 |
1 | 2146-2149 | 8 A slab of material of dielectric constant K has the same
area as the plates of a parallel-plate capacitor but has a thickness
(3/4)d, where d is the separation of the plates How is the capacitance
changed when the slab is inserted between the plates Solution Let E0 = V0/d be the electric field between the plates when
there is no dielectric and the potential difference is V0 If the dielectric
is now inserted, the electric field in the dielectric will be E = E0/K |
1 | 2147-2150 | How is the capacitance
changed when the slab is inserted between the plates Solution Let E0 = V0/d be the electric field between the plates when
there is no dielectric and the potential difference is V0 If the dielectric
is now inserted, the electric field in the dielectric will be E = E0/K The potential difference will then be
0
0
1
3
(
)
(
)
4
4
E
V
E
d
d
K
=
+
0
0
1
3
3
(
)
4
4
K4
E d
V
K
+K
=
+
=
The potential difference decreases by the factor (K + 3)/4K while the
free charge Q0 on the plates remains unchanged |
1 | 2148-2151 | Solution Let E0 = V0/d be the electric field between the plates when
there is no dielectric and the potential difference is V0 If the dielectric
is now inserted, the electric field in the dielectric will be E = E0/K The potential difference will then be
0
0
1
3
(
)
(
)
4
4
E
V
E
d
d
K
=
+
0
0
1
3
3
(
)
4
4
K4
E d
V
K
+K
=
+
=
The potential difference decreases by the factor (K + 3)/4K while the
free charge Q0 on the plates remains unchanged The capacitance
thus increases
0
0
0
0
4
4
3
3
Q
Q
K
K
C
C
V
K
V
K
=
=
=
+
+
2 |
1 | 2149-2152 | If the dielectric
is now inserted, the electric field in the dielectric will be E = E0/K The potential difference will then be
0
0
1
3
(
)
(
)
4
4
E
V
E
d
d
K
=
+
0
0
1
3
3
(
)
4
4
K4
E d
V
K
+K
=
+
=
The potential difference decreases by the factor (K + 3)/4K while the
free charge Q0 on the plates remains unchanged The capacitance
thus increases
0
0
0
0
4
4
3
3
Q
Q
K
K
C
C
V
K
V
K
=
=
=
+
+
2 14 COMBINATION OF CAPACITORS
We can combine several capacitors of
capacitance C1, C2,…, Cn to obtain a system with
some effective capacitance C |
1 | 2150-2153 | The potential difference will then be
0
0
1
3
(
)
(
)
4
4
E
V
E
d
d
K
=
+
0
0
1
3
3
(
)
4
4
K4
E d
V
K
+K
=
+
=
The potential difference decreases by the factor (K + 3)/4K while the
free charge Q0 on the plates remains unchanged The capacitance
thus increases
0
0
0
0
4
4
3
3
Q
Q
K
K
C
C
V
K
V
K
=
=
=
+
+
2 14 COMBINATION OF CAPACITORS
We can combine several capacitors of
capacitance C1, C2,…, Cn to obtain a system with
some effective capacitance C The effective
capacitance depends on the way the individual
capacitors are combined |
1 | 2151-2154 | The capacitance
thus increases
0
0
0
0
4
4
3
3
Q
Q
K
K
C
C
V
K
V
K
=
=
=
+
+
2 14 COMBINATION OF CAPACITORS
We can combine several capacitors of
capacitance C1, C2,…, Cn to obtain a system with
some effective capacitance C The effective
capacitance depends on the way the individual
capacitors are combined Two simple
possibilities are discussed below |
1 | 2152-2155 | 14 COMBINATION OF CAPACITORS
We can combine several capacitors of
capacitance C1, C2,…, Cn to obtain a system with
some effective capacitance C The effective
capacitance depends on the way the individual
capacitors are combined Two simple
possibilities are discussed below 2 |
1 | 2153-2156 | The effective
capacitance depends on the way the individual
capacitors are combined Two simple
possibilities are discussed below 2 14 |
1 | 2154-2157 | Two simple
possibilities are discussed below 2 14 1 Capacitors in series
Figure 2 |
1 | 2155-2158 | 2 14 1 Capacitors in series
Figure 2 26 shows capacitors C1 and C2
combined in series |
1 | 2156-2159 | 14 1 Capacitors in series
Figure 2 26 shows capacitors C1 and C2
combined in series The left plate of C1 and the right plate of C2
are connected to two terminals of a battery and
have charges Q and –Q , respectively |
1 | 2157-2160 | 1 Capacitors in series
Figure 2 26 shows capacitors C1 and C2
combined in series The left plate of C1 and the right plate of C2
are connected to two terminals of a battery and
have charges Q and –Q , respectively It then
follows that the right plate of C1 has charge –Q
and the left plate of C2 has charge Q |
1 | 2158-2161 | 26 shows capacitors C1 and C2
combined in series The left plate of C1 and the right plate of C2
are connected to two terminals of a battery and
have charges Q and –Q , respectively It then
follows that the right plate of C1 has charge –Q
and the left plate of C2 has charge Q If this was
not so, the net charge on each capacitor would
not be zero |
1 | 2159-2162 | The left plate of C1 and the right plate of C2
are connected to two terminals of a battery and
have charges Q and –Q , respectively It then
follows that the right plate of C1 has charge –Q
and the left plate of C2 has charge Q If this was
not so, the net charge on each capacitor would
not be zero This would result in an electric field
in the conductor connecting C1and C2 |
1 | 2160-2163 | It then
follows that the right plate of C1 has charge –Q
and the left plate of C2 has charge Q If this was
not so, the net charge on each capacitor would
not be zero This would result in an electric field
in the conductor connecting C1and C2 Charge
would flow until the net charge on both C1 and
C2 is zero and there is no electric field in the
conductor connecting C1 and C2 |
1 | 2161-2164 | If this was
not so, the net charge on each capacitor would
not be zero This would result in an electric field
in the conductor connecting C1and C2 Charge
would flow until the net charge on both C1 and
C2 is zero and there is no electric field in the
conductor connecting C1 and C2 Thus, in the
series combination, charges on the two plates
(±Q) are the same on each capacitor |
1 | 2162-2165 | This would result in an electric field
in the conductor connecting C1and C2 Charge
would flow until the net charge on both C1 and
C2 is zero and there is no electric field in the
conductor connecting C1 and C2 Thus, in the
series combination, charges on the two plates
(±Q) are the same on each capacitor The total
FIGURE 2 |
1 | 2163-2166 | Charge
would flow until the net charge on both C1 and
C2 is zero and there is no electric field in the
conductor connecting C1 and C2 Thus, in the
series combination, charges on the two plates
(±Q) are the same on each capacitor The total
FIGURE 2 26 Combination of two
capacitors in series |
1 | 2164-2167 | Thus, in the
series combination, charges on the two plates
(±Q) are the same on each capacitor The total
FIGURE 2 26 Combination of two
capacitors in series FIGURE 2 |
1 | 2165-2168 | The total
FIGURE 2 26 Combination of two
capacitors in series FIGURE 2 27 Combination of n
capacitors in series |
1 | 2166-2169 | 26 Combination of two
capacitors in series FIGURE 2 27 Combination of n
capacitors in series Rationalised 2023-24
Physics
72
potential drop V across the combination is the sum of the potential drops
V1 and V2 across C1 and C2, respectively |
1 | 2167-2170 | FIGURE 2 27 Combination of n
capacitors in series Rationalised 2023-24
Physics
72
potential drop V across the combination is the sum of the potential drops
V1 and V2 across C1 and C2, respectively V = V1 + V2 =
1
2
Q
Q
C
+C
(2 |
1 | 2168-2171 | 27 Combination of n
capacitors in series Rationalised 2023-24
Physics
72
potential drop V across the combination is the sum of the potential drops
V1 and V2 across C1 and C2, respectively V = V1 + V2 =
1
2
Q
Q
C
+C
(2 55)
i |
1 | 2169-2172 | Rationalised 2023-24
Physics
72
potential drop V across the combination is the sum of the potential drops
V1 and V2 across C1 and C2, respectively V = V1 + V2 =
1
2
Q
Q
C
+C
(2 55)
i e |
1 | 2170-2173 | V = V1 + V2 =
1
2
Q
Q
C
+C
(2 55)
i e ,
1
2
1
1
QV
C
C
=
+
,
(2 |
1 | 2171-2174 | 55)
i e ,
1
2
1
1
QV
C
C
=
+
,
(2 56)
Now we can regard the combination as an effective capacitor with
charge Q and potential difference V |
1 | 2172-2175 | e ,
1
2
1
1
QV
C
C
=
+
,
(2 56)
Now we can regard the combination as an effective capacitor with
charge Q and potential difference V The effective capacitance of the
combination is
Q
C
=V
(2 |
1 | 2173-2176 | ,
1
2
1
1
QV
C
C
=
+
,
(2 56)
Now we can regard the combination as an effective capacitor with
charge Q and potential difference V The effective capacitance of the
combination is
Q
C
=V
(2 57)
We compare Eq |
1 | 2174-2177 | 56)
Now we can regard the combination as an effective capacitor with
charge Q and potential difference V The effective capacitance of the
combination is
Q
C
=V
(2 57)
We compare Eq (2 |
1 | 2175-2178 | The effective capacitance of the
combination is
Q
C
=V
(2 57)
We compare Eq (2 57) with Eq |
1 | 2176-2179 | 57)
We compare Eq (2 57) with Eq (2 |
1 | 2177-2180 | (2 57) with Eq (2 56), and obtain
1
2
1
1
1
C
C
C
=
+
(2 |
1 | 2178-2181 | 57) with Eq (2 56), and obtain
1
2
1
1
1
C
C
C
=
+
(2 58)
The proof clearly goes through for any number of capacitors arranged
in a similar way |
1 | 2179-2182 | (2 56), and obtain
1
2
1
1
1
C
C
C
=
+
(2 58)
The proof clearly goes through for any number of capacitors arranged
in a similar way Equation (2 |
1 | 2180-2183 | 56), and obtain
1
2
1
1
1
C
C
C
=
+
(2 58)
The proof clearly goes through for any number of capacitors arranged
in a similar way Equation (2 55), for n capacitors arranged in series,
generalises to
1
2
n
1
2
n |
1 | 2181-2184 | 58)
The proof clearly goes through for any number of capacitors arranged
in a similar way Equation (2 55), for n capacitors arranged in series,
generalises to
1
2
n
1
2
n Q
Q
Q
V
V
V
V
C
C
C
=
+
+
+
=
+
+
+
(2 |
1 | 2182-2185 | Equation (2 55), for n capacitors arranged in series,
generalises to
1
2
n
1
2
n Q
Q
Q
V
V
V
V
C
C
C
=
+
+
+
=
+
+
+
(2 59)
Following the same steps as for the case of two
capacitors, we get the general formula for effective
capacitance of a series combination of n capacitors:
1
2
3
n
1
1
1
1
1 |
1 | 2183-2186 | 55), for n capacitors arranged in series,
generalises to
1
2
n
1
2
n Q
Q
Q
V
V
V
V
C
C
C
=
+
+
+
=
+
+
+
(2 59)
Following the same steps as for the case of two
capacitors, we get the general formula for effective
capacitance of a series combination of n capacitors:
1
2
3
n
1
1
1
1
1 C
C
C
C
C
=
+
+
+
+
(2 |
1 | 2184-2187 | Q
Q
Q
V
V
V
V
C
C
C
=
+
+
+
=
+
+
+
(2 59)
Following the same steps as for the case of two
capacitors, we get the general formula for effective
capacitance of a series combination of n capacitors:
1
2
3
n
1
1
1
1
1 C
C
C
C
C
=
+
+
+
+
(2 60)
2 |
1 | 2185-2188 | 59)
Following the same steps as for the case of two
capacitors, we get the general formula for effective
capacitance of a series combination of n capacitors:
1
2
3
n
1
1
1
1
1 C
C
C
C
C
=
+
+
+
+
(2 60)
2 14 |
1 | 2186-2189 | C
C
C
C
C
=
+
+
+
+
(2 60)
2 14 2 Capacitors in parallel
Figure 2 |
1 | 2187-2190 | 60)
2 14 2 Capacitors in parallel
Figure 2 28 (a) shows two capacitors arranged in
parallel |
1 | 2188-2191 | 14 2 Capacitors in parallel
Figure 2 28 (a) shows two capacitors arranged in
parallel In this case, the same potential difference is
applied across both the capacitors |
1 | 2189-2192 | 2 Capacitors in parallel
Figure 2 28 (a) shows two capacitors arranged in
parallel In this case, the same potential difference is
applied across both the capacitors But the plate charges
(±Q1) on capacitor 1 and the plate charges (±Q2) on the
capacitor 2 are not necessarily the same:
Q1 = C1V, Q2 = C2V
(2 |
Subsets and Splits
No saved queries yet
Save your SQL queries to embed, download, and access them later. Queries will appear here once saved.