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1 | 2190-2193 | 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 61)
The equivalent capacitor is one with charge
Q = Q1 + Q2
(2 |
1 | 2191-2194 | 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 61)
The equivalent capacitor is one with charge
Q = Q1 + Q2
(2 62)
and potential difference V |
1 | 2192-2195 | 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 61)
The equivalent capacitor is one with charge
Q = Q1 + Q2
(2 62)
and potential difference V Q = CV = C1V + C2V
(2 |
1 | 2193-2196 | 61)
The equivalent capacitor is one with charge
Q = Q1 + Q2
(2 62)
and potential difference V Q = CV = C1V + C2V
(2 63)
The effective capacitance C is, from Eq |
1 | 2194-2197 | 62)
and potential difference V Q = CV = C1V + C2V
(2 63)
The effective capacitance C is, from Eq (2 |
1 | 2195-2198 | Q = CV = C1V + C2V
(2 63)
The effective capacitance C is, from Eq (2 63),
C = C1 + C2
(2 |
1 | 2196-2199 | 63)
The effective capacitance C is, from Eq (2 63),
C = C1 + C2
(2 64)
The general formula for effective capacitance C for
parallel combination of n capacitors [Fig |
1 | 2197-2200 | (2 63),
C = C1 + C2
(2 64)
The general formula for effective capacitance C for
parallel combination of n capacitors [Fig 2 |
1 | 2198-2201 | 63),
C = C1 + C2
(2 64)
The general formula for effective capacitance C for
parallel combination of n capacitors [Fig 2 28 (b)]
follows similarly,
Q = Q1 + Q2 + |
1 | 2199-2202 | 64)
The general formula for effective capacitance C for
parallel combination of n capacitors [Fig 2 28 (b)]
follows similarly,
Q = Q1 + Q2 + + Qn
(2 |
1 | 2200-2203 | 2 28 (b)]
follows similarly,
Q = Q1 + Q2 + + Qn
(2 65)
i |
1 | 2201-2204 | 28 (b)]
follows similarly,
Q = Q1 + Q2 + + Qn
(2 65)
i e |
1 | 2202-2205 | + Qn
(2 65)
i e , CV = C1V + C2V + |
1 | 2203-2206 | 65)
i e , CV = C1V + C2V + CnV(2 |
1 | 2204-2207 | e , CV = C1V + C2V + CnV(2 66)
which gives
C = C1 + C2 + |
1 | 2205-2208 | , CV = C1V + C2V + CnV(2 66)
which gives
C = C1 + C2 + Cn
(2 |
1 | 2206-2209 | CnV(2 66)
which gives
C = C1 + C2 + Cn
(2 67)
FIGURE 2 |
1 | 2207-2210 | 66)
which gives
C = C1 + C2 + Cn
(2 67)
FIGURE 2 28 Parallel combination of
(a) two capacitors, (b) n capacitors |
1 | 2208-2211 | Cn
(2 67)
FIGURE 2 28 Parallel combination of
(a) two capacitors, (b) n capacitors Rationalised 2023-24
Electrostatic Potential
and Capacitance
73
EXAMPLE 2 |
1 | 2209-2212 | 67)
FIGURE 2 28 Parallel combination of
(a) two capacitors, (b) n capacitors Rationalised 2023-24
Electrostatic Potential
and Capacitance
73
EXAMPLE 2 9
FIGURE 2 |
1 | 2210-2213 | 28 Parallel combination of
(a) two capacitors, (b) n capacitors Rationalised 2023-24
Electrostatic Potential
and Capacitance
73
EXAMPLE 2 9
FIGURE 2 29
Example 2 |
1 | 2211-2214 | Rationalised 2023-24
Electrostatic Potential
and Capacitance
73
EXAMPLE 2 9
FIGURE 2 29
Example 2 9 A network of four 10 mF capacitors is connected to a 500 V
supply, as shown in Fig |
1 | 2212-2215 | 9
FIGURE 2 29
Example 2 9 A network of four 10 mF capacitors is connected to a 500 V
supply, as shown in Fig 2 |
1 | 2213-2216 | 29
Example 2 9 A network of four 10 mF capacitors is connected to a 500 V
supply, as shown in Fig 2 29 |
1 | 2214-2217 | 9 A network of four 10 mF capacitors is connected to a 500 V
supply, as shown in Fig 2 29 Determine (a) the equivalent capacitance
of the network and (b) the charge on each capacitor |
1 | 2215-2218 | 2 29 Determine (a) the equivalent capacitance
of the network and (b) the charge on each capacitor (Note, the charge on
a capacitor is the charge on the plate with higher potential, equal and
opposite to the charge on the plate with lower potential |
1 | 2216-2219 | 29 Determine (a) the equivalent capacitance
of the network and (b) the charge on each capacitor (Note, the charge on
a capacitor is the charge on the plate with higher potential, equal and
opposite to the charge on the plate with lower potential )
Solution
(a) In the given network, C1, C2 and C3 are connected in series |
1 | 2217-2220 | Determine (a) the equivalent capacitance
of the network and (b) the charge on each capacitor (Note, the charge on
a capacitor is the charge on the plate with higher potential, equal and
opposite to the charge on the plate with lower potential )
Solution
(a) In the given network, C1, C2 and C3 are connected in series The
effective capacitance C¢ of these three capacitors is given by
1
2
3
1
1
1
1
C
C
C
C
=
+
+
′
For C1 = C2 = C3 = 10 mF, C¢ = (10/3) mF |
1 | 2218-2221 | (Note, the charge on
a capacitor is the charge on the plate with higher potential, equal and
opposite to the charge on the plate with lower potential )
Solution
(a) In the given network, C1, C2 and C3 are connected in series The
effective capacitance C¢ of these three capacitors is given by
1
2
3
1
1
1
1
C
C
C
C
=
+
+
′
For C1 = C2 = C3 = 10 mF, C¢ = (10/3) mF The network has C¢ and C4
connected in parallel |
1 | 2219-2222 | )
Solution
(a) In the given network, C1, C2 and C3 are connected in series The
effective capacitance C¢ of these three capacitors is given by
1
2
3
1
1
1
1
C
C
C
C
=
+
+
′
For C1 = C2 = C3 = 10 mF, C¢ = (10/3) mF The network has C¢ and C4
connected in parallel Thus, the equivalent capacitance C of the
network is
C = C¢ + C4 = 10
3
+10
mF =13 |
1 | 2220-2223 | The
effective capacitance C¢ of these three capacitors is given by
1
2
3
1
1
1
1
C
C
C
C
=
+
+
′
For C1 = C2 = C3 = 10 mF, C¢ = (10/3) mF The network has C¢ and C4
connected in parallel Thus, the equivalent capacitance C of the
network is
C = C¢ + C4 = 10
3
+10
mF =13 3mF
(b) Clearly, from the figure, the charge on each of the capacitors, C1,
C2 and C3 is the same, say Q |
1 | 2221-2224 | The network has C¢ and C4
connected in parallel Thus, the equivalent capacitance C of the
network is
C = C¢ + C4 = 10
3
+10
mF =13 3mF
(b) Clearly, from the figure, the charge on each of the capacitors, C1,
C2 and C3 is the same, say Q Let the charge on C4 be Q¢ |
1 | 2222-2225 | Thus, the equivalent capacitance C of the
network is
C = C¢ + C4 = 10
3
+10
mF =13 3mF
(b) Clearly, from the figure, the charge on each of the capacitors, C1,
C2 and C3 is the same, say Q Let the charge on C4 be Q¢ Now, since
the potential difference across AB is Q/C1, across BC is Q/C2, across
CD is Q/C3 , we have
1
2
3
500 V
Q
Q
Q
C
C
C
+
+
= |
1 | 2223-2226 | 3mF
(b) Clearly, from the figure, the charge on each of the capacitors, C1,
C2 and C3 is the same, say Q Let the charge on C4 be Q¢ Now, since
the potential difference across AB is Q/C1, across BC is Q/C2, across
CD is Q/C3 , we have
1
2
3
500 V
Q
Q
Q
C
C
C
+
+
= Also, Q¢/C4 = 500 V |
1 | 2224-2227 | Let the charge on C4 be Q¢ Now, since
the potential difference across AB is Q/C1, across BC is Q/C2, across
CD is Q/C3 , we have
1
2
3
500 V
Q
Q
Q
C
C
C
+
+
= Also, Q¢/C4 = 500 V This gives for the given value of the capacitances,
3
10
500
F
1 |
1 | 2225-2228 | Now, since
the potential difference across AB is Q/C1, across BC is Q/C2, across
CD is Q/C3 , we have
1
2
3
500 V
Q
Q
Q
C
C
C
+
+
= Also, Q¢/C4 = 500 V This gives for the given value of the capacitances,
3
10
500
F
1 7
10
C
3
Q
V
−
=
×
µ
=
×
and
3
500
10 F
5 |
1 | 2226-2229 | Also, Q¢/C4 = 500 V This gives for the given value of the capacitances,
3
10
500
F
1 7
10
C
3
Q
V
−
=
×
µ
=
×
and
3
500
10 F
5 0
10
C
Q
V
−
=
×
µ
=
×
′
2 |
1 | 2227-2230 | This gives for the given value of the capacitances,
3
10
500
F
1 7
10
C
3
Q
V
−
=
×
µ
=
×
and
3
500
10 F
5 0
10
C
Q
V
−
=
×
µ
=
×
′
2 15 ENERGY STORED IN A CAPACITOR
A capacitor, as we have seen above, is a system of two conductors with
charge Q and –Q |
1 | 2228-2231 | 7
10
C
3
Q
V
−
=
×
µ
=
×
and
3
500
10 F
5 0
10
C
Q
V
−
=
×
µ
=
×
′
2 15 ENERGY STORED IN A CAPACITOR
A capacitor, as we have seen above, is a system of two conductors with
charge Q and –Q To determine the energy stored in this configuration,
consider initially two uncharged conductors 1 and 2 |
1 | 2229-2232 | 0
10
C
Q
V
−
=
×
µ
=
×
′
2 15 ENERGY STORED IN A CAPACITOR
A capacitor, as we have seen above, is a system of two conductors with
charge Q and –Q To determine the energy stored in this configuration,
consider initially two uncharged conductors 1 and 2 Imagine next a
process of transferring charge from conductor 2 to conductor 1 bit by
Rationalised 2023-24
Physics
74
bit, so that at the end, conductor 1 gets charge Q |
1 | 2230-2233 | 15 ENERGY STORED IN A CAPACITOR
A capacitor, as we have seen above, is a system of two conductors with
charge Q and –Q To determine the energy stored in this configuration,
consider initially two uncharged conductors 1 and 2 Imagine next a
process of transferring charge from conductor 2 to conductor 1 bit by
Rationalised 2023-24
Physics
74
bit, so that at the end, conductor 1 gets charge Q By
charge conservation, conductor 2 has charge –Q at
the end (Fig 2 |
1 | 2231-2234 | To determine the energy stored in this configuration,
consider initially two uncharged conductors 1 and 2 Imagine next a
process of transferring charge from conductor 2 to conductor 1 bit by
Rationalised 2023-24
Physics
74
bit, so that at the end, conductor 1 gets charge Q By
charge conservation, conductor 2 has charge –Q at
the end (Fig 2 30 ) |
1 | 2232-2235 | Imagine next a
process of transferring charge from conductor 2 to conductor 1 bit by
Rationalised 2023-24
Physics
74
bit, so that at the end, conductor 1 gets charge Q By
charge conservation, conductor 2 has charge –Q at
the end (Fig 2 30 ) In transferring positive charge from conductor 2
to conductor 1, work will be done externally, since at
any stage conductor 1 is at a higher potential than
conductor 2 |
1 | 2233-2236 | By
charge conservation, conductor 2 has charge –Q at
the end (Fig 2 30 ) In transferring positive charge from conductor 2
to conductor 1, work will be done externally, since at
any stage conductor 1 is at a higher potential than
conductor 2 To calculate the total work done, we first
calculate the work done in a small step involving
transfer of an infinitesimal (i |
1 | 2234-2237 | 30 ) In transferring positive charge from conductor 2
to conductor 1, work will be done externally, since at
any stage conductor 1 is at a higher potential than
conductor 2 To calculate the total work done, we first
calculate the work done in a small step involving
transfer of an infinitesimal (i e |
1 | 2235-2238 | In transferring positive charge from conductor 2
to conductor 1, work will be done externally, since at
any stage conductor 1 is at a higher potential than
conductor 2 To calculate the total work done, we first
calculate the work done in a small step involving
transfer of an infinitesimal (i e , vanishingly small)
amount of charge |
1 | 2236-2239 | To calculate the total work done, we first
calculate the work done in a small step involving
transfer of an infinitesimal (i e , vanishingly small)
amount of charge Consider the intermediate situation
when the conductors 1 and 2 have charges Q¢ and
–Q¢ respectively |
1 | 2237-2240 | e , vanishingly small)
amount of charge Consider the intermediate situation
when the conductors 1 and 2 have charges Q¢ and
–Q¢ respectively At this stage, the potential difference
V¢ between conductors 1 to 2 is Q¢/C, where C is the
capacitance of the system |
1 | 2238-2241 | , vanishingly small)
amount of charge Consider the intermediate situation
when the conductors 1 and 2 have charges Q¢ and
–Q¢ respectively At this stage, the potential difference
V¢ between conductors 1 to 2 is Q¢/C, where C is the
capacitance of the system Next imagine that a small
charge d Q¢ is transferred from conductor 2 to 1 |
1 | 2239-2242 | Consider the intermediate situation
when the conductors 1 and 2 have charges Q¢ and
–Q¢ respectively At this stage, the potential difference
V¢ between conductors 1 to 2 is Q¢/C, where C is the
capacitance of the system Next imagine that a small
charge d Q¢ is transferred from conductor 2 to 1 Work
done in this step (d W), resulting in charge Q¢ on
conductor 1 increasing to Q¢+ d Q¢, is given by
Q
W
V
Q
Q
C
δ
δ
′δ
=
=
′
′
′
(2 |
1 | 2240-2243 | At this stage, the potential difference
V¢ between conductors 1 to 2 is Q¢/C, where C is the
capacitance of the system Next imagine that a small
charge d Q¢ is transferred from conductor 2 to 1 Work
done in this step (d W), resulting in charge Q¢ on
conductor 1 increasing to Q¢+ d Q¢, is given by
Q
W
V
Q
Q
C
δ
δ
′δ
=
=
′
′
′
(2 68)
Integrating eq |
1 | 2241-2244 | Next imagine that a small
charge d Q¢ is transferred from conductor 2 to 1 Work
done in this step (d W), resulting in charge Q¢ on
conductor 1 increasing to Q¢+ d Q¢, is given by
Q
W
V
Q
Q
C
δ
δ
′δ
=
=
′
′
′
(2 68)
Integrating eq (2 |
1 | 2242-2245 | Work
done in this step (d W), resulting in charge Q¢ on
conductor 1 increasing to Q¢+ d Q¢, is given by
Q
W
V
Q
Q
C
δ
δ
′δ
=
=
′
′
′
(2 68)
Integrating eq (2 68)
W
Q
C
Q
C
Q
Q
C
Q
Q
=
′
=
′
=
∫
0
2
0
2
1
2
2
δ
’
We can write the final result, in different ways
2
2
1
1
2
2
2
Q
W
CV
QV
C
=
=
=
(2 |
1 | 2243-2246 | 68)
Integrating eq (2 68)
W
Q
C
Q
C
Q
Q
C
Q
Q
=
′
=
′
=
∫
0
2
0
2
1
2
2
δ
’
We can write the final result, in different ways
2
2
1
1
2
2
2
Q
W
CV
QV
C
=
=
=
(2 69)
Since electrostatic force is conservative, this work is stored in the form
of potential energy of the system |
1 | 2244-2247 | (2 68)
W
Q
C
Q
C
Q
Q
C
Q
Q
=
′
=
′
=
∫
0
2
0
2
1
2
2
δ
’
We can write the final result, in different ways
2
2
1
1
2
2
2
Q
W
CV
QV
C
=
=
=
(2 69)
Since electrostatic force is conservative, this work is stored in the form
of potential energy of the system For the same reason, the final result for
potential energy [Eq |
1 | 2245-2248 | 68)
W
Q
C
Q
C
Q
Q
C
Q
Q
=
′
=
′
=
∫
0
2
0
2
1
2
2
δ
’
We can write the final result, in different ways
2
2
1
1
2
2
2
Q
W
CV
QV
C
=
=
=
(2 69)
Since electrostatic force is conservative, this work is stored in the form
of potential energy of the system For the same reason, the final result for
potential energy [Eq (2 |
1 | 2246-2249 | 69)
Since electrostatic force is conservative, this work is stored in the form
of potential energy of the system For the same reason, the final result for
potential energy [Eq (2 69)] is independent of the manner in which the
charge configuration of the capacitor is built up |
1 | 2247-2250 | For the same reason, the final result for
potential energy [Eq (2 69)] is independent of the manner in which the
charge configuration of the capacitor is built up When the capacitor
discharges, this stored-up energy is released |
1 | 2248-2251 | (2 69)] is independent of the manner in which the
charge configuration of the capacitor is built up When the capacitor
discharges, this stored-up energy is released It is possible to view the
potential energy of the capacitor as ‘stored’ in the electric field between
the plates |
1 | 2249-2252 | 69)] is independent of the manner in which the
charge configuration of the capacitor is built up When the capacitor
discharges, this stored-up energy is released It is possible to view the
potential energy of the capacitor as ‘stored’ in the electric field between
the plates To see this, consider for simplicity, a parallel plate capacitor
[of area A (of each plate) and separation d between the plates] |
1 | 2250-2253 | When the capacitor
discharges, this stored-up energy is released It is possible to view the
potential energy of the capacitor as ‘stored’ in the electric field between
the plates To see this, consider for simplicity, a parallel plate capacitor
[of area A (of each plate) and separation d between the plates] Energy stored in the capacitor
=
2
2
0
1
(
)
2
2
Q
A
d
C
A
σ
ε
=
×
(2 |
1 | 2251-2254 | It is possible to view the
potential energy of the capacitor as ‘stored’ in the electric field between
the plates To see this, consider for simplicity, a parallel plate capacitor
[of area A (of each plate) and separation d between the plates] Energy stored in the capacitor
=
2
2
0
1
(
)
2
2
Q
A
d
C
A
σ
ε
=
×
(2 70)
The surface charge density s is related to the electric field E between
the plates,
0
E
=εσ
(2 |
1 | 2252-2255 | To see this, consider for simplicity, a parallel plate capacitor
[of area A (of each plate) and separation d between the plates] Energy stored in the capacitor
=
2
2
0
1
(
)
2
2
Q
A
d
C
A
σ
ε
=
×
(2 70)
The surface charge density s is related to the electric field E between
the plates,
0
E
=εσ
(2 71)
From Eqs |
1 | 2253-2256 | Energy stored in the capacitor
=
2
2
0
1
(
)
2
2
Q
A
d
C
A
σ
ε
=
×
(2 70)
The surface charge density s is related to the electric field E between
the plates,
0
E
=εσ
(2 71)
From Eqs (2 |
1 | 2254-2257 | 70)
The surface charge density s is related to the electric field E between
the plates,
0
E
=εσ
(2 71)
From Eqs (2 70) and (2 |
1 | 2255-2258 | 71)
From Eqs (2 70) and (2 71) , we get
Energy stored in the capacitor
U = (
)
2
0
1/2
E
A d
ε
×
(2 |
1 | 2256-2259 | (2 70) and (2 71) , we get
Energy stored in the capacitor
U = (
)
2
0
1/2
E
A d
ε
×
(2 72)
FIGURE 2 |
1 | 2257-2260 | 70) and (2 71) , we get
Energy stored in the capacitor
U = (
)
2
0
1/2
E
A d
ε
×
(2 72)
FIGURE 2 30 (a) Work done in a small
step of building charge on conductor 1
from Q¢ to Q¢ + d Q¢ |
1 | 2258-2261 | 71) , we get
Energy stored in the capacitor
U = (
)
2
0
1/2
E
A d
ε
×
(2 72)
FIGURE 2 30 (a) Work done in a small
step of building charge on conductor 1
from Q¢ to Q¢ + d Q¢ (b) Total work done
in charging the capacitor may be
viewed as stored in the energy of
electric field between the plates |
1 | 2259-2262 | 72)
FIGURE 2 30 (a) Work done in a small
step of building charge on conductor 1
from Q¢ to Q¢ + d Q¢ (b) Total work done
in charging the capacitor may be
viewed as stored in the energy of
electric field between the plates Rationalised 2023-24
Electrostatic Potential
and Capacitance
75
EXAMPLE 2 |
1 | 2260-2263 | 30 (a) Work done in a small
step of building charge on conductor 1
from Q¢ to Q¢ + d Q¢ (b) Total work done
in charging the capacitor may be
viewed as stored in the energy of
electric field between the plates Rationalised 2023-24
Electrostatic Potential
and Capacitance
75
EXAMPLE 2 10
Note that Ad is the volume of the region between the plates (where
electric field alone exists) |
1 | 2261-2264 | (b) Total work done
in charging the capacitor may be
viewed as stored in the energy of
electric field between the plates Rationalised 2023-24
Electrostatic Potential
and Capacitance
75
EXAMPLE 2 10
Note that Ad is the volume of the region between the plates (where
electric field alone exists) If we define energy density as energy stored
per unit volume of space, Eq (2 |
1 | 2262-2265 | Rationalised 2023-24
Electrostatic Potential
and Capacitance
75
EXAMPLE 2 10
Note that Ad is the volume of the region between the plates (where
electric field alone exists) If we define energy density as energy stored
per unit volume of space, Eq (2 72) shows that
Energy density of electric field,
u =(1/2)e0E 2
(2 |
1 | 2263-2266 | 10
Note that Ad is the volume of the region between the plates (where
electric field alone exists) If we define energy density as energy stored
per unit volume of space, Eq (2 72) shows that
Energy density of electric field,
u =(1/2)e0E 2
(2 73)
Though we derived Eq |
1 | 2264-2267 | If we define energy density as energy stored
per unit volume of space, Eq (2 72) shows that
Energy density of electric field,
u =(1/2)e0E 2
(2 73)
Though we derived Eq (2 |
1 | 2265-2268 | 72) shows that
Energy density of electric field,
u =(1/2)e0E 2
(2 73)
Though we derived Eq (2 73) for the case of a parallel plate
capacitor, the result on energy density of an electric field is, in fact,
very general and holds true for electric field due to any configuration
of charges |
1 | 2266-2269 | 73)
Though we derived Eq (2 73) for the case of a parallel plate
capacitor, the result on energy density of an electric field is, in fact,
very general and holds true for electric field due to any configuration
of charges Example 2 |
1 | 2267-2270 | (2 73) for the case of a parallel plate
capacitor, the result on energy density of an electric field is, in fact,
very general and holds true for electric field due to any configuration
of charges Example 2 10 (a) A 900 pF capacitor is charged by 100 V battery
[Fig |
1 | 2268-2271 | 73) for the case of a parallel plate
capacitor, the result on energy density of an electric field is, in fact,
very general and holds true for electric field due to any configuration
of charges Example 2 10 (a) A 900 pF capacitor is charged by 100 V battery
[Fig 2 |
1 | 2269-2272 | Example 2 10 (a) A 900 pF capacitor is charged by 100 V battery
[Fig 2 31(a)] |
1 | 2270-2273 | 10 (a) A 900 pF capacitor is charged by 100 V battery
[Fig 2 31(a)] How much electrostatic energy is stored by the capacitor |
1 | 2271-2274 | 2 31(a)] How much electrostatic energy is stored by the capacitor (b) The capacitor is disconnected from the battery and connected to
another 900 pF capacitor [Fig |
1 | 2272-2275 | 31(a)] How much electrostatic energy is stored by the capacitor (b) The capacitor is disconnected from the battery and connected to
another 900 pF capacitor [Fig 2 |
1 | 2273-2276 | How much electrostatic energy is stored by the capacitor (b) The capacitor is disconnected from the battery and connected to
another 900 pF capacitor [Fig 2 31(b)] |
1 | 2274-2277 | (b) The capacitor is disconnected from the battery and connected to
another 900 pF capacitor [Fig 2 31(b)] What is the electrostatic
energy stored by the system |
1 | 2275-2278 | 2 31(b)] What is the electrostatic
energy stored by the system FIGURE 2 |
1 | 2276-2279 | 31(b)] What is the electrostatic
energy stored by the system FIGURE 2 31
Solution
(a) The charge on the capacitor is
Q = CV = 900 × 10–12 F × 100 V = 9 × 10–8 C
The energy stored by the capacitor is
= (1/2) CV 2 = (1/2) QV
= (1/2) × 9 × 10–8C × 100 V = 4 |
1 | 2277-2280 | What is the electrostatic
energy stored by the system FIGURE 2 31
Solution
(a) The charge on the capacitor is
Q = CV = 900 × 10–12 F × 100 V = 9 × 10–8 C
The energy stored by the capacitor is
= (1/2) CV 2 = (1/2) QV
= (1/2) × 9 × 10–8C × 100 V = 4 5 × 10–6 J
(b) In the steady situation, the two capacitors have their positive
plates at the same potential, and their negative plates at the
same potential |
1 | 2278-2281 | FIGURE 2 31
Solution
(a) The charge on the capacitor is
Q = CV = 900 × 10–12 F × 100 V = 9 × 10–8 C
The energy stored by the capacitor is
= (1/2) CV 2 = (1/2) QV
= (1/2) × 9 × 10–8C × 100 V = 4 5 × 10–6 J
(b) In the steady situation, the two capacitors have their positive
plates at the same potential, and their negative plates at the
same potential Let the common potential difference be V¢ |
1 | 2279-2282 | 31
Solution
(a) The charge on the capacitor is
Q = CV = 900 × 10–12 F × 100 V = 9 × 10–8 C
The energy stored by the capacitor is
= (1/2) CV 2 = (1/2) QV
= (1/2) × 9 × 10–8C × 100 V = 4 5 × 10–6 J
(b) In the steady situation, the two capacitors have their positive
plates at the same potential, and their negative plates at the
same potential Let the common potential difference be V¢ The
Rationalised 2023-24
Physics
76
charge on each capacitor is then Q¢ = CV¢ |
1 | 2280-2283 | 5 × 10–6 J
(b) In the steady situation, the two capacitors have their positive
plates at the same potential, and their negative plates at the
same potential Let the common potential difference be V¢ The
Rationalised 2023-24
Physics
76
charge on each capacitor is then Q¢ = CV¢ By charge conservation,
Q¢ = Q/2 |
1 | 2281-2284 | Let the common potential difference be V¢ The
Rationalised 2023-24
Physics
76
charge on each capacitor is then Q¢ = CV¢ By charge conservation,
Q¢ = Q/2 This implies V¢ = V/2 |
1 | 2282-2285 | The
Rationalised 2023-24
Physics
76
charge on each capacitor is then Q¢ = CV¢ By charge conservation,
Q¢ = Q/2 This implies V¢ = V/2 The total energy of the system is
6
1
1
2
'
'
2 |
1 | 2283-2286 | By charge conservation,
Q¢ = Q/2 This implies V¢ = V/2 The total energy of the system is
6
1
1
2
'
'
2 25
10
J
2
4
Q V
QV
−
=
×
=
=
×
Thus in going from (a) to (b), though no charge is lost; the final
energy is only half the initial energy |
1 | 2284-2287 | This implies V¢ = V/2 The total energy of the system is
6
1
1
2
'
'
2 25
10
J
2
4
Q V
QV
−
=
×
=
=
×
Thus in going from (a) to (b), though no charge is lost; the final
energy is only half the initial energy Where has the remaining energy
gone |
1 | 2285-2288 | The total energy of the system is
6
1
1
2
'
'
2 25
10
J
2
4
Q V
QV
−
=
×
=
=
×
Thus in going from (a) to (b), though no charge is lost; the final
energy is only half the initial energy Where has the remaining energy
gone There is a transient period before the system settles to the
situation (b) |
1 | 2286-2289 | 25
10
J
2
4
Q V
QV
−
=
×
=
=
×
Thus in going from (a) to (b), though no charge is lost; the final
energy is only half the initial energy Where has the remaining energy
gone There is a transient period before the system settles to the
situation (b) During this period, a transient current flows from
the first capacitor to the second |
1 | 2287-2290 | Where has the remaining energy
gone There is a transient period before the system settles to the
situation (b) During this period, a transient current flows from
the first capacitor to the second Energy is lost during this time in
the form of heat and electromagnetic radiation |
1 | 2288-2291 | There is a transient period before the system settles to the
situation (b) During this period, a transient current flows from
the first capacitor to the second Energy is lost during this time in
the form of heat and electromagnetic radiation EXAMPLE 2 |
1 | 2289-2292 | During this period, a transient current flows from
the first capacitor to the second Energy is lost during this time in
the form of heat and electromagnetic radiation EXAMPLE 2 10
SUMMARY
1 |
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