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CAPACITORS
Prepared by :-
Dinesh Sharma (047)
SHARDA UNIVERSITY
Capacitance
Two
conductors
with equal
but opposite
charges
Electric
Field
Potential
Difference
Between
the
Charges
V
Q
C
∆
≡ (C/V=F)
Calculating Capacitance
Assume a
charge q on
the plates
Calculate the
e-field using
Gauss’ Law
Calculate
the
potential
Find the
capacitance
qAdE =⋅∫

0ε
qEA =0ε
∫ ⋅−=−
f
i
if sdEVV

∫
+
−
=∆ EdsV
V
q
C
∆
=
Parallel Plate Capacitor
 A parallel-plate capacitor consists of two parallel
conducting plates, each of area A, separated by a
distance d. When the capacitor is charged, the plates
carry equal amounts of charge. One plate carries
positive charge, and the other carries negative charge.
Parallel Plate Capacitor
Two parallel metallic plates of equal area
A separated by a distance d as shown.
One plate carries a charge Q and the
other carries a charge –Q. And surface
charge density of each plate is σ = Q/A.
In the outer region of both the plates, the
electric field due to the two charged plates
cancel out. So the net field is zero.
E = σ − σ = 0
εo εo
 In the inner region between the two capacitor plates, the
electric field due to the two charged plates add up. The
net field is :
E = σ + σ = σ
2εo 2εo εo
Potential difference between the plates
= electric field X distance between the plates
=> V= Ed = σd
εo
What is a lightning discharge ?
 Friction forces between the air
molecules within a cloud result in
positively charged molecules moving
to the lower surface, and the negative
charges moving to the upper surface .
 The lower surface induces a high
concentration of negative charges in
the earth beneath.
 The resultant electric field is very
strong, containing large amounts of
charge and energy.
 If the electric field is greater than than
breakdown strength of air, a lightning
discharge occurs, in which air
molecules are ripped apart, forming a
conducting path between the cloud
and the earth.
Cylindrical Capacitor






=






=
a
b
k
l
C
a
b
k
l
C
e
e
ln2
1
ln2
lQ λ=
r
k
E eλ2
=






−=−=−=− ∫∫ a
b
k
r
dr
kEdrVV e
b
a
e
b
a
ab ln22 λλ






=
∆
=
a
b
k
l
V
Q
C
e ln2 λ
λ
Spherical Capacitor
)( abk
ab
C
e −
=
2
r
Q
kE e=






−==−=− ∫∫ ba
Qk
r
dr
EdrVV e
b
a
b
a
ab
11
2






−
=
∆
=
ba
Qk
Q
V
Q
C
e
11
Circuits with Capacitors
 A battery drives
charges to accumulate
on the capacitor plates.
 A switch controls the
flow of charge.
∆V
C,Q
+
Derive series, parallel
formulae on board.
Parallel Combination
VCVCQ
VCVCQ
∆=∆=
∆=∆=
2222
1111
21
21
VVV
QQQ
∆=∆=∆
+=
VCQ eq∆=
21 CCCeq +=
For parallel capacitors
∑=
=
n
j
jeq CC
1
Parallel Connections
All are parallel connections
This is not
Series Combination
21
21
QQQ
VVV
==
∆+∆=∆
22
2
2
11
1
1
C
Q
C
Q
V
C
Q
C
Q
V
==∆
==∆
eqC
Q
V =∆
21
111
CCCeq
+=
∑=
=
n
j jeq CC 1
11For series capacitors
Series and Parallel
 Consider circuit on board.
 1. Name all pairs of capacitors in series.
 2. Name all pairs of capacitors in series.
Key ideas:
1. For capacitors in series, the
charges are all the same.
2. For capacitors in parallel, the
potential differences are all the
same.
Mixed Combination of Capacitors
Energy Stored in a Capacitor
 Since capacitors can carry charge after a battery
is disconnected, they can store energy.
 To calculate this energy, consider the work done
to move a small amount of charge from one plate
of the capacitor to the other.
 As more and more charges are moved, the work
required increases.
dq
C
q
VdqdW =∆=
C
Q
dqq
C
dq
C
q
W
QQ
2
1 2
00
=== ∫∫
2
2
)(
2
1
2
1
2
VCVQ
C
Q
U ∆=∆==
Concept Question
Consider a simple parallel-plate capacitor whose plates
are given equal and opposite charges and are separated
by a distance d. Suppose the plates are pulled apart until
they are separated by a distance D > d. The electrostatic
energy stored in the capacitor is
1. greater than
2. the same as
3. smaller than
before the plates were pulled apart.
Energy Density
 Defined as stored energy per unit volume.
 Consider the parallel plate capacitor.
2
)(
2
1
VCU ∆=
d
A0ε 2
)(Ed
2
0 )(
2
1
EAdU ε=
2
0
2
02
1
2
1
)(
Eu
Ad
EAd
Vol
U
u
E
E
ε
ε
=
==
Dielectrics
 Dielectrics are insulators like glass, rubber
or wood.
 They are used to increase the charge
storage capacity and maximum voltage in
capacitors while providing mechanical
support.
 They are characterized by a dielectric
constant, κ, and are limited by the dielectric
strength.
Capacitors With Dielectrics
κ
0V
V
∆
=∆
0
00
V
Q
V
Q
C
∆
=
∆
= κ
0CC κ=
An Atomic View
 As we apply a voltage to the plates, the
dielectric in between is polarized.
 This creates a field in the opposite direction
to the applied field and therefore reduces
the effective voltage.
Concept Question
A parallel-plate capacitor is attached to a
battery that maintains a constant potential
difference V between the plates. While the
battery is still connected, a glass slab is
inserted so as to just fill the space between the
plates. The stored energy
1. increases.
2. decreases.
3. remains the same.
Metallic Slab Inside a Capacitor
ad
A
C
−
= 0ε
Summary
 Capacitors are geometries that can hold
charges.
 Use Gauss’ Law symmetries to calculate
capacitance.
 Series and parallel connections.
 Dielectrics increase capacitance.
Thank You 

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CAPACITORS

  • 1. CAPACITORS Prepared by :- Dinesh Sharma (047) SHARDA UNIVERSITY
  • 3. Calculating Capacitance Assume a charge q on the plates Calculate the e-field using Gauss’ Law Calculate the potential Find the capacitance qAdE =⋅∫  0ε qEA =0ε ∫ ⋅−=− f i if sdEVV  ∫ + − =∆ EdsV V q C ∆ =
  • 4. Parallel Plate Capacitor  A parallel-plate capacitor consists of two parallel conducting plates, each of area A, separated by a distance d. When the capacitor is charged, the plates carry equal amounts of charge. One plate carries positive charge, and the other carries negative charge.
  • 5. Parallel Plate Capacitor Two parallel metallic plates of equal area A separated by a distance d as shown. One plate carries a charge Q and the other carries a charge –Q. And surface charge density of each plate is σ = Q/A. In the outer region of both the plates, the electric field due to the two charged plates cancel out. So the net field is zero. E = σ − σ = 0 εo εo
  • 6.  In the inner region between the two capacitor plates, the electric field due to the two charged plates add up. The net field is : E = σ + σ = σ 2εo 2εo εo Potential difference between the plates = electric field X distance between the plates => V= Ed = σd εo
  • 7. What is a lightning discharge ?  Friction forces between the air molecules within a cloud result in positively charged molecules moving to the lower surface, and the negative charges moving to the upper surface .  The lower surface induces a high concentration of negative charges in the earth beneath.  The resultant electric field is very strong, containing large amounts of charge and energy.  If the electric field is greater than than breakdown strength of air, a lightning discharge occurs, in which air molecules are ripped apart, forming a conducting path between the cloud and the earth.
  • 8. Cylindrical Capacitor       =       = a b k l C a b k l C e e ln2 1 ln2 lQ λ= r k E eλ2 =       −=−=−=− ∫∫ a b k r dr kEdrVV e b a e b a ab ln22 λλ       = ∆ = a b k l V Q C e ln2 λ λ
  • 9. Spherical Capacitor )( abk ab C e − = 2 r Q kE e=       −==−=− ∫∫ ba Qk r dr EdrVV e b a b a ab 11 2       − = ∆ = ba Qk Q V Q C e 11
  • 10. Circuits with Capacitors  A battery drives charges to accumulate on the capacitor plates.  A switch controls the flow of charge. ∆V C,Q + Derive series, parallel formulae on board.
  • 12. Parallel Connections All are parallel connections This is not
  • 14. Series and Parallel  Consider circuit on board.  1. Name all pairs of capacitors in series.  2. Name all pairs of capacitors in series. Key ideas: 1. For capacitors in series, the charges are all the same. 2. For capacitors in parallel, the potential differences are all the same.
  • 15. Mixed Combination of Capacitors
  • 16. Energy Stored in a Capacitor  Since capacitors can carry charge after a battery is disconnected, they can store energy.  To calculate this energy, consider the work done to move a small amount of charge from one plate of the capacitor to the other.  As more and more charges are moved, the work required increases. dq C q VdqdW =∆= C Q dqq C dq C q W QQ 2 1 2 00 === ∫∫ 2 2 )( 2 1 2 1 2 VCVQ C Q U ∆=∆==
  • 17. Concept Question Consider a simple parallel-plate capacitor whose plates are given equal and opposite charges and are separated by a distance d. Suppose the plates are pulled apart until they are separated by a distance D > d. The electrostatic energy stored in the capacitor is 1. greater than 2. the same as 3. smaller than before the plates were pulled apart.
  • 18. Energy Density  Defined as stored energy per unit volume.  Consider the parallel plate capacitor. 2 )( 2 1 VCU ∆= d A0ε 2 )(Ed 2 0 )( 2 1 EAdU ε= 2 0 2 02 1 2 1 )( Eu Ad EAd Vol U u E E ε ε = ==
  • 19. Dielectrics  Dielectrics are insulators like glass, rubber or wood.  They are used to increase the charge storage capacity and maximum voltage in capacitors while providing mechanical support.  They are characterized by a dielectric constant, κ, and are limited by the dielectric strength.
  • 21. An Atomic View  As we apply a voltage to the plates, the dielectric in between is polarized.  This creates a field in the opposite direction to the applied field and therefore reduces the effective voltage.
  • 22. Concept Question A parallel-plate capacitor is attached to a battery that maintains a constant potential difference V between the plates. While the battery is still connected, a glass slab is inserted so as to just fill the space between the plates. The stored energy 1. increases. 2. decreases. 3. remains the same.
  • 23. Metallic Slab Inside a Capacitor ad A C − = 0ε
  • 24. Summary  Capacitors are geometries that can hold charges.  Use Gauss’ Law symmetries to calculate capacitance.  Series and parallel connections.  Dielectrics increase capacitance.