EMF 2005
Handout 5: Capacitance
1
CAPACITANCE
Definition of capacitance
Recall: For a point charge or a charged sphere
V=
Q
4πε o r
In general, POTENTIAL ∝ CHARGE for any size or shape of
conductor.
Definition: The constant of proportionality between V and Q is
called CAPACITANCE, C:
Q
C =
V
Coulombs
≡
Volts
CV −1
Units of capacitance:
C ≡
Definition:
1 Farad (F) ≡ 1 CV-1
Capacitance is a measure of the ability of a conductor or a system of
conductors to store charge (and hence to store energy).
Recall: For a charged conductor
⇒
U tot
1
= CV 2
2
or
U tot
U tot =
1 Q2
=
2 C
1
QV
2
Relationship between
capacitance and energy
Alternative units for εo:
E =
Q
Coulombs
Volts
≡
≡
metres
ε ometres 2
4πε or 2
⇒
εo ≡
⇒
εo ≡ Fm-1
Coulombs
Farads
≡
( Volts)(metres )
metres
These are the units in which εo is usually quoted.
EMF 2005
Handout 5: Capacitance
2
Procedure for finding capacitance
1.
Imagine a charge Q on the conductor (if it’s a single conductor) or
charges of ±Q (for a pair of conductors)
2.
Find E (e.g., use Gauss’s Law)
3.
Find the potential (difference) using
∆V =
b
∫ E ⋅ dL
a
(never mind the sign).
4.
Put C = Q/V
[Q always cancels out]
Note:
1.
C depends only on the geometry - the size and shape of the
conductors and the distance between them.
2.
C is independent of Q [because V ∝ Q]
Examples:
1.
2.
3.
4.
Capacitance of an isolated sphere
Parallel plate capacitor
Capacitance of two concentric spheres
Capacitance per unit length of a co-axial cable
See lecture notes
EMF 2005
Handout 5: Capacitance
Capacitor
3
C
An electrical component designed
to have a particular value of C.
Capacitors in parallel
∆V is the same for both:
∆V1 = ∆V2 = ∆V
∆V1 = ∆V1 = ∆Vtot
C1 +Q1 C2 +Q2
-Q1
+Qtot
∆V
-Qtot
∆Vtot = ∆V
Qtot = Q1 + Q2
C tot =
Therefore
≡
-Q2
Ctot
Q1
Q
+ 2
∆V
∆V
C tot = C1 + C 2
⇒
Capacitors in series
C1
-Q
+Q
≡
C2
+Q
∆V1
-Q
Ctot
∆Vtot = ∆V1 + ∆V2
∆V2
In this case Q is the same for both and ∆V is different.
C1 =
Q
∆V1
C2 =
∆V
∆V 2
1
= 1 +
C tot
Q
Q
Q
∆V2
⇒
C tot =
1
C tot
=
Q tot
∆Vtot
1
1
+
C1
C2
=
Q
∆V1 + ∆V2
EMF 2005
Handout 5: Capacitance
4
Dielectrics and polarisation
Until now, we have assumed that the space between charges,
conductors, etc. is empty (a vacuum). What if it’s filled with some
insulating material?
Recall: A DIELECTRIC (insulator) is electrically neutral.
But it contains many +ve and -ve charges in its atoms or molecules.
Because it’s an insulator, the charges can’t move around.
BUT: some molecules have a natural DIPOLE MOMENT
⇒
when placed in an external
electric field Eo , the dipoles
tend to align with it.
Torque
+
Edipole
Eo
-
Even if no INTRINSIC dipoles exist, they
can be INDUCED by an applied field
Eo , producing the same effect.
-
+
Eo
Consider a parallel plate capacitor, with positive and negative charges
on the plates. This creates an electric field Eo between the plates.
If there is a dielectric between the plates, then due to dipole alignment,
or (POLARISATION) the distributions of +ve and -ve charge do not
overlap exactly:
E
Excess +ve charge on the right.
Excess -ve charge on the left.
The induced (POLARISED) field, Ei ,
tends to OPPOSE Eo .
The resultant field is LESS than
the applied field:
Etot = Eo - Ei
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
o
+
+
+
+
+
+
+
+
+
-
Eo
-
+
+
+
+
+
+
+
+
+
-
Ei
EMF 2005
Handout 5: Capacitance
5
Dielectric constant
Let E tot =
Etot = Eo - Ei
1
Eo
K
(K ≥ 1)
K (sometimes written as the greek letter κ - kappa) is the DIELECTRIC
CONSTANT or RELATIVE PERMITTIVITY of the material.
Let σo = charge density on capacitor plates:
Let σi = induced charge density on surfaces of dielectric:
Therefore
Eo = σo/εo
Ei = σi/εo
1

σi = ε oEi = ε o [E o − E tot ] = ε oE o 1 − 
K

So the induced surface charge density is
1

σi = σ o 1 − 
K

The effect of dielectric constant on capacitance
Consider the case of a parallel plate capacitor.
+Q
E1
C1 =
d
No dielectric:
d
With dielectric: C 2 =
ε A
Q
Q
=
= o
E1d
d
∆V1
-Q
+Q
E2
Kε o A
Q
Q
KQ
=
=
=
E 2d
E1d
d
∆V2
-Q
Capacitance with dielectric = K( Capacitance without dielectric)
So, capacitance (i.e., the ability to store charge and hence energy) is
increased by the use of a dielectric.
EMF 2005
Handout 5: Capacitance
6
Typical values of dielectic constant K
Air
Polythene
Glass
Germanium
Water
A perfect conductor
1.00059 (not much different from a vacuum)
2.3
5 - 10
16
80
?
Gauss's Law in dielectrics
Just replace εo with Kεo :
Φ =
∫
E ⋅ dA =
Q enclosed
Kε o
Dielectric breakdown
If E > some limit, the BREAKDOWN FIELD STRENGTH of the material,
then the bonds between the electrons and the atoms are broken and the
material becomes a CONDUCTOR ( → short circuit in a cable or
capacitor).
EMF 2005
Handout 5: Capacitance
7
Capacitance per unit length of a pair of
co-axial cylinders (e.g., a co-axial cable)
a
b
E
E
Step 1: Let charge per unit length be + λ on inner conductor
- λ on outer conductor
Step 2: Find E :
Field pattern: Field lines go from +ve charges on inner
conductor to -ve charges on outer conductor
E = 0 for r < a and r > b (A Gaussian cylinder encloses
no charge)
λ
Apply Gauss’s Law
E =
→
2 πε or
(similar to a previous example)
Step 3: Find ∆V:
∆V =
∫
E ⋅ dL
E
Path: Integrate along a field line → d L = d r
r = b
dL
r = a
and d L is parallel to E so E ⋅ d L = Edr
b
∆V =
∫
a
λ
λ
b
dr =
ln 
2πε or
2πε o  a 
We take ∆V to be positive when finding capacitance.
Step 4: Put C = Q/V → C =
2πε o
b
ln 
a
(Capacitance/unit length)
r = 0