PH504 – Part 6
1. Capacitance; definition
In a conductor, electrons are free to roam. Any net charge resides
on the surface – the electric field is zero inside.
When an isolated, finite size conductor is given a charge Q, its
potential (with respect to a zero at infinity) is V. It can be shown
that for any body that Q is proportional to V and the constant of
proportionality is known as the capacitance (C) of the conductor.
C = Q/V
The capacitance can be thought of expressing the amount of
charge the conductor can carry for a given potential V.
Farad: The units of capacitance are the farad (symbol F).
The Farad, F, is the SI unit for capacitance, and from the definition of
capacitance is seen to be equal to a Coulomb/Volt.
The capacitance of a body is a property of its shape and size.
Example: It can be shown (see below) that a conducting sphere of
radius a and carrying a charge Q has a potential
.
Hence from C = Q / V 
……… just depending on its size! With:
For a = 1cm, C = 0.01 x 1.11x10-10, this is about 1 pF.
For the planet Earth: about 709 µF.
This is the self-capacitance.
1
Practical capacitors generally consist of two conductors, in
operation one carries a charge +Q the other a charge –Q. The
definition of (mutual) capacitance is still C = Q / V but now V is the
potential difference between the bodies.
2. Calculating Capacitance
Place a charge +Q on one conductor and –Q on the other.
Find the potential difference between the conductors by
(a)
using a suitable equation for the potential appropriate to the
symmetry of the problem or
(b)
find the form of the E-field in the region between the two
conductors and then integrate E with respect to a suitable
spatial co-ordinate to find the potential difference.
The result will be an equation for V in terms of Q and the spatial
dimensions of the conductors. Finally use the definition C=Q/V to
find an expression for C.
Example 1: 2 Spheres (HyperPhysics website)
Spherical Capacitor
The capacitance for spherical or cylindrical
conductors can be obtained by evaluating the
voltage difference between the conductors for a
given charge on each.
By applying Gauss' law to a charged conducting
sphere, the magnitude of the electric field outside
it is found to be
Voltage between the two spheres: integrating the electric field along
2
a radial line:
The capacitance is:
Does an isolated charged sphere have capacitance?
Isolated Sphere
Capacitor?
An isolated charged conducting sphere has
capacitance.
Applications for such a capacitor may not be
immediately evident, but it does illustrate that a
charged sphere has stored some energy as a
result of being charged. Taking the concentric
sphere capacitance expression:
and taking the limits as b goes to infinity, and a = R, gives
Further confirmation of this comes from examining the potential of a
charged conducting sphere:
Example 2: Coaxial Cable
3
Consider a cylindrical surface of radius R.
Take a charge per unit length 
Then Gauss's Law yields
E 2r = o E = 2ro) for r > R.
E=0
for r < R
Suppose the permittivity within a coaxial cable is ko
k = relative permittivity of the dielectric material between the
plates.
k=1 for free space, k>1 for all media, approximately =1 for air.
Cylindrical
Capacitor
For a cylindrical geometry like a
coaxial cable, the capacitance
is usually stated as a
capacitance per unit length.
The charge resides on the outer
surface of the inner conductor
and the inner wall of the outer
conductor.
By applying Gauss' law to an
infinite cylinder, the electric field
outside a charged cylinder is
found to be
The voltage between the cylinders (in a vacuum)can be found by
integrating the electric field along a radial line:
4
The capacitance per unit length ( with rel. permittivity k) is:
The capacitance is proportion to the length, of course.
Example 3 :
2 parallel plates
charge Q & -Q, area A, separation d:
Apply Gauss’s law: E = Q/(oA)
- constant
Potential difference: V = d Q/(oA)
and so
C = Ao/d
3. Energy stored by a capacitor
In charging a capacitor from zero to a finite charge Q, work
must be done (i.e. to remove electrons from the positive plate
and overcome the repulsion from the negative plate).
5
If at some point the charge is q and the potential is v (hence
q=Cv) then to add an additional charge dq work dW=vdq must be
done.
Hence total work in charging from 0 to Q is given by an integral:
or
This is the work done in charging the capacitor and hence also
equals the potential energy U stored by a charged capacitor
assuming that zero potential energy corresponds to zero charge.
From the definition of voltage as the energy per unit charge, one
might expect that the energy stored on this ideal capacitor would
be just QV. That is, all the work done on the charge in moving it
from one plate to the other would appear as energy stored.
But in fact, the expression above shows that just half of that
work appears as energy stored in the capacitor.
For a finite resistance, one can show that half of the energy
supplied by the battery for the charging of the capacitor is
dissipated as heat in the resistor, regardless of the size of the
resistor.
6
Charging a capacitor
The capacitor (C) in the circuit diagram is
being charged from a supply voltage (Vs) with
the current passing through a resistor (R). The
voltage across the capacitor (Vc) is initially
zero but it increases as the capacitor charges.
The capacitor is fully charged when Vc = Vs.
The charging current (I) is determined by the
voltage across the resistor (Vs - Vc):
Charging current, I = (Vs - Vc) / R (note that
Vc is increasing)
At first Vc = 0 V
so the initial current, Io = Vs / R
Vc increases as soon as charge (Q) starts to build up (Vc = Q/C), this reduces
the voltage across the resistor and therefore reduces the charging current.
This means that the rate of charging becomes progressively slower.
time constant = R × C
time constant is in seconds (s)
R = resistance in ohms ( )
C = capacitance in farads (F)
For example:
If R = 47k and C = 22µF,
the time constant, RC = 47k
If R = 33k
× 22µF = 1.0s.
and C = 1µF,
the time constant, RC = 33k × 1µF = 33ms.
A large time constant means the capacitor charges slowly. Note
that the time constant is a property of the circuit containing the
capacitance and resistance, it is not a property of a capacitor
alone.
Stored energy in terms of the E-field
7
The energy is stored in the electric field.
For a parallel plate capacitor we can show that
and
where C is the capacitance, A is the area of the plates, d their
separation and E is the E-field between the plates (the only
region where E is non-zero).
From above, the potential energy U is given by
where E has been used to eliminate Q.
The final result is simply (1/2)0E2 x (volume between plates)
This result suggests a general one that the potential energy is
given by (1/2)0E2 multiplied by the volume over which E is nonzero or if E is not constant
4. Capacitor Combinations
Capacitors in parallel add ... simply increasing Area A
C = C1 + C2 + ….
8
Capacitors in series combine as reciprocals ... since they
share the voltage V: Q/C = V = V1 + V2 + V3 + ….
Since charge cannot be added or taken away from the conductor
between series capacitors, the net charge there remains zero.
You store less charge on series capacitors than you would on
either one of them alone with the same voltage!
Does it ever make sense to put capacitors in series? You get less
capacitance and less charge storage than with either alone.
It is sometimes done in electronics practice because capacitors have
maximum working voltages, and with two "600 volt maximum" capacitors in
series, you can increase the working voltage to 1200 volts.
Conclusions


Capacitance – definition and determination
Potential energy stored in a capacitor
9
10