CAPACITORS AND CURRENTS

1. Two conductors isolated from one another and from their surroundings, form a .
These conductors may be of any shape and size, and at any distance from each other. If
a potential diff
capacitor
erence is created between the conductors (say, by connecting the terminals
of a battery to them), then there is an electric field in the space between them. The electric
field comes from the charges that have been pushed to the plates by the battery. The
amount of charge pushed on to the conductors is proportional to the potential difference
between the battery terminals (which is the same as between the capacitor plates). Hence,
. To convert this into an equality, we write . This provides the definition of
capacitance, .
2. Using the above definition, let us c
Q V Q CV
C Q
V
∝ =
=
alculate the capacitance of two parallel plates separated
by a distance d as in the figure below.
0
0
Recall Gauss's Law: . Draw any Gaussian surface. Since the electric
field is zero above the top plate, the flux through the area A of the plate is ,
where is
enclosed E dA q
EA Q
Q
ε
ε
Φ ≡ ⋅ =
Φ = =
∫
􀁇 􀁇
0
0
the total charge on the plate. Thus, is the electric field in the gap
between the plates. The potential difference is , and so . You can see
that the capacitance will be larg
E Q
A
V E C Q A
d V d
ε
ε
=
= = =
e if the plates are close to each other, and if the plates
have a large area. We have simplified the calculation here by assuming that the electric
field is strictly directed downwards. This is only true if the plates are infinitely long. But
we can usually neglect the side effects. Note that any arrangement with two plates forms
a capacitor: plane, cylindrical, spherical, etc. The capacitance depends upon the geometry,
the size of plates and the gap between them.
− − − − − − − − −
+q
−q
eq C
V
1 C 2 C
V
1 V2 V
eq C
V
1 C
2 C
3. One can take two (or more) capacitors in various ways and thus change the amount of
charge they can contain. Consider first two capacitors connected in parallel with each
other. The same vol 1 1 2 2
1 2 1 2
tage exists across both. For each capacitor, , where
is the potential between terminals and . The total charge is:
q CV q CV
V a b
Q q q CV CV
= =
= + = + =
( )
1 2
1 2
( )
Now, let us define an "effective" or "equivalent" capacitance as . Then we can
immediately see that for 2 capacitors , and for capacitors .
eq
eq eq n
C C V
C Q
V
C C C C C n
+
=
= + = Σ
1 2 1 2
4. We can repeat the analysis above when the capacitors are put in series. Here the difference
is that now we must start with , where and are the voltages across the two.
Clearly th
V=V +V V V
1 2
1 2 1 2
1 2
e same charge had to cross both the capacitors. Hence,
(1 1).
From our definition, , it follows that 1 1 1. The total capac eq
eq
V V V Q Q Q
C C C C
C Q
V C C C
= + = + = +
= = +
( )
itance is now
less than if they were in parallel. In general, 1 1 for capacitors .
eq n
n
C C
= Σ
5. When a battery is connected to a capacitor, positive and negative charges appear on the
opposite plates. Some energy has been transferred from the battery to the capacitor, and
now been stored in it. When the capacitor is discharged, the energy is recovered. Now
let us calculate the energy required to charge a capacitor from zero to volts.
Begin: the amount of energy required
V
to transfer a small charge to the plates is
v , where v is the voltage at a time when the charge is v. As time goes on,
the total charge increases until it reaches the final charge
dq
dU=dq q=C
Q (at which point the voltage
becomes V ). So,
v 1 .
2 2
But where in the capacitor is the energy stored? Answer, it is present in the electric field
in the volume between the two plates. We can calcula
Q dU dqqdq U dU qdq Q CV
C C C
= = ⇒ =∫ =∫ = =
2 2
0 2
0
0
te the energy density:
1
energy stored in capacitor 2 1 .
volume of capacitor 2 2
In the above we have used , derived earlier. The impor
u U CV V E
Ad Ad d
C A
d
ε
ε
ε
= = = = ⎛⎜ ⎞⎟ =
⎝ ⎠
=
2
0
tant result here is that
. Turning it around, wherever there is an electric field, there is energy available.
6. Consider a free charge . Around it is an electric field, 1
4
u E
Q E
πε ε
∝
Dielectrics. + = 2 .
Now suppose this charge is placed among water molecules. These molecules will polarise,
i.e. the centre of positive charge and centre of negative charge will be slightly displaced.
Th
r
Q
r
2
0
e negative part of the water molecule will be attracted toward the positive charge .
So, in effect, the electric field is weakened by 1 and becomes, 1 . Here I have
4
introduced a n
r r
Q
Q
ε πε ε r
+
ew quantity called "dielectric constant". This is a number that is usually
bigger than one and measures the strength of the polarization induced in the material. For
air, 1.0003 while 8
r
r r
ε
ε = ε ≈
0 0
0 for pure water. The effect of a dielectric is to increase the
capacitance of a capacitor: if the air between the plates of a capacitor is replaced by a
dielectric, . r
C A A
d d
ε ε
= →ε
V
I I
R