Unformatted text preview: 6. Lecture 6
6.1 Energy contained in a capacitor
We said that a capacitor stores energy. We now compute how much energy it actually
contains. A way to do that is to consider how much work we need to separate the two
plates that form the capacitor. Suppose the left plate is ﬁxed and we move the right
one by a distance ∆d (see ﬁg.31). The electric ﬁeld produced by the other plate is
E  = 2σ0 where σ = Q . The force is then
ε
A
F  = σ
Q
2 ε0 (6.1) The work we need to do to separates the plates by a distance ∆d is
σ
W = ∆U = Uﬁnal − Uinitial = F ∆d =
Q∆ d
2 ε0 (6.2) From here we ﬁnd that
U= σ
1
d
1 Q2
Qd = Q2
=
2 ε0
2 Aε0
2C (6.3) where we remembered that the capacity is
C= Aε0
d (6.4) Using that Q = C ∆V we ﬁnd the equivalent expressions for the energy contained in
the capacitor:
1 Q2
1
1
U=
= C ( ∆ V ) 2 = Q∆ V
(6.5)
2C
2
2
6.2 Dielectrics
We mentioned before that between the plates in a capacitor we include an insulator.
This can actually modify the properties of the capacitor and increase the capacity.
The reason is that insulators are generically dielectric. This means that their electrical
properties can be understood as if they are made out of tiny dipoles. The dipoles
align themselves with the electric ﬁeld decreasing its value in a phenomenon known as
screening. If we consider a single charge inside a dielectric the screening phenomenon
is illustrated in ﬁgure 32. Notice that outside the dielectric the ﬁeld is the same from
symmetry and Gauss law. Experimentally the result is that the electric ﬁeld inside is – 40 – d ∆d V2 V1
Q E −Q Figure 31: The energy stored in a capacitor can be computed by doing a small displacement
of the plates by ∆d and computing the work needed to do that. suppressed by a factor ε0 /ε where is known as the permitivity of the medium. The
electric ﬁeld inside the dielectric is then
1Q
E=
r
ˆ
4πε r2 (6.6) outside is again 1Q
r
ˆ
(6.7)
4πε0 r2
In the case of the capacitor it works the same, we just need to replace ε0 → ε. Therefore
the capacity when there is a dielectric in between the plates is
E= C= Aε
d Typical values for ε/ε0 are
ε
Material
ε0
Vacuum
1 (deﬁnition)
Air
1.0006
Teﬂon
2.1
Polyethylene
2.25
Paper
3.5
Water
80 – 41 – (6.8) We see that air and vacuum are very close to each other which is why we ignore
the diﬀerence. It also shows that if we want to store more energy in a capacitor, at the
same voltage it is convenient to use a dielectric of high dielectric constant.
+ + − − +
+ + + +
+
− + −
− + + + − − + − − − + + − − + + + − − − −
− −
−
− + + + + − + + − + + − − − + +
+ − − + − + + + − − − + + − − + + −
− + + −
+ − +q +
+ + +
− − − −
−
− + −
+ + − − − + − + − + + + − − −
+ + + − − −
− −
− − +
− +
−
+ +
+ − + −
+
+
− + + − Figure 32: A charge in a dielectric is partially screened by the dipoles aligning with the
electric ﬁeld. Due to temperature and other variables the alignments is partially random.
As a result the ﬁeld inside the dielectric is smaller that without the medium. Outside the
medium, in this case, the electric ﬁeld is the same with or without the dielectric. This is
because the conﬁguration is spherically symmetric and using Gauss’s law, the electric ﬁeld is
given by the total charge enclosed in a surface (the dielectric has no net charge). 6.3 RC circuits
Now that we know what a capacitor is we can see how to charge it and what applications
we can ﬁnd for it. The simplest way to charge it is with a battery and through a resistor.
One might think the resistor is not necessary but not matter what we do there is some
resistance in the circuit so we need to include it. Also, sometimes we do not want to
charge the capacitor as fast s we can and, as we are going to see, a resistor makes
the charging process slower. When the contact is closed, current starts to circulate
but it cannot go through the capacitor so charge starts to accumulate in it. Initially – 42 – the current is just I = ∆V but as the capacitor charges a potential diﬀerence appears
R
and the current is smaller. In the ﬁnal stage, the capacitor is completely charged, the
potential diﬀerence between its terminals is the same as the battery (but opposes it)
and no more current circulates. An estimate of the time the capacitor takes to charge
is obtained by dividing the total charge by the initial current:
τ= Q
C ∆V
=
= RC
I
∆V /R (6.9) This is a characteristic time of the circuit. The larger the resistance and the capacity
the larger the time it takes to charge. We can then discharge the capacitor through a
smaller resistance for example and it will release its energy in a much shorter interval.
An application is the ﬂash in a photographic camera. The battery with relatively large
internal resistance charges a capacitor and then it is suddenly discharged though the
ﬂash. Another typical application is in timing circuits. For example oscillators etc. A
resistor of variable resistance is inserted and the time of charge and discharge is used
to determine the period of the oscillations. It is not very precise so it can be used only
when the exact timing is not crucial. Another important application is in computer
memories. A charged capacitor for example can represent a 1 and a discharged one
a 0. In that case we want the capacity small to be able to charge it and discharge it
fast. However if we do not read the memory current will leak and the capacitor will be
discharged. For that reason when the computer is turned oﬀ what was in the memory
disappears. When it is working a special circuit continuously reads the memory and
rewrites it before the capacitors have time to discharge (these memories are called
DRAM and are presently the most common ones).
We can analyze the circuit in slightly more detail by deﬁning some points (a), (b),
(c) as in ﬁgure 33 and considering the voltages at those points Va , Vb , Vc . Using Ohm’s
law ∆V = IR for the resistor and Q = C ∆V for the capacitor we have
Vb − Vc = ∆Vresistor = IR
Q
Va − Vb = ∆Vcapacitor =
C
Va − Vc = ∆Vbattery (6.10)
(6.11)
(6.12) If we add the ﬁrst two we have
Q
= Va − Vc = ∆Vbattery
C
So the current circulating through the circuit is
1
Q
I=
∆Vbattery −
R
C
Vb − Vc + Va − Vb = IR + – 43 – (6.13) (6.14) Since ∆VBattery is ﬁxed we see that initially, when the capacitor is not charged (Q = 0)
1
the current is I = R ∆VBattery . The current starts to charge the capacitor, Q increases
and I decreases but keeps charging the capacitor until Q = C ∆Vbattery which gives
I = 0. If we plot the voltage across the capacitor and the current we get a plot as in
ﬁg.34. C
+
+
+
+ −
−
−
− (a) I
+ (b) −
(c) ∆V R Battery Figure 33: A capacitor is usually charged with a battery and through a resistor. The time
it takes to charge depends on the capacity and the resistor. It is given by ∆t = RC and is a
characteristic time of this type of circuit. – 44 – Figure 34: Time dependence of the voltage and current across the capacitor for the RC
circuit in arbitrary units. It illustrates the general behavior of these quantities. – 45 – ...
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This note was uploaded on 12/07/2011 for the course PHY 219 taught by Professor Na during the Fall '11 term at Purdue.
 Fall '11
 NA
 Energy, Work

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