Download CHAPTER 24: CAPACITANCE AND DIELECTRICS • Suppose you

CHAPTER 24: CAPACITANCE AND DIELECTRICS
 Suppose you have two isolated conductors that are insulated from each other
and initially they are both neutral
 Now imagine moving a small quantity of charge from one conductor to the
other so that one has a net positive charge and the other a net negative
charge
 This creates an electric field and, therefore, a potential difference between
the two conductors
 Now, if we transfer more charge between the conductors we must do work
to move charge through this potential difference
 The more charge we transfer, the more difficult it gets to move additional
charge
 The total work done in moving all of the charge is stored as electrical
potential energy of the resulting charge configuration
 Devices that store electrical potential energy using a pair of oppositely
charged conductors are called capacitors and are used in most electronic
devices such as radios, TVs, computers, etc. When you tune a radio to
change stations, you are adjusting a capacitor.
 Typically capacitors are used for short-term energy storage where it is
necessary to release electrical energy very quickly such as in the flash of a
camera
 For a particular capacitor, it turns out that the ratio of the amount of charge
on each conductor to the potential difference between the conductors is a
constant called the capacitance. This ratio depends on the geometry of the
capacitor as well as the type of insulating material (could be a vacuum)
between the charged conductors.
CAPACITORS AND CAPACITANCE
 In an uncharged capacitor both conductors are neutral. You charge the
capacitor by transferring charge (usually electrons) from one conductor to
the other. An electric field is created as a result of the separation of positive
and negative charge and potential energy is said to be “stored in the field”.
 The two conductors have the same amount of excess charge but with
opposite sign – the net charge of the whole capacitor is zero
 Since there is an electric field in charged capacitor there is a potential
difference between the two conductors
 As the amount of charge on the conductors increases, the electric field and
potential difference between them also increases. Since 𝐸 ∝ 𝑄 and,
therefore, ∆𝑉 ∝ 𝑄, the ratio of the amount of charge on the conductors to the
potential difference between them is constant – the value of this constant is
the capacitance C, and quantifies the ability of a capacitor to store energy.
𝑄
 𝐶 = |∆𝑉| is the definition of capacitance for a given capacitor (units: C/V
which is defined to be a farad, abbreviated F)
Calculating Capacitance
 We will start by considering capacitors where the empty space between the
conductors can be treated as a vacuum and then we will consider the effects
of inserting an insulating material between them
 The simplest and most important capacitor design is the parallel-plate
configuration. This arrangement consists of two parallel conducting plates
with area A and separated by a distance d, where 𝑑 ≪ √𝐴.
 Using Gauss’s law one can show that the electric field between two
conducting plates with planar symmetry is E = σ/ε0 (see Example 22.8, p
736) and the charge is distributed uniformly over the surface of both plates
so σ = Q/A and E = Q/(ε0A)
 The field is uniform between the plates so the potential difference is |∆𝑉| =
𝐸𝑑 =
𝑄𝑑
𝜀0 𝐴
𝑄
. And the capacitance is 𝐶 = |∆𝑉| =
𝜀0 𝐴
𝑑
. As claimed earlier, C only
depends on the geometry and is independent of Q or |ΔV|. This equation also
shows that we can also express the units for the constant 𝜀0 as F/m.
**SEE EXAMPLE #1**
CONNECTING CAPACITORS
 Often a single capacitor with a desired combination of capacitance and
working voltage is not available. But this combination can be achieved by
connecting two or more capacitors in series and/or parallel.
Capacitors in Series
 Figure (a) above shows two capacitors connected in series (one after the
other with no junction in between)
 Imagine a battery with voltage |∆𝑉| is connected to them completing the
circuit. The end of the battery at the higher potential puts +Q on the upper
plate of C1 and, as a result, the lower plate of C2 acquires a net charge of –Q
because of the principle of conservation of charge.
 The positive charge on the upper plate of C1 attracts –Q to the lower plate of
C1 and the negative charge on the lower plate of C2 attracts +Q to the upper
plate of C2. Therefore we can conclude that…
 Capacitors connected in series carry the same magnitude of charge on
each plate
 Capacitors in series have the same amount of charge on all of their plates but
the potential difference across each capacitor is not necessarily the same
unless the capacitors are identical (same value of C)
 I will denote the potential difference across C1 as V1 and the potential
difference across C2 as V2. The electric fields in the two capacitors point in
the same direction so the total voltage across the series combination (which
is the potential difference across the equivalent capacitor) is V = V1 + V2.
From the definition of capacitance, we also have V1 = Q/C1 and V2 = Q/C2.
 Therefore, V = Q/C1 + Q/C2 = Q(1/C1 + 1/C2) and V/Q = 1/C = 1/C1 + 1/C2
 So, the equivalent capacitance (shown in Figure (b) above) of two capacitors
connected in series is given by:
1
𝐶𝑒𝑞
=
1
𝐶1
+
1
𝐶2
 This result extends to any number of capacitors connected in series
1
1
1
1
= + + +⋯
𝐶𝑒𝑞 𝐶1 𝐶2 𝐶3
 This equation means that when capacitors are connected in series, the
equivalent capacitance is always less than any individual capacitance
 Physically, two or more capacitors with equal plate areas connected in series
effectively adds their plate separations, producing a smaller overall
capacitance
Capacitors in Parallel
 Figure (a) above shows two capacitors connected in parallel. Two capacitors
are connected in parallel if they share two common junctions. Or, another
way of saying this – if two capacitors are connected at both ends, then they
are connected in parallel
 We are treating the wires as ideal conductors so there is no potential
difference across them. Therefore the potential difference across each
capacitor is equal to the applied potential difference V on the left (e.g., a
battery) because they are both connected directly to V. However the charges
Q1 and Q2 on the two capacitors are not necessarily equal unless the
capacitors are identical (same value of C)
 Capacitors connected in parallel have the same potential difference
across their respective plates and this is equal to the potential difference
of the source voltage that is charging them
 The magnitudes of the charges on the plates of the two capacitors are given
by the definition of capacitance: Q1 = C1V and Q2 = C2V
 The total charge Q of the parallel combination (which is the charge on the
equivalent capacitor) is Q = Q1 + Q2 = (C1 + C2)V and Q/V = C = C1 + C2
 So, the equivalent capacitance (shown in Figure (b) above) of two capacitors
connected in parallel is given by: 𝐶𝑒𝑞 = 𝐶1 + 𝐶2
 This result extends to any number of capacitors connected in parallel
𝐶𝑒𝑞 = 𝐶1 + 𝐶2 + 𝐶3 + ⋯
 This equation means that when capacitors are connected in parallel, the
equivalent capacitance is always greater than any individual capacitance.
 Physically, two or more capacitors with equal spacing between their plates
connected in parallel amounts to adding their plate areas, producing a larger
overall capacitance
**SEE EXAMPLE #2**
Energy Stored in a Capacitor
 The electric potential energy U stored in a charged capacitor is equal to the
amount of work W needed to charge it
 When the capacitor is discharged the stored energy is converted into work
done by electrical forces
 We need to calculate the work needed to assemble charges from 0 to Q
increasing the potential difference from 0 to V across the two conductors
 The potential difference across the capacitor is proportional the amount of
charge currently on the capacitor: V(q) = (1/C)q
 dU = V(q)dq = (1/C)qdq
1
𝑄2
𝑄
1
1
 𝑈 = ∫ 𝑉(𝑞)𝑑𝑞 = ∫0 𝑞𝑑𝑞 = = 𝐶𝑉 2 = 𝑄𝑉 where the last two
𝐶
2𝐶
2
2
equalities follow from the definition of capacitance 𝑄 = 𝐶𝑉
 These relations indicate that capacitance is a measure of the capacitor’s
ability to store both energy as well as charge
**SEE EXAMPLE #3**
Energy Stored in the Electric Field
 Moving charges from one plate to the other in a parallel-plate capacitor
(charging the capacitor) requires doing work against the electric field
between the plates and this is converted into potential energy which, as they
say, is “stored in the field”.
 In general, if the work done in moving the charges is positive, then the new
field configuration this produces contains more energy; if the work done in
moving the charges is negative, then the new field configuration this
produces contains less energy
 Earlier we concluded that since the field between the plates is uniform, the
potential difference is 𝑉 = 𝐸𝑑 =
𝑞𝑑
𝜀0 𝐴
. As we did above, the energy stored in
moving the charge against the field is given by integrating the electric
potential over the charge being transferred: 𝑈 = ∫ 𝑉(𝑞)𝑑𝑞 =
𝑑
2𝜀0 𝐴
𝑄
𝑑
𝜀0
𝑄
∫ 𝑞 𝑑𝑞 =
𝐴 0
2
 Every electric field represents stored energy and the amount of energy
should depend on the field strength. Since the magnitude of the electric field
can vary with position, the amount of energy is usually expressed in terms of
an energy density u, which is stored energy per unit volume (units: J/m^3)
 For a parallel-plate capacitor, the electric field strength is given by 𝐸 =
so 𝑄 = 𝜀0 𝐴𝐸 and 𝑈 =
𝑑
2𝜀0 𝐴
1
(𝜀0 𝐴𝐸)2 = 𝜀0 𝐸 2 𝐴𝑑
2
𝑄
𝜀0 𝐴
,
 The volume of the region between the plates is the plate area times their
separation, which is 𝐴𝑑. Thus, the energy density is 𝑢 =
1
 The electric energy density (in a vacuum) is: 𝑢 = 𝜀0 𝐸
𝑈
𝐴𝑑
2
2
 Even though this result was obtained for the specific case of a parallel-plate
capacitor, it is a universal expression that is valid for any electric field in a
vacuum
**SEE EXAMPLE #4**
DIELECTRICS
 So far we have been treating the space between the conductors as a vacuum.
However, most capacitors have a nonconducting material (called a
dielectric) between their plates. A dielectric is an insulator that can be
polarized by an applied electric field – the molecules making up these
materials behave as dipoles in the presence of a field.
 Inserting a slab of dielectric material between the plates of a capacitor serves
a few purposes – it maintains the physical separation between the plates, it
minimizes any charge leakage, its molecular properties influence the
capacitance, and it allows the capacitor to be able to sustain a higher
potential difference between the plates which allows it to store larger
amounts of charge and energy than without the presence of the dielectric
 In Chapter 21, we learned that a dipole in a uniform electric field
experiences no net force, however the field does exert a net torque on the
dipole, causing the dipole moment vector to rotate to the stable equilibrium
position where it is aligned with the direction of the field
 Therefore, when a dielectric is inserted between two charged capacitor
plates, the dipole moments associated with the molecules of the dielectric
tend to align with the electric field between the plates (see the rightmost
panel above)
 This dipole moment is either intrinsic to the molecules themselves or it is
due to a separation of charge induced in them by the field between the
plates.
 In Chapter 21, we defined the dipole moment vector as having magnitude qd
and points from the negative charge towards the positive charge, therefore
the electric field between the charges of the dipole points in the opposite
direction. This direction is also opposite the electric field between the
charged plates of the capacitor.
 The result is a net reduction in the electric field between the plates. V = Ed
for a uniform field, so there is corresponding reduction in the electric
potential as well.
 If the capacitor is not connected to anything, the charges on the plates have
nowhere to go, which means Q remains constant and therefore the
capacitance C = Q/V must increase
 The capacitance with nothing between the plates (vacuum – see Figure (a)
above) will be denoted C0 = Q/V0 and the capacitance with the dielectric
present is C = Q/V (see Figure (b)).
 As we have just argued, C must be greater than C0. We now define what is
known as the dielectric constant of the material, 𝐾 =
𝐶
𝐶0
. Defined in this
way, K > 1 and K = 1 when no dielectric is present (vacuum).
 When the charge on the plates is constant, Q = C0V0 = CV so C/C0 = V0/V.
Therefore, 𝑉 =
𝑉0
𝐾
when Q is constant – the electric field is reduced by the
same factor since the field is uniform (𝐸 =
𝐸0
𝐾
).
 When a dielectric (insulator) is placed in an electric field, its charges are
immobile (as discussed in Chapter 21), nonetheless each atom polarizes
slightly
 Charge polarization occurs because each atom’s negative electron cloud and
positive nucleus shift very slightly in opposite directions in response to the
applied electric field
 As a result, an induced positive charge builds up on one surface of the
insulator, and an induced negative charge builds up on the other surface.
This is shown in the figure above.
 To summarize, a capacitor’s electric field polarizes the dielectric which
causes the dielectric to create an electric field of its own, which is opposite
in direction to the capacitor’s field, resulting a net reduction of the field and
potential inside the capacitor
 For most materials, as long as the applied field is not too strong, the induced
surface charge on the dielectric is directly proportional to the field induced
inside of it
 Let σi be the magnitude of the induced surface charge density on the
dielectric and σ is the surface charge density already on the conductor plates.
Then the magnitude of the net surface charge density on either side of the
capacitor is σnet = (σ – σi). As in Example 22.8, p 736, the field between two
conducting plates is given by E = σnet/ε0. So, without the presence of the
dielectric, the electric field between the plates is E0 = σ/ε0. When the
dielectric is inserted between the plates, the field becomes E = (σ – σi)/ε0.
 Using the fact that 𝐸 =
𝐸0
𝐾
, we can relate the surface charge density on the
capacitor plates to the induced charge density on the dielectric:
1
𝜎i = 𝜎 (1 − )
𝐾
 So, when the dielectric constant is very large the two charge densities are
approximately equal and the field inside the capacitor is very nearly
cancelled by the field that it induces inside the dielectric
 We now define the permittivity of the dielectric as ε = Kε0 (remember that
ε0 is the permittivity of a vacuum)
𝜎
 Since 𝜎 − 𝜎i = , the field between the plates in the presence of the
𝐾
dielectric is given by 𝐸 =
𝜎
𝐾𝜀0
=
𝜎
𝜀
 Using ε we can also write the capacitance in the presence of the dielectric as
C = KC0 = Kε0(A/d) = ε(A/d)
 Similarly, we can write the electric energy density as u = ½εE2
 In a vacuum, ε = ε0 and these relationships reduce their vacuum counterparts
**SEE EXAMPLE #5**
 So, for a given capacitor, as you increase the amount of charge on the plates,
there is a corresponding increase in the potential difference such that their
ratio (the capacitance C) remains constant. However, these two quantities
cannot increase indefinitely, for two reasons.
 A capacitor cannot store arbitrarily large amounts of charge because all
capacitors “leak” charge due to their imperfect insulation.
 Additionally, every capacitor has some maximum voltage rating called the
capacitor’s “working voltage”. Beyond this value, the electric field becomes
strong enough that it can cause dielectric breakdown – electrons are torn
from their molecules and collide with other molecules knocking out even
more electrons and the dielectric essentially becomes a conductor.
 Lightning is an example of dielectric breakdown in the air
**SKIP NEXT SECTION**
Gauss’s Law in Dielectrics
 The figure above shows a close up of the left hand side of the capacitor. To
apply Gauss’s law, we choose a closed rectangular prism, shown in the
“perspective view” above, as our gaussian surface. We position it such that it
lies half in the conductor and half in the dielectric as shown in the “side
view” above.
 The net electric field is perpendicular to the conducting plates, so there is no
flux through four of the sides of the gaussian surface, only through the left
and right ends. In fact, the left end is inside the conductor where the electric
field is zero, so there is only flux through the right end which has area A.
 The electric field is everywhere uniform and parallel to the area vector on
this end, so the total flux through the gaussian surface is EA
 To compute the total charge enclosed by this surface, we need to multiply
the net surface charge density by the cross-sectional area of the gaussian
surface, A
 So, Qenc = (σ – σi)A and Gauss’s law says that EA = (σ – σi)A/ε0. σi is the
charge density induced on the dielectric which is an unknown and E is the
net field between the plates which, without knowing σi, is an unknown as
well. We have two unknowns so we need another equation.
 Earlier we derived a relation between the charge density on the plates and
𝜎
the induced charge density on the dielectric: 𝜎 − 𝜎1 =
 Substituting this into Gauss’s law, we have 𝐸𝐴 =
𝜎𝐴
𝐾𝜀0
𝐾
or 𝐾𝐸𝐴 =
𝜎𝐴
𝜀0
 What the 2 relationship says is that, instead of the flux of E, it is the flux of
KE that is proportional, not to the enclosed net charge but, rather, the
enclosed free charge
 The free charge is associated with σ, the existing charge density on the
conducting plates. While the bound charge is associated with σ1, the charge
density induced on the dielectric.
 Thus, in the presence of an insulating material (not a vacuum), Gauss’s law
⃗⃗ ⋅ 𝑑𝑨
⃗⃗ = (𝑄𝑓𝑟𝑒𝑒 )𝑒𝑛𝑐
should be altered as follows: ∮ 𝐾𝑬
nd
𝜀0
 (𝑄𝑓𝑟𝑒𝑒 )𝑒𝑛𝑐 is the free charge on the conductor(s) that is enclosed by the
gaussian surface
⃗⃗ ⋅ 𝑑𝑨
⃗⃗ = (𝑄𝑓𝑟𝑒𝑒)𝑒𝑛𝑐 is the way I have usually seen Gauss’s law for
 ∮𝑬
𝜀
dielectrics written, which agrees with the equation above since 𝐾𝜀0 = 𝜀