I-7 Electric Energy Storage. Dielectrics.

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I-7 Electric Energy Storage. Dielectrics.
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Main Topics

• Electric Energy Storage.
• Inserting a Conductor into a Capacitor.
• Inserting a Dielectric into a Capacitor.
• Microscopic Description of Dielectrics
• Concluding Remarks to Electrostatics.

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Electric Energy Storage I

• We have to do work to charge a capacitor.
• This work is stored as a potential energy and all
  (if neglecting the losses) can be used at later
  time (e.g. faster to gain power).
• If we do any changes to a charged capacitor we do
  or the field does work. It has to be
  distinguished whether the power source is
  connected or not during the change.

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Electric Energy Storage II

• Charging a capacitor means to take a positive
  charge from the negative electrode and move it to
  the positive electrode or to take a negative
  charge from the positive electrode and move it to
  the negative electrode.
• In both cases (we can take any path) we are doing
  work against the field and thereby increasing its
  potential energy. Charge should not physically
  pass through the gap between the electrodes of
  the capacitor!

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Electric Energy Storage III

• A capacitor with the capacitance C charged by the
  charge Q or to the voltage V has the energy
• Up Q2/2C CV2/2 QV/2
• The factor ½ in the formulas reveals higher
  complexity then one might expect. By moving a
  charge between the electrodes we also change Q, V
  so we must integrate.

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Electric Energy Storage IV

• The energy density
• Let us have a parallel plate capacitor A,d,C,
  charged to some voltage V
• Since Ad is volume of the capacitor we can treat
  ?0E2/2 as the density of (potential) energy
• In uniform field each volume contains the same
  energy.

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Electric Energy Storage V

• In non-uniform fields energy has to be integrated
  over volume elements where E is (approximately)
  constant.
• In the case of charged sphere these volumes would
  be concentric spherical shells ( r gt ri )

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Electric Energy Storage VI

• Integrating from some R ? ri to infinity we get
• For R ri we get the same energy as from a
  formula for spherical capacitor.
• We can also see, for instance, that half of the
  total energy is in the interval ri lt r lt 2ri or
  99 of the total energy would be in the interval
  ri lt r lt 99ri

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Inserting a Conductor Into a Capacitor I

• Let us insert a conductive slab of area A and
  thickness ? lt d into the gap between the plates
  of a parallel plate capacitor A,d, ?0,?.
• The conductive slab contains enough free charge
  to form on its edges a charge density ?p equal to
  the original ?. So the original field is exactly
  compensated in the slab.
• Effectively the gap changed to d - ?.

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A Guiz

• Inserting a conductive slab of area A and
  thickness ? lt d into the gap between the plates
  of a parallel plate capacitor A,d, ?0,? will
  increase its capacitance.
• Where should we put the slab to maximize the
  capacitance ?
• A) next to one of the plates.
• B) to the plane of symmetry.
• C) it doesnt matter.

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C It doesnt matter !

• Let us insert the slab some distance x from the
  left plate. Then we effectively have a serial
  connection of two capacitors, both with the same
  A. One has the gap x and the other d-x-?. So we
  have

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Inserting a Conductor Into a Capacitor II

• The capacitance has increased.
• In the case of disconnected power source the
  charge is conserved and the energy decreases
  the slab would be pulled in.
• In the case of connected power source the voltage
  is conserved and the energy increases we do
  work to push the slab in and also the source does
  work to put some more charge in.

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Inserting a Dielectric Into a Capacitor I

• Let us charge a capacitor, disconnect it from the
  power source and measure the voltage across it.
  When we insert a dielectric slab we shall notice
  that
• The voltage has dropped by a ratio K V0/V
• The slab was pulled in by the field
• We call K the dielectric constant or the relative
  permitivity (?r) of the dielectric.
• ?r depends on various qualities T, f!

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Inserting a Dielectric Into a Capacitor II

• What has happened Since the inserted plate is a
  dielectric it contains no free charges to form a
  charge density on its edges, which would be
  sufficient to compensate the original field.
• But the field orientates electric dipoles. That
  effectively leads to induced surface charge
  densities which weaken the original field and
  thereby increase the capacitance.

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Inserting a Dielectric Into a Capacitor III

• The field orientates electric dipoles their
  charges compensate everywhere except on the edges
  next to the capacitor plates, where some charge
  density ?p lt ? remains.
• The field in the dielectric is then a
  superposition of the field generated by the
  original ? and the induced ?p charge densities.
• In the case of homogeneous polarization the
  induced charge density ?p P which is so called
  polarization or the density of dipole moments.

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Inserting a Dielectric Into a Capacitor IV

• Inserting dielectrics is actually the most
  effective way to increase the capacitance. Since
  the electric field decreases, the absolute
  breakdown charge increases.
• Moreover for most dielectrics their breakdown
  intensity (or dielectric strength) is higher than
  that of air. They are better insulators!

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Energy Density in Dielectrics

• If we define the permitivity of a material as
• ? K?0 ?r?0
• and use it in all formulas where ?0 appears .
• For instance the energy density can be written as
  ?E2/2.
• If dielectrics are non-linear or/and non-uniform
  their description is considerably more
  complicated!

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Capacitor Partly Filled with a Dielectrics

• If we neglect the effects near the edges of the
  dielectrics, we can treat the system as a serial
  or/and parallel combination of capacitors,
  depending on the particular situation.

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Concluding Remarks To Electrostatics

• We have illustrated most of things on very
  simplified examples.
• Now we know relatively deeply all the important
  qualitative principles of the whole
  electrostatics.
• This should help us to understand easier the
  following parts ad well as functioning of any
  device based on electrostatics!

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Homework

• The homework from yesterday is due tomorrow!

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Things to Read

• This lecture covers
• Chapter 24 4, 5, 6
• Advance reading
• Chapter 25 1, 2, 3, 5, 6

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Charging a Capacitor

• Let at some point of charging the capacitor C
  have some voltage V(q) which depends on the
  current charge q. To move a charge dq across V(q)
  we do work dEp V(q)dq. So the total work to
  reach the charge Q is

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Polarization ? Dipole Moment Density I

• Let us take some volume V which is small in the
  macroscopic scale but large in the microscopic
  scale so it is representative of the whole sample

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Polarization ? Induced Surface Charge Density II

• Let a single dipole moment p lq be confined in
  a prism of the volume v al. A volume V of the
  uniformly polarized dielectric is built of the
  same prisms, so the polarization must be the same
  as in any of them

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Polarization III

• The result field in the dielectric

We can express the original charge density
So the original field is distributed to the
result field and the polarization according to
the ability of the dielectric to be polarized.
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Polarization IV

• In linear dielectrics is proportional to the
  result field . They are related by the
  dielectric susceptibility ?

The result field E is K times weaker than the
original field E0 and can also define the
permitivity of a dielectric material as ?.