Chapter 26 Capacitance and Dielectrics

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Chapter 26 Capacitance and Dielectrics

• PHYS 2326-11

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Concepts to Know

• Capacitors
• Capacitance
• Energy Density
• Dielectric
• Dielectric Constant
• Permittivity
• Series and Parallel Circuits

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Capacitance

• Capacitance is the ability for an object to hold
  a charge for a given potential energy
• C Q/?V
• C is the capacitance in Farads
• that can hold a charge Q with an increase in
  potential of ?V volts.
• 1 Farad 1 coulomb of charge per Volt
• This is a huge value

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Capacitors

• Youre responsible for knowing
• Parallel Plate Capacitors
• Spherical Capacitors
• Cylindrical Capacitors
• Typical values range from a few pico farads to a
  few thousand micro farads

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Parallel Plate Capacitor

• most common type of component although most are
  rolled up or contain many plates
• C eo A/d
• Capacitance is proportional to area /inverse
  proportional to separation

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Energy Stored in Capacitor

• Work done to charge a capacitor given by eqn
  26.11 which is the integral of the dw required
  for each dq of charge added
• U ½ QV ½ C V2 Q2 /2C
• Eqn 26.13 is the concept of energy density of the
  electric field
• u ½ eo E2

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Dielectrics

• When nonconducting material is inserted in a
  capacitor, in between the plates the voltage
  drops.
• That means since CQ/V that C increases
• C ?Co
• e ? eo permittivity of free space
• The equations for capacitors are the same except
  using the permittivity of the material

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Dielectric Materials

• Dielectrics have either polar or non polar
  molecules. In the electric field the non polar
  molecules tend to become polarized
• The electric field is E Eo/? s/?eo between 2
  plane sheets
• E becomes the difference between the original Eo
  and E induced (opposite direction to Eo )
• Study examples 26.7 8

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Capacitor Circuits

• 2 capacitors in parallel ?V is the same and the
  total charge is the sum of each
• C C1 C2 .
• 2 capacitors in series Q is the same and ?V is
  the sum of the individual ?Vs.
• 1/C 1/C1 1/C2
• Parallel Series

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Solving for Equivalents

• Reduce the number of real capacitors down to
  equivalent capacitors, combining simple parallel
  and series combinations

C6
C1
C2
C5
a
c
a
a
C3

C3
b

C4
b
b
a
C4

C4
C7
b
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Example 1

• 4 capacitors arranged in previous slide
• potential is 10V above ground for a, ground for
  b. Find a) voltage and charge on each capacitor
  b) find voltage on pt c wrt ground.
• C1 12nF, C2 24nF, C3 22nF, C415nf

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Example 1

• C1 and C2 in series combine to a C5 1/C5
  1/C1 1/C2 8 nF
• C3 and C5 are parallel combine to C6 C6 C5
  C3 30nF
• C6 and C4 are series combine to C7 1/C7 1/C6
  1/C5 10nF
• V for C7 10V, QCV (10nF)(10) 100nC

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Example 1

• A) V7 V Q7 C7V7 Q7 100nC
• Q6 Q7 Q6 C6V6 V6 3.33V
• Q4Q7 Q4C4V4 V46.667V
• V3 V6 Q3C3V3 Q3 73.33nC
• V5 V6 Q5C5V5 Q526.67nC
• Q2Q5 Q2C2V2 V21.111V
• Q1Q5 Q1C1V1 V12.22V
• b) VcV2V4

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Example 2

• A parallel plate capacitor made from 2 squares of
  metal, 2mm thick and 20cm on a side separated by
  1mm with 1000V between them
• Find
• a) capacitance b)charge per plate c) charge
  density d)electric field e) energy stored f)
  energy density

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Example 2

• Given
• L 20cm 0.2m
• d 1mm 1E-3m
• V 1E3 V
• epsilon 8.85E-12 F/m
• Equations
• Ceo A/d, AL2 , QCV, s Q/A, V-Ed, U1/2
  QV, u U/Ad 1/2 eo E2

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Example 2

• a) C eo A/d (8.85E-12)(0.2)2/(1E-3) 0.354nF
• b) QCV (0.354nF)(1E3V) 0.354 µC
• c) s Q/A (0.354 µC)/(0.04 m2 ) 8.85E-6 C/
  m2
• d) VEd, E V/d 1000/0.001 1.0E6 V/m (this
  is a magnitude)
• e) U (1/2)(0.354E-6)(1000)1.77E-4J
• f) u (1/2)(8.85E-12)(1.0E6) 2 4.425J/m3

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