23. Electrostatic Energy and Capacitors

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Topics

Capacitors

Electrostatic Energy

Using Capacitors

Capacitors

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Capacitors

A capacitor is a device that stores charge –Q on one conductor and charge +Q on the other conductor

The stored chargecreates an electricfield, and therefore,a potential difference between the conductors

–Q+Q

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Capacitance

QC

V

The capacitance of a device is defined by

The unit is the farad (F) = Coulomb/VoltBut since the farad is such a huge unit, the morecommonly used units are F = 10-6 F or pF = 10-12 F

Q is the charge stored on a conductorV is the potential difference betweenthe conductors

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CapacitanceSpherical Conductor

QV k

R

The potential on the surface of aspherical conductor of radius R is

04Q R

C RV k

Therefore, its capacitance is

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CapacitanceParallel Plate Capacitors

If two conducting plates ofarea A are separated by a small distance d the electric field between them will beapproximately constant and of magnitude

0 0/ /( )E Q A

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CapacitanceParallel Plate Capacitors

0 0( / ) /( )

V E

Q

d

d Ad

Since the electric field isconstant, the potential difference between the plates is simply

so the capacitance is

0Cd

A

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CapacitanceCylindrical Capacitors

0

12

2

ln( / )RC

R

L

A coaxial cable of length L is an example of a cylindrical capacitor

R2

R1

Electrostatic Energy

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Electrostatic Energy

q1

q2

q3

Total work done1 3 2 31 2

2 3

kq q kq qkq qW W W

a a a

2 2 1 21W q q

kqV

a a

1 23 3 1 2 3( ) ( )W q q

kq kqV V

a a

aa

Work is required to assemble a charge distribution

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Electrostatic Energy

dW dqV

W dW Vdq

dq

The work dW required to add an element of charge dqto an existing charge distribution is

where V is the potential at the finallocation of the charge element. The

total work required is thereforeSince the electric is conservative, the work is stored as electrostatic energy,U.

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Storage of Electrostatic Energy

Work must be done to move positive chargefrom a negatively charged conductor to onethat is positively charged. Or to move negative charge in the reverse direction.

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Storage of Electrostatic Energy

In moving charge dq, the electrostaticenergy of the capacitor is increased by dU Vdq

2

0

1

21

2

Q q QU dq

C C

QV

Therefore,

21

2U CV

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Energy Density of Electric Field

1

2U VQ

0/( )E Q A

20 0

1 1( )( ) ( )

2 2EdE dA AU E

V Ed

Potential energy

Electric field

Electric potential

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/( )E

energyu U

vold

umeA

The energy density uE

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1

2Eu EThis expressionholds true forany electric field

Energy Density of Electric Field

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Example – A Thunderstorm

How much electrical energy is stored in a typical thundercloud?

Assume a cloud of

height h = 10 km,

radius r = 10 km,

with a uniform

electric field

E = 105 V/m.

http://redcrossggr.files.wordpress.com/2008/06/thunderstorm.jpg

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Example – A Thunderstorm

Narrative

The problem is about stored electric energy.

Since the electric field, E, is uniform, so to is the energy density uE = ½ 0 E2 in the cloud. Therefore, the electric energy stored in the cloud is just the electric energy density times the volume of the cloud.

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Example – A Thunderstorm

Diagram

Thundercloudh = 10 km

r = 10 km

volume = r2 h

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Example – A Thunderstorm

Calculation

The electric energy density in the cloud is

uE = ½ 0 E2 = 4.4 x 10-2 J/m3.

The volume of the cloud is, v = r2 h, that is,

v = 3.1 x 1012 m3.

Therefore, the total electric energy stored in the cloud is U = uEv = 140 GJ.

Using Capacitors

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The Effect of Dielectrics

Michael Faraday1791 – 1867

wikimedia

Michael Faraday discovered that the capacitance increases when the space between conductors is

replaced by a dielectric.

Today, we understand this to be a consequence of the polarization of molecules.

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The Effect of Dielectrics

The polarized molecules of the dielectric tend toalign themselves parallel to the electric field createdby the charges on the conductors

b b

--------

++++++++

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The Effect of Dielectrics

The bound charge b induced on the surface of the dielectric creates an electric field opposed to the electric field of the free chargef on the conductors,thereby reducing thefield between them.

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The Effect of Dielectrics

The reduction in electric field strength from theinitial field E0 to the reduced field E is quantifiedby the dielectric constant (kappa)

0EE

The dielectric increases the capacitance by the same factor .

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The Effect of Dielectrics

For a parallel plate capacitor, with a dielectricbetween the plates, the electric field is

0 is called thepermittivity

The product of the dielectric constant and thepermittivity of free space 0

0/( )E Q A

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Capacitors in Parallel

At equilibrium, the potential across each capacitor is the same, namely, 12 V

same potential

same potential

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The twocapacitorsare equivalentto a single capacitor withcapacitance

1 2C C C

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Capacitors in Series

The sum of thepotentials across both capacitors will be equal to 12 V

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The potential V1

across C1 plus thepotential V2 across C2 is equal to thepotential difference V between points a and b:V = V1 + V2

1 21/ 1/ 1/C C C

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Summary

Capacitance C = Q / V (farad)Parallel plate C = 0 A/d

CapacitorsIn parallel C = C1 + C2

In series 1/C = 1/C1 + 1/C2

Stored energy U = ½ QV Energy density uE = ½ 0 E2

Effect of dielectric E = E0 /