Transcript Electric Potential and Capacitance

Electric Potential and
Capacitance
What’s a volt anyway?
Why Potential?
Energy makes some problems easier to
solve
 Energy is a scalar
 The concept of a potential is
useful to have in our physics
toolbox
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Define Change in Electric
Potential Energy
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When a positive charge released near the
positive plate moves to the right it loses
potential energy equal in magnitude to the
work done by the field on the charge
Positive
plate
Negative
plate
b
a
Electrical Potential Energy
The change in potential energy equals the
negative of the work done by the field
 Electrical PE changes to kinetic
 The positive charge has its greatest PE near
positive plate
 Only differences in potential energy are
measurable
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Electrical Potential
Potential is potential energy per unit charge
 Analogous to field which is force per unit
charge
 Symbol of potential is V;
Va = PEa/q
 Only potential differences are measurable;
zero point of potential is arbitrary
 Vab = Va – Vb = - Wba/q
 Wba is work done to move q from b to a
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Units
Unit of electric potential is the volt
 Abbreviation V
 1 V = 1 Joule/Coulomb = 1 J/C
 Thus electrical work = qV
 Potential difference is called voltage
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V = 0 Arbitrary
Usually ground is zero point of potential
 Sometimes potential chosen zero at infinity
 + terminal of 12V battery is
said to be at 12V higher
potential than – terminal
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Electric Potential and Potential
Energy
DPE = PEb – PEa = qVba
If object with charge q moves through a potential
difference Vba its potential energy changes by qVba
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Example: What is the gain of electrical PE when 1
C of charge moves between the terminals of a 12
volt battery? 12 joules
Water Analogy:
Voltage is like water pressure
Example: Electron in Computer
Monitor
An electron is accelerated from rest through
a potential difference of 5000 volts
 Find its change in potential energy
 Find its speed after acceleration
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Vba = 5000 v = Vb - Va
a
b
Example: Electron in Computer
Monitor
An electron is accelerated from rest through
a potential difference of 5000 volts
 Find its change in potential energy
 Find its speed after acceleration
 PE = qV = 1.6x 10-19 C x 5 x 103 V =
8 x 10-16 J = ½ m v2
 m = 9.11 x 10-31 kg ; v = 4.2 x 107 m/s
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Questions on Preceding Example
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Does the energy depend on the particle’s
mass?
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Does the final speed depend on the mass?
Electric Potential and Electric
Field
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Can describe charge distribution in terms field or
potential. Consider uniform field:
F = Eq
W = qVba
W = Fd = qE d
Thus Vba = Ed or E = Vba/d
Alternate units for E: volts per meter v/m
Instead of? N/C
Equipotential Lines
An equipotential surface is perpendicular to
the electric field at any point
 If not charges would move along the surface
 This contradicts the idea of an equipotential
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Describe the equipotentials…
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…between large oppositely charged
conducting plates
+
-
Electric Potential Due to Point
Charges
V = k Q/r derived from Calculus
 Here V = 0 at r = infinity; V represents
potential difference between r and infinity
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Example: Work to force two
point charges together
What work is needed to bring a 2mC charge
from far away to within 10cm of a 5 mc
charge?
 Work required = change in potential energy
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W=
0.90 J
qVba = q{kQ/rb – kQ/ra} = qkQ/rb
Potential at an Arbitrary Point
Near Several Point Charges
Add the potentials due to each point charge
 Use the right sign for the charge
 Relax; potential isn’t a vector
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What is true about the mid-plane between
two equal point charges of opposite sign?
Potential everywhere zero
Find…
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The potential due to a microcoulomb charge
at a point one meter away
9000 volts
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The potential energy of two microcoulomb
charges one meter apart
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9 x 10-3 J
The Electron Volt
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The energy acquired by a particle carrying a
charge equal to that of the electron when
accelerated through one volt
1 eV = 1.6 x 10-19 joules
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It’s about ENERGY
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1 KeV = 1000 eV = 1.6 x 10-16 joules
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1 MeV = 106 eV
1 Gev = 109 eV
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Capacitance
Stores charge
 Two conducting plates
 NOT touching
 May have insulating
material between
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Q = CV
Capacitance
Symbol C
 Unit: coulombs per volt = farad
 1 pf = 1 picofarad = 10-12 farad
 1nf = 1 nanofarad = 10-9 f
 1mf = 1 microfarad = 10-6 f
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C is Constant for a Given
Capacitor
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Does not depend on Q or V
Proportional to area
Inversely proportional to distance between plates
C
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= e0 A/d
If dielectric like oil or paper between plates use e
= Ke0; K is called dielectric constant
 e is called permittivity. e0 is permittivity of free
space
Find the Capacitance
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A capacitor can hold 5 mC of charge at a
potential difference of 100 volts. What is its
capacitance
Q = CV
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C = Q/V = 5 x 10-6 C/ 100V =
5 x 10-8 f = 50000 pf = 0.05 mf
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Why is the Electric Field inside the dielectric
less than outside?
Less since fewer field lines
Capacitors
Photos courtesy Illinois Capacitor, Inc
Supercapacitors -
Supercapcitors use double layer electrolyte
technology, usually with activated carbon and
sulfuric acid. The carbon has huge surface area
Applications
In automotive ignitions
 In strobe lights
 In electronic flash
 In power supplies
 In nearly all electronics
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Supercapacitor applications
Power motor vehicles for short bursts
 Backup power for computers
 Operate emergency doors and slides in
commercial aircraft
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Lower energy density but greater power
density than batteries
A Capacitor Stores Electric
Energy
A battery produces electric energy bit by bit
 A capacitor is NOT a type of battery
 A battery can be used to charge a capacitor
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Energy in a Capacitor Holding
Charge Q at Voltage V
U
= QV/2 = CV2/2 = Q2/2C
Derivation: the work needed to charge a
capacitor by bringing charge onto a plate
when some is already there (use W =QV)
 Initially V = 0
 Average voltage during the charging
process is V/2
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Example
A 20 mf capacitor stores 20 millijoules of
energy. What is the voltage across it?
 U = CV2/2
 V = (2U/C)1/2 = (2x20 x 10-3J/20x10-6F)1/2 =
44.7 V
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