Chapter 24 Electric Potential - USNA
Transcript of Chapter 24 Electric Potential - USNA
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• Electrical potential energy U
• If a test charge q0 is put in electric field E,
then q0 experiences a force F = q0E
• This force is conservative (i.e., path-
independent)
• ds is an infinitesimal displacement vector
oriented tangent to any spatial path
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• Electric U, 2
• Work done by E-field is F•ds = q0E•ds
• As field does this work, charge-field
system’s potential energy U changes by
!U = -q0E•ds
• For finite displacement of charge from
A"B,
(Eqs. 24-17 & 24-18, p. 633)
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• Electric U, 3
• Since q0E is conservative, line integral
doesn’t depend on charge’s path
• Thus !U = change in system’s PE
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• Electric potential V
• PE per unit charge U/q0 is the electric
potential, which is:
• independent of q0’s value
• defined at every point in an E-field
• Define electric potential V = U/q0
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• Electric V, 2
• V is a scalar quantity since energy is a
scalar
• As a charged particle moves in an E-
field, it experiences change in potential
(Eq. 24-18, p. 633)
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• Electric V, 3
• Only V differences (i.e., !V) are meaningful
• We often set V = 0 at some convenientreference point in E-field
• V is a scalar characteristic of an E-field,independent of any test charges in that field
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• Work & electric V
• Assume a charge q is moved in an E-
field without changing charge’s KE
• Then work Wapp done by an external
agent on charge q is: Wapp = !U = q !V(Eqs. 24-13 & 24-14, p. 631)
• Contrast this with work W done by field
itself on charge q: W = –!U = –q !V(Eq. 24-1, p. 628)
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• Units of !V
• 1 V = 1 J/C (Eq. 24-9, p. 630)
• V is a volt
• 1 joule of work is needed to move a 1-C chargethrough a !V = 1 volt
• Also, 1 N/C = 1 V/m {= 1 J/(C*m) = 1 N*m/(C*m)}(see Eq. 24-10, p. 630)
• Can interpret E-field as measure of spatialgradient of V
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• Electron volt
• An energy unit commonly used in atomic &
nuclear physics is electron-volt (eV)
• 1 eV is energy that a charge-field system
gains or loses when a charge of |e| (electron
or proton) moves through !V = 1 volt
• So 1 eV = 1.60 x 10-19 C•(1V) = 1.60 x 10-19 J
since 1 J = 1 C•V (see p. 630)
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• !V in a uniform E-field
• Can simplify equations for !V if E-field
is uniform: VB–VA = !V = – E•ds
= –E ds = –Ed, where d = path
distance from A"B (see Sample Problem, p. 634)
• – sign shows V at point B < V at point A
#B
A
#B
A
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• Energy & E-field direction
• When E-field points $, point
B is at lower V than point A
• If a +test charge moves fromA"B, then charge-field
system loses PE (i.e., !V < 0)
(SJ 2008Fig. 25.2,p. 695)
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• Energy & E-field direction, 2
• System of +q0 & E-field loses electric U (i.e.,!U < 0) when +q0 moves with field
• True since E-field does work on +q0 when itmoves with field
• A moving +q0 gains KE = PE lost by charge-field system (i.e., energy conservation)
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• Energy & E-field direction, 3
• If q0 is –, then !U > 0
• System consisting of –q0 & E-field gains
U when –q0 moves in field’s direction
• For –q0 to move with E-field, external
agent must do +work on the charge
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• Equipotentials
• Point B is at lower V than
point A
• Points B & C are at same V
• An equipotential surface is
any surface with continuous
distribution of points at
constant V
(compare Fig. 24-5, p. 634)
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• Charged particle in uniform E-field
• +q0 is released from rest& moves with E-field
• Then !V < 0 & !U < 0
• Fe & a on +q0 are in E-field’s direction
(SJ 2008 Fig. 25.6, p. 696)
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• !V & point charges
• +point charge " E-field
pointing radially outward
• Then !V from A"B is:
(SJ 2008 Eq. 25.10, p. 697)
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• !V & point charges, 2
• !V is independent of A"B path
• Pick reference V = 0 at rA = %
• Then V at point r is V = keq/r(Eq. 24-26, p. 635)
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• V for a point charge
• Consider V in plane around
a point charge
• Red line shows V ’s 1/rdependence
(compare Fig. 24-7, p. 635)
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• V for multiple charges
• Net V from several point charges is & ofVs from individual charges (superpositionprinciple)
• Then (Eq. 24-27, p. 636)
where V = 0 at reference distance r = %
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• V for a dipole
• Consider V for an electric
dipole (along y-axis)
• Steep slope between +/-
charges shows strong
E-field here
• V = ke*p*cos(')/r2 where
' = angle from dipole axis(Eq. 24-30, p. 638) (SJ 2008 Fig. 25.8, p. 698)
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• E & V for infinite sheet of charge
• Equipotential lines are
dashed blue lines
• E-field lines are brown lines
• Note equipotential lines are
everywhere ( E-field lines
(SJ 2008 Fig. 25.12, p. 701)
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• E & V for a point charge
• Equipotential lines are dashed
blue lines
• E-field lines are brown lines
• As before, equipotential lines
are everywhere ( E-field lines
• Note that equipotentials are
closer together as V-gradient
steepens near charge(SJ 2008 Fig. 25.12, p. 701)
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• E & V for a dipole
• Equipotential lines are dashed
blue lines
• E-field lines are brown lines
• As before, equipotential lines
are everywhere ( E-field lines
(SJ 2008 Fig. 25.12, p. 699)
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• V for continuous charge distribution
• Treat small charge element dq
as a point charge on an object
of finite size & arbitrary shape
• Then potential dV at any point
due to dq is dV = kedq/r(Eq. 24-31, p. 639)
(SJ 2008 Fig. 25.14, p. 703)
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• V for continuous charge distribution, 2
• To find total V, must integrate to include
contributions from all elements dq:
V = ke #dq/r (Eq. 24-32, p. 639)
• assumes reference value V(r =%) = 0 for
P infinitely far away from charged object
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• V for a uniformly charged ring
• P is on ( central axis of
uniformly charged ring of
radius a & total charge Q
(SJ 2008 Fig.25.15, p. 704)
(SJ 2008 Eq. 25.21, p. 705)
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• V for a uniformly charged disk
• Ring of radius a &surface charge density !
(compare Fig. 24-13, p. 640)
(compare Eq. 24-37, p. 640)
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• V for finite line of charge
• Rod of length l has
total charge Q & linearcharge density "
(compare Fig.24-12, p. 639)
(compare Eq. 24-35, p. 640)
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• V for uniformly charged sphere
• Solid insulating sphere of radius
R & total charge Q
• For r > R, V = keQ/r
• For r < R {e.g., inner radius D},
VD = keQ(3–r2/R2)
2R
(SJ 2004 Fig. 25.19, p. 777)
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• V for uniformly charged sphere, 2
• Parabolic VD curve is forpotential inside sphere, & itsmoothly joins VB curve
• Note that ED = keQr/R3, justas found earlier
• Hyperbolic VB curve is forpotential outside sphere
(SJ 2004 Fig.25.20, p. 777)
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• Finding E from V
• Assume E has only an x-component, so that
Ex = – dV/dx (Eq. 24-41, p. 641)
• Similar statements apply to Ey & Ez components
• Equipotential surfaces are always ( E-field linespassing through them
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• E-field from V, general
• In general, V varies in all 3 dimensions
• Given V(x,y,z), you can find Ex, Ey, & Ez
as partial derivatives of V:
(Eq. 24-41, p. 641)
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• U for multiple charges
• For 2 charged particles, system’s
PE (or U) is U = keq1q2/r12
(see Eq. 24-43, p. 643)
• If 2 charges have same sign,
U > 0 & must do work to bringthem together (i.e., to ) U)
• If 2 charges have opposite signs,
U < 0 & must do work to movethem apart () U again)
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• U for multiple charges, 2
• If > 2 charges, find U for eachcharge pair & add these Us
• So for 3 charges:
• Result is independent of order ofmoving charges to givenpositions
(see Sample Problem, p. 643)
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• V from charged conductor
• Consider 2 points A & B on surface ofarbitrary charged conductor
• E = 0 inside conductor & onconductor’s surface, have E ( surface
• Thus always have E ( displacementds along surface, so E•ds = 0
• As a result, !V between A & B = 0
(SJ 2008 Fig. 25.18, p. 707)
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• V from charged conductor, 2
• V = constant everywhere on surface of chargedconductor in equilibrium (i.e., !V = 0 betweenany 2 points on surface)
• Surface of any charged conductor inelectrostatic equilibrium is equipotential surface
• Since E = 0 inside conductor, we know thatV = Vsurface (a constant) throughout interior
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• E compared to V
• V(r) * 1/r, but E(r) * 1/r 2
• In space surrounding a charge, itsets up:• vector E-field related to force
• scalar potential V related to energy
V(r) & E(r) for conducting sphere (SJ 2008 Fig. 25.19, p. 707)
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• Irregularly shaped objects
• Charge density + is high where
radius of curvature is small & low
where radius of curvature is large
• Model conductor as collection of
point charges with E(r) = keq/r 2
• So for nearby uncharged surfaces,distance r to conductor variesmore for its small-curvature parts
• E-field is large near convex points
with small radii of curvature & is
largest at sharp points (SJ 2004 Fig. 25.23, p. 779)
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• Irregularly shaped objects, 2
• E-field lines are everywhere ( conducting surface
• Equipotential surfaces are everywhere ( E-field
lines
(SJ 2004 Fig.25.24, p. 779)
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• Cavity in a conductor
(SJ 2008 Fig. 25.26, p. 780)
• Assume:
(1) an irregularly shaped cavity
is inside a conductor
(2) no charges within cavity
• Know that E = 0 inside
conductor. Why is this so?
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• Cavity in a conductor, 2
• E-field inside doesn’t depend on chargedistribution on conductor’s exterior
• For all paths between points A & B, potential
difference VB–VA = – E•ds = 0 (i.e., all points
on conductor’s cavity wall are at same V)
• Since this holds for all paths on inner wall,must have E = 0 N/C everywhere inside cavity
• Thus a cavity with (a) conducting walls &(b) no enclosed charges is a field-free region
#B
A
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• Corona discharge
• If E-field near a conductor is large, free
electrons from random ionizations of air
molecules accelerate away from these
“parent” molecules
• These free electrons
can ionize additional
molecules near
conductor
60 kV Tesla coil discharge
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• Corona discharge, 2
• This " more free electrons
• Corona discharge is glow that resultsfrom recombination of these freeelectrons with ionized air molecules
• Ionization & corona discharge are mostlikely to occur near very sharp pointswhere + is largest
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Millikan Oil-Drop Experiment
, Robert Millikan measured e, the
magnitude of the elementary charge on
the electron
, He also demonstrated the quantized
nature of this charge
, Oil droplets pass through a small hole &
are illuminated by a light
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Oil-Drop Experiment, 2
, With no electric field
between the plates,
the gravitational
force & the drag
force (viscous) act
on the electron
, The drop reaches
terminal velocity with
FD = mg
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Oil-Drop Experiment, 3
, When an electric fieldis set up between theplates, The upper plate has a
higher potential
, The drop reaches anew terminal velocitywhen the electricalforce equals the sumof the drag force &gravity
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Oil-Drop Experiment, final
, The drop can be raised & allowed to fall
numerous times by turning the electric
field on & off
, After many experiments, Millikan
determined:
, q = ne where n = 1, 2, 3, …
, e = 1.60 x 10-19 C
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Van de Graaff
Generator
, Charge is delivered continuously toa high-potential electrode by meansof a moving belt of insulatingmaterial
, The high-voltage electrode is ahollow metal dome mounted on aninsulated column
, Large potentials can be developedby repeated trips of the belt
, Protons accelerated through suchlarge potentials receive enoughenergy to initiate nuclear reactions
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Electrostatic Precipitator
, An application of electrical discharge in
gases is the electrostatic precipitator
, It removes particulate matter from
combustible gases
, The air to be cleaned enters the duct &
moves near the wire
, As the electrons & negative ions created by
the discharge are accelerated toward the
outer wall by the electric field, the dirt
particles become charged
, Most of the dirt particles are negatively
charged & are drawn to the walls by the
electric field
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Application – Xerographic
Copiers
, The process of xerography is used for
making photocopies
, Uses photoconductive materials
, A photoconductive material is a poor
conductor of electricity in the dark but
becomes a good electric conductor when
exposed to light
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Application – Laser Printer
, The steps for producing a document on a laserprinter is similar to the steps in the xerographicprocess, Steps a, c, & d are the same
, The major difference is the way the image forms onthe selenium-coated drum
, A rotating mirror inside the printer causes the beam of thelaser to sweep across the selenium-coated drum
, The electrical signals form the desired letter in positivecharges on the selenium-coated drum
, Toner is applied & the process continues as in thexerographic process