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W04D1 Electric Potential and Gauss’ Law
Equipotential Lines
Today’s Reading Assignment Course Notes: Sections 3.3-3.4, 4.4-4.6. 4.10.5
Announcements
Exam One Thursday Feb 28 7:30-9:30 pm Room Assignments (See Stellar webpage announcements) Review Tuesday Feb 26 from 9-11 pm in 26-152 PS 3 due Tuesday Tues Feb 26 at 9 pm in boxes outside 32-082 or 26-152
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Outline
Review E and V Deriving E from V
Using Gauss’s Law to find V from E
Equipotential Surfaces
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Electric Potential and Electric Field
Set of discrete charges: Continuous charges: If you already know electric field (Gauss’ Law) compute electric potential difference using
V (!r) !V (") = ke
d #q!r - !#rsource
$
VB !VA = !
!E
A
B
" # d !s
V (!r)!V (") = ke
qi!r - !rii#
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E Field and Potential: Effects
!F = q
!E
If you put a charged particle, (charge q), in a field:
Wext =!U = q!V
To move a charged particle, (charge q), in a field and the particle does not change its kinetic energy then:
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Concept Question: Two Point Charges
The work done in moving a positively charged object that starts from rest at infinity and ends at rest at the point P midway between two charges of magnitude +Q and –Q 1. is positive. 2. is negative. 3. is zero. 4. can not be determined – not enough info is given.
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E Field and Potential: Creating
A point charge q creates a field and potential around it: Use superposition for systems of charges
V (!r) !V (") = ke
d #q!r - !#rsource
$
!E = ke
qr 2 r
V (r) !V (") = ke
qr
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Using Gauss’s Law to find Electric Potential from Electric Field
If the charge distribution has a lot of symmetry, we use Gauss’s Law to calculate the electric field and then calculate the electric potential V using
B
B AA
V V d− = − ⋅∫E s
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Group Problem: Coaxial Cylinders
A very long thin uniformly charged cylindrical shell (length h and radius a) carrying a positive charge +Q is surrounded by a thin uniformly charged cylindrical shell (length h and radius a ) with negative charge -Q , as shown in the figure. You may ignore edge effects. Find V(b) – V(a).
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Worked Example: Spherical Shells These two spherical shells have equal but opposite charge. Find for the regions (i) b < r (ii) a < r < b (iii) 0 < r < a Choose V (!) = 0
V (r) !V (")
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Electric Potential for Nested Shells
!E =
Q4!"0r
2 r, a < r < b
0, elsewhere
#
$%
&%%
From Gauss’s Law
V (r) !V (")
= 0! = ! 0
"
r
# dr = 0
VB !VA = !
!E
A
B
" # d !sUse
Region 1: r > b
r
No field No change in V!
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Electric Potential for Nested Shells Region 2: a < r < b
V (r) !V (r = b)= 0
!"# $# = ! dr Q4"#0r
2b
r
$
= Q
4!"0r r=b
r
= 1
4!"0
Q 1r# 1
b$%&
'()
r
Electric field is just a point charge. Electric potential is DIFFERENT – surroundings matter
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Electric Potential for Nested Shells Region 3: r < a
V (r) ! V (a)
= kQ 1a!
1b
"#$
%&'
! = ! dr 0a
r
( = 0
V (r) =V (a) = Q
4!"0
1a# 1
b$%&
'()
r
Again, potential is CONSTANT since E = 0, but the potential is NOT ZERO for r < a.
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Group Problem: Charge Slab
Infinite slab of thickness 2d, centered at x = 0 with uniform charge density . Find !
V (xA ) !V (0) ; xA > d
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Deriving E from V
!!s = !x i
A = (x,y,z), B=(x+Δx,y,z)
Ex ! " #V
#x$ " %V
%xEx = Rate of change in V with
y and z held constant
!V = "
!E
A
B
# $d !s
!V = "
!E
( x ,y ,z )
( x+!x ,y ,z )
# $d !s % "!E $ !!s = "
!E $(!x i) = "Ex!x
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Gradient (del) operator:
!! " #
#xi + #
#yj + #
#zk
If we do all coordinates:
!E = !
!"V
Deriving E from V
!E = ! "V
"xi + "V
"yj + "V
"zk
#$%
&'(
= ! "
"xi + "
"yj + "
"zk
#$%
&'(
V
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Concept Question: E from V Consider the point-like charged objects arranged in the figure below. The electric potential difference between the point P and infinity and is
V (P) = ! kQ aFrom that can you derive E(P)?
1. Yes, its kQ/a2 (up) 2. Yes, its kQ/a2 (down) 3. Yes in theory, but I don’t know how to take a
gradient 4. No, you can’t get E(P) from V(P)
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Group Problem: E from V Consider two point like charged objects with charge –Q located at the origin and +Q located at the point (0,a). (a) Find the electric potential V(x,y)
at the point P located at (x,y).
(b) Find the x-and y-components of the electric field at the point P using
Ex (x, y) = ! "V
"x, Ey (x, y) = ! "V
"y
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Concept Question: E from V
The graph above shows a potential V as a function of x. The magnitude of the electric field for x > 0 is
1. larger than that for x < 0 2. smaller than that for x < 0 3. equal to that for x < 0
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Concept Question: E from V
The above shows potential V(x). Which is true?
1. Ex > 0 is positive and Ex < 0 is positive 2. Ex > 0 is positive and Ex < 0 is negative 3. Ex > 0 is negative and Ex < 0 is negative 4. Ex > 0 is negative and Ex < 0 is positive
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Group Problem: E from V
A potential V(x,y,z) is plotted above. It does not depend on x or y. What is the electric field everywhere? Are there charges anywhere? What sign?
-5 0 50
5
10Po
tent
ial (
V)
Z Position (mm)
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Equipotential Curves: Two Dimensions
All points on equipotential curve are at same potential. Each curve represented by V(x,y) = constant
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Direction of Electric Field E
E is perpendicular to all equipotentials
Constant E field Point Charge Electric dipole
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Direction of Electric Field E
http://web.mit.edu/viz/EM/visualizations/electrostatics/InteractingCharges/zoo/zoo.htm
E is perpendicular to all equipotentials Field of 4 charges Equipotentials of 4 charges
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Properties of Equipotentials
E field lines point from high to low potential E field lines perpendicular to equipotentials
E field has no component along equipotential The electrostatic force does zero work to
move a charged particle along equipotential
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