Download PHY481: Electrostatics Introductory E&M review (2) Course web site: www.pa.msu.edu/courses/phy481

PHY481: Electrostatics
Introductory E&M review (2)
Course web site: www.pa.msu.edu/courses/phy481
All homework is due AT 12:30 pm (no later)
HW1 due one week from today
Lecture 2
Carl Bromberg - Prof. of Physics
Field of a charged spherical shell - Gauss’s Law
 Consider a radius R spherical shell with surface charge density σ.
– A spherical Gaussian surface with radius r < R, is inside the charge
surface, and encloses no charge -> Einside = 0.
– A spherical Gaussian surface with radius r > R, has the electric field
normal to its surface.
$
2$
2
E
!
dA
=
E
r
"S
" sin# d# " d%
0
0
q
2
E4$ r =
&0
q = ! 4" R 2
q
q
E=
; E=
r̂
2
2
4$& 0 r
4$& 0 r
Same electric field as
charge q at the origin
Lecture 2
Carl Bromberg - Prof. of Physics
1
Uniform charge density sphere - Gauss’s Law
 Find the electric field inside of a sphere, radius R, with a
uniform charge density ρ throughout the volume.
– Pick a spherical Gaussian surface, radius r,
inside the charged sphere
$
2$
2
E
!
dA
=
Er
"S
" sin# d# " d%
0
0
3
q
Qr
encl
E4$ r 2 =
=
&0
& 0 R3
!=Q/
Qr
Qr
E=
; E=
r̂
3
3
4$& 0 R
4$& 0 R
qencl = !
(
(
3
4
"
R
3
4
"r3
3
)
)
r3
=Q 3
R
– Electric field outside of a sphere
E=
Lecture 2
Q
r̂
2
4!" 0 r
Same electric field as
charge Q at the origin
Carl Bromberg - Prof. of Physics
2
Infinite sheet (again) - Gauss’s Law
 Infinite sheet of charge with surface density σ.
– Pick a cylindrical Gaussian surface, radius r, passing through
the sheet.
– The dot product E•dA is non zero only on TWO the ends.
R
2$
"S E ! dA = 2E " rdr "
0
qencl
2(E$ R ) =
%0
2
qencl = ! (" R 2 )
d#
0
& ($ R 2 )
=
%0
&
&
E=
; E=±
k̂
2% 0
2% 0
Lecture 2
+ above
– below
Carl Bromberg - Prof. of Physics
3
Electric Potential Energy U(x)
 Potential energy change ΔU of charge q’ in known E-field.
– Calculate the (negative) of the work done by the field along a path
r2
r2
U is a scalar !
!U = " $ Felec # ds = " q% $ E # ds
r1
r1
 Example: parallel plate capacitor
x
!U = q"E # dx " = q ' Ex
0
E = ! E î
ds = dx " î
 Move q’ across entire capacitor
!U = U (d) " U (0) = q ' Ed
Lecture 2
U = 0 at negative plate.
Carl Bromberg - Prof. of Physics
4
Electric potential V(x)
 Potential V(x) is potential energy/unit test charge q’
– The potential V is defined without the test charge q’
PE of test charge
U&V
!U
=#
are scalars ! !V =
q"
r2
% E $ ds
r1
U (x) + arbitary constant
V (x) =
q!
 Example: parallel plate capacitor
Set V(0) = 0
constant = 0
U (x)
V (x) =
= Ex V (d) = Ed
q!
 Equipotential surfaces
– Electric field lines always cross
equipotential surfaces at 90°.
– In parallel plate capacitors equipotential
surfaces are planes parallel to plates
Lecture 2
Carl Bromberg - Prof. of Physics
5
Batteries, capacitors, and energy storage
 A battery moves charge Q between plates of area A
– Battery moves electrons to create charge densities σ.
– We have two expressions for electric field E !
E=
!
E=
"0
VB
d
– Find expression relating Q and V
!0 A
Q=
VB = CVB
d
!=
Q
A
!0 A
C=
d
– Find energy stored while charging plates to Q
U = ! dU =
Lecture 2
Q
Q
! Vdq = !
0
0
2
q
Q
dq =
C
2C
2
1
U = CVB
2
Energy density
1
Show ! u = 2 ! 0 E
Carl Bromberg - Prof. of Physics
2
6
Potential of a point charge
 Move test charge q’ toward charge q.
– ΔU is negative of work done by field in this motion
r2
r2
qq% dr
!U = " $ Felec # ds = "
4&' 0 $ r 2
r1
r1
qq%
=
4&' 0
r2
(1 1+
* r " r - (> 0)
) 2 1,
q
– Potential of a point charge
Lecture 2
r1
dr is negative
ds = dr r̂
E=
!U
!V =
= V (r2 ) # V (r1 )
q"
V is a scalar !
q#
q
V (!) = 0
V (r) =
4!" 0 r
q
4!"0 r
2
r̂
+
dV
q
E=!
r̂ =
r̂
2
dr
4"# 0 r
becomes -Grad (V)
Carl Bromberg - Prof. of Physics
7
Potential due to a charge distribution
 Two ways to get potential of a charge distribution
– Line integral of a known electric field
– Integration of point charge potentials
!V = " $ E # ds
V (r) =
1
4%& 0
 Example: spherical shell with charge density σ
– Electric field known from Gauss’s Law
(same as point charge r > R, and zero r < R)
– Potential obtained from line integral along E
!V = "
outside
point charge
V (r) =
potential
Lecture 2
Q
4#$ 0
Q
4!" 0 r
r
dr
% r2
R
=
E=
dq
$r
Q
4!" 0 r
2
r̂
Q &1 1 )
" +
(
4#$ 0 ' r R *
Q
r # R V (r) =
4!" 0 R
r < R inside potential
is constant
Carl Bromberg - Prof. of Physics
8
Potential by integration over point charges dq
 Potential on axis from ring with charge density λ
2"
1
dq
!R
V (r) =
=
d$
1 %
#
4!" 0 r
2
2 2
4"# 0 ( z + R ) 0
1
Q ( 2
2 #2
=
z +R )
4!" 0
!=
Q
2" R
 Obtain E on axis from (–)Gradient of V
3
dV
Q ( 2
2 !2
E=!
k̂ =
z z + R ) k̂
dz
4"# 0
Note:
z(z + R
2
)
1
2 !2
= cos"
 Why V ? More tools available to determine V than E
Lecture 2
Carl Bromberg - Prof. of Physics
9
Conductors and static electric fields
e–
Move electrons from pebble
to a metal surface. Pebble
becomes charged +q.
e–
 When charges stop moving, the electric field within the conductor is
zero, charge is “pulled” to the surface. Also, Gauss’s Law requires
that the charge density within this conductor is zero.
 When charges stop moving, the components of the electric field
parallel to the surface, E|| = zero. Also, Gauss’s Law requires that at
the surface the electric field normal component, E⊥ = σ /ε0 .
 The electric potential is a constant throughout the conductor.
 Later we will learn “method of images” to determine fields & charges
Lecture 2
Carl Bromberg - Prof. of Physics
10
Magnetic fields
 A charge q moves at a velocity v in magnetic field B.
– Force on the charge (use right hand rule for + charges)
F = qv ! B
– Circular motion for v perpendicular to B
2
F = mv r = qvB
 A current I flows in a thin wire
– Force on small segment or on length L
dF = I d! ! B
F= IL!B
– Force between parallel straight wires
µ0 I1 I 2 L
#7
F=
; µ0 = 4! " 10 T $ m/A
4! d
Attractive for same I’s
Lecture 2
straight wire B
µ0 I ˆ
B=
"
2! r
right hand rule
Carl Bromberg - Prof. of Physics
11
Magnetic fields from currents
 Ampere’s Law
– Closed path integral around current I
" B ! ds = µ 0 I
Choose a
circular path!
– Example: long straight wire carrying current I
# B ! ds = 2"rB = µ I
0
µ0 I
B=
2"r
µ0 I ^
φ
B=
2"r
 Biot-Savart Law
– Magnetic field from small current element dI
µ 0 I d! " r̂
dB =
4! r 2
Lecture 2
Carl Bromberg - Prof. of Physics
12