Download PHY481: Electrostatics Introductory E&M review (3) Lecture 3

PHY481: Electrostatics
Introductory E&M review (3)
Lecture 3
Carl Bromberg - Prof. of Physics
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.
qencl = σ (π R 2 )
⎛ R 2π ⎞
∫S E ⋅ dA = 2E ⎜⎜ ∫ rdr ∫ dφ ⎟⎟
⎝0
⎠
0
2)
(
q
σ
π
R
2E(π R 2 ) = encl =
ε0
ε0
σ
σ
E=
; E=±
k̂
2ε 0
2ε 0
Lecture 3
+ above
– below
Carl Bromberg - Prof. of Physics
1
Electric Potential Energy U(x)
 Potential energy change ΔU of charge q’ in known field.
– Obtained by calculating the –work done by the field along path
r2
r2
U is a scalar !
ΔU = − ∫ Felec ⋅ ds = − q′ ∫ E ⋅ ds
r1
r1
 Example: parallel plate capacitor
x
ΔU = q′E ∫ dx ′ = q ' Ex
0
E = − E î
ds = dx ′ î
 Move q’ across entire capacitor
ΔU = U (d) − U (0) = q ' Ed
Lecture 3
U = 0 at negative plate.
Carl Bromberg - Prof. of Physics
2
Electric potential V(x)
 Potential V(x) is potential energy/unit test charge q’
– The potential V is defined without the test charge q’
r2
ΔU
ΔV =
= − ∫ E ⋅ ds
q′
r1
PE of test charge
U (x) = q′V (x)
U & V are a scalars !
 Example: parallel plate capacitor
U (x)
V (x) =
= Ex
q′
V (d) = Ed
 Equipotential surfaces
– Electric field lines always cross
equipotential surfaces at 90°.
– In parallel plate capacitors equipotential
surfaces are planes parallel to plates
Lecture 3
Carl Bromberg - Prof. of Physics
3
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=
σ
E=
ε0
VB
d
σ=
– Find expression relating Q and V
ε0 A
Q=
VB = CVB
d
Q
A
ε0 A
C=
d
– Find energy stored while charging plates to Q
U = ∫ dU =
Lecture 3
Q
Q
∫ Vdq = ∫
0
0
2
q
Q
dq =
C
2C
2
1
U = CVB
2
Energy density
1
Show ! u = 2 ε 0 E
Carl Bromberg - Prof. of Physics
2
4
Potential of a point charge
 Move test charge q’ in known field of charge q.
– ΔU is –work done by field in this motion
r1
r
r
2
qq′ 2 dr
ΔU = − ∫ Felec ⋅ ds = −
4πε 0 ∫ r 2
r1
r1
qq ′
=
4πε 0
⎡1 1⎤
⎢ r − r ⎥ (> 0)
⎣ 2 1⎦
– Potential of a point charge
q
V (∞) = 0
V is a scalar ! V (r) =
4πε 0 r
Lecture 3
ds = dr r̂
r2
E=
q
ΔU
ΔV =
= V (r2 ) − V (r1 )
q′
q′
dr is negative
q
4πε0 r
2
r̂
+
dV
q
E=−
r̂ =
r̂
2
dr
4πε 0 r
becomes -Grad (V)
Carl Bromberg - Prof. of Physics
5
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
ΔV = − ∫ E ⋅ ds
V (r) =
1
4πε 0
dq
∫r
 Example: spherical shell with charge density σ
– Electric field known from Gauss’s Law
(same as point charge r > R, and zero r < R)
E=
Q
4πε 0 r
2
r̂
– Potential obtained from line integral along E
ΔV = −
outside
point charge V (r) =
potential
Lecture 3
Q
4πε 0
Q
4πε 0 r
r
dr
∫ r2
R
=
Q ⎡1 1 ⎤
− ⎥
⎢
4πε 0 ⎣ r R ⎦
Q
r ≥ R V (r) =
4πε 0 R
r < R inside potential
is constant
Carl Bromberg - Prof. of Physics
6
Potential by integration over point charges dq
 Potential on axis from ring with charge density λ
2π
1
dq
λR
V (r) =
=
dφ
1 ∫
∫
4πε 0 r
2
2 2
4πε 0 ( z + R ) 0
1
Q ( 2
2 −2
=
z +R )
4πε 0
Q = λ 2π R
 Obtain E on axis from –Gradient ofV
3
dV
Q ( 2
2 −2
E=−
k̂ =
z z + R ) k̂
dz
4πε 0
Note:
z(z + R
2
)
1
2 −2
= cosθ
 Why V ? More tools available to determine V than E
Lecture 3
Carl Bromberg - Prof. of Physics
7
Conductors and static electric fields
Move charge -q from small object
to the surface of a metal. Object
becomes charged +q.
 When charges stop moving, the electric field within the
conductor is zero, with charge only on 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, Eperp = σ /ε0 .
 The electric potential is a constant throughout the conductor.
Lecture 3
Carl Bromberg - Prof. of Physics
8
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 = 4π × 10 T ⋅ m/A
4π d
Attractive for same I’s
Lecture 3
straight wire B
µ0 I
B=
φ̂
2π r
right hand rule
Carl Bromberg - Prof. of Physics
9
Magnetic fields from currents
 Ampere’s Law
– Closed path integral around current I
∫ B ⋅ ds = µ0 I
– Example: long straight wire carrying current I
∫ B ⋅ ds = 2π rB = µ0 I
 Biot-Savart Law
µ0 I
µ0 I
B=
; B=
φ̂
2π r
2π r
– Magnetic field from small current element dI
µ 0 I d × r̂
dB =
4π r 2
Lecture 3
Carl Bromberg - Prof. of Physics
10