Download PHY481: Electromagnetism Exam 1 Solutions Charge on a conductor Lecture 14

PHY481: Electromagnetism
Exam 1 Solutions
Charge on a conductor
Lecture 14
Carl Bromberg - Prof. of Physics
Exam 1
1) Stokes’s theorem:
∫ S (∇ × F ) ⋅ n̂ dA = ∫C F ⋅ d
z
F(x) = î × (xî + yĵ + zk̂)
1) d = ĵdy, z = 0, 2) d = k̂dz, y = 1
4) d = k̂dz, y = 0
1
a)
d 2 = k̂dz
d 4 = k̂ dz
1
0
0
0
1
1
x
1
y
d 1 = ĵdy
⎛ ∂Fz ∂Fy ⎞ ⎛ ∂Fx ∂Fz ⎞
⎛ ∂Fy ∂Fx ⎞
∇ × F = î ⎜
−
+ ĵ ⎜⎝
−
−
= î(2) + ĵ( 0 ) + k̂(0)
⎟⎠ + k̂ ⎜
⎟
⎟
⎝ ∂y
⎝ ∂x
∂z ⎠
∂z
∂x
∂y ⎠
c) n̂ = î; dA = dxdy ;
∫ (∇ × F ) ⋅ n̂ dA = 2
S
Lecture 14
F1 = yk̂ − 0 ĵ
0
∫3 F ⋅ d = −1∫ dy = 1; ∫4 F ⋅ d = −0 ∫ dz = 0
C
F2 = k̂ − zĵ
F4 = 0k̂ − zĵ
∫1F ⋅ d = −0 ∫ dy = 0 ; ∫2 F ⋅ d = 1∫ dz =1;
b) ∫ F ⋅ d = 2
d  3 = ĵ dy
1
= yk̂ − zĵ
3) d = ĵdy, z = 1,
F3 = yk̂ − ĵ
∴
∫ (∇ × F ) ⋅ n̂ dA = ∫ F ⋅ d = 2
S
Carl Bromberg - Prof. of Physics
C
1
Exam 1 (cont’d)
2)
a) integral to obtain scalar potential from charge distribution
1
V (x) =
4πε 0
3
ρ( x ′ )d x ′
∫ x − x′
b) differential equation relating charge density and the electric field
∇ ⋅ E(x) =
ρ(x)
ε0
c) differential equation relating electric field and potential
E(x) = −∇V (x)
d) integral to obtain electric field from the charge distribution
E(x) =
Lecture 14
1
4πε 0
x − x′
∫ x − x′
3
ρ( x ′ )d x ′
3
Carl Bromberg - Prof. of Physics
2
Exam 1 (cont’d)
3a) Gauss’s law:
b)
∫ S E ⋅ dA = qencl
ε0
ET (x) = E0k̂; E B (x) = − E0k̂
θ
n̂T = k̂; n̂ B = −k̂
∫ E ⋅ dA = ∫ E0k̂ ⋅k̂ dA + ∫ ( − E0k̂ ) ⋅( −k̂ dA)
S
T
R
E = E0k̂
n̂ = r̂
n̂ = k̂
σ
σ
B
2E0 A = qencl ε 0 = σ A ε 0
c)
n̂ = − k̂
E0 = σ 2ε 0
E = − E0 k̂
2
n̂ = r̂; dA = R sin θ dθ dφ ; qencl = σπ R
2
⎡ π2
⎤
π
2
∫ E ⋅ dA = ∫ ( E0k̂ ) ⋅r̂ dA + ∫ ( − E0k̂ ) ⋅r̂ dA = 2π E0 R ⎢⎢ ∫ dθ cosθ sinθ − ∫π dθ cosθ sinθ ⎥⎥
S
T
B
⎣0
⎦
2
⎡ 2 1⎤
1
0
⎡
⎤
x ⎥
2
2
2
= 2π E0 R ⎢ ∫ xdx − ∫ xdx ⎥ = 2π E0 R ⎢ 2
= 2π E0 R
⎢⎣ 2 0 ⎥⎦
⎣0
⎦
1
2
2
2π E0 R = qencl ε 0 = σπ R / ε 0 ∴ E0 = σ 2ε 0
Lecture 14
Carl Bromberg - Prof. of Physics
3
Exam 1 (cont’d)
4a)
3
q = ∫ ρ( x ′ )d x ′ = ∫ λ d =
b)
E(x) =
x − x′
1
4πε 0
∫ x − x′
π 2
∫
Rdφ = π Rλ
−π 2
3
ρ( x ′ )d x ′
3
2
2 32
(
x − x ′ = zk̂ − R ( î cos φ + ĵsin φ ) = z + R )
1
zk̂ − R ( î cos φ + ĵsin φ )
E(x) =
λ Rdφ
∫
3
2
4πε 0
( z 2 + R2 )
c)
3
3
1 ⎛
=
⎜
4πε 0 ⎝ (
=
1
4πε 0
⎞
( î cos φ + ĵsin φ ) dφ ⎟
π zk̂ +
∫
3
2
3
2
2
2
⎠
( z 2 + R2 ) −π 2
z +R )
λR
λR
λR
( z2 + R )
d) Ez (x) =
Lecture 14
2 32
1
4πε 0
(π zk̂ − 2Rî )
R z → 0 (i.e., z >> R)
πλ Rz
(z
2
π 2
2
+R
)
2 32
=
q
4πε 0
z
(
z 1+ ( R z)
3
)
2 32
Ez (x) ≈
q
4πε 0 z
2
Carl Bromberg - Prof. of Physics
4
Exam 1 (cont’d)
5) Identity for
∇ ⋅ ( A × B ) using ∇ × A and ∇ × B
C = A × B where Ci = ε ijk Aj Bk
∇ ⋅ (C) =
∂Aj
∂B
∂
=
ε ijk Aj Bk = Bk ε ijk
+ Aj ε ijk k
∂xi ∂xi
∂xi
∂xi
∂Ci
(
)
Make replacements ε kij = ε ijk and − ε jik = ε ijk
∇ ⋅ ( C ) = Bk ( ε kij )
∂Aj
∂xi
(
+ Aj −ε jik
∂Bk
) ∂x
= Bk ( ∇ × A )k − Aj ( ∇ × B ) j
i
= B ⋅ (∇ × A ) − A ⋅ (∇ × B )
Lecture 14
Carl Bromberg - Prof. of Physics
5
Exam 1 (cont’d)
6a) Dipole moment
3
p = ∫ rρ(x)d x
r̂ = k̂ cosθ + sin θ ( î cos φ + jsin φ )
2
dA = R sin θ dθ dφ
p = 2π R
3
∫ r̂ (σ 0 cosθ ) sinθ dθ
π
2π
⎡π
⎤
2
3
2
= 2π R σ 0 ⎢ ∫ k̂ cos θ sin θ dθ ⎥ + R σ 0 ∫ cosθ sin θ dθ ∫ dφ ( î cos φ + jsin φ )
⎣0
⎦
0
0
3
−1
⎡ x3 ⎤
3
2
3
= 2π R σ 0k̂ ∫ −x dx = 2π R σ 0k̂ ⎢ − ⎥
⎣ 3 ⎦1
1
−1
3
=
4π R σ 0
3
k̂
b) Top is positive, bottom is negative, the same as a dipole
c) From far away, potential is that of a point dipole : V (r,θ ) =
Lecture 14
Carl Bromberg - Prof. of Physics
p ⋅ r̂
4πε 0 r
2
6
Exam 1 (cont’d)
7a) Prove
∞
f (− b a)
3
(d x should be dx, answer OK)
a
∫ δ (ax + b) f (x)dx =
−∞
nd
x ′ = ax + b; dx = dx ′ a ; 2 integral is f at x ′ = 0
∞
∞
1
f (− b a)
δ
(ax
+
b)
f
(x)dx
=
δ
(
x
)
f
(
x
a
−
b
a)d
x
=
′
∫
∫ ′ ′
a
a
−∞
−∞
∞
b) Prove
3
−1
3
∫ δ (Ax + b) f (x)d x = f (−A b) / det A
Jacobian
det A
−∞
−1
−1
x = A x′ − A b
3
3
d x = d x ′ Det A
∞
3
∫ δ (Ax + b) f (x)d x =
−∞
∞
−1
−1
3
∫ δ (x′) f (A x′ − A b) d x′ Det A
−∞
−1
= f (−A b) Det A
Lecture 14
Carl Bromberg - Prof. of Physics
7
Conductors in static equilibrium
Why these statements are true ?
– Electric field E inside a conductor is zero.
– Potential V inside a conductor is a constant.
– Inside a conductor the charge density ρ is zero.
– Charge Q on a conductor resides only on the surface.
– A net charge Qnet here is always paired with charge -Qnet elsewhere.
– On a conductor either V or Qnet, but not both, can be specified.
– At a conductor’s surface, field component Et = 0, component En = σ /ε0 ,
but in force calculations, En = σ /2ε0 due to only “distant” charges.
– In an empty cavity in a conductor, E = 0 and surface σcavity = 0.
– The potential V of a conductor can be set by a battery VB.
– The earth is an “infinite” source of charge at a constant V .
– V or Qnet of conductors (and Qexternal) determine V everywhere.
Lecture 14
Carl Bromberg - Prof. of Physics
8
Boundary value problems: 1-parallel plates
Parallel plates (find potential, field & surface charge)
The ground is a constant potential, so let V1 = 0 .
Battery draws electrons from plate 2 and puts
them on plate 1. Process “continues” until V2 = VB .
Potential between the plates
Between the plates the potential depends only on x.
Laplace’s equation: ∇ 2V ( x ) = 0
Boundary conditions:
V (0) = 0; V (d) = VB
General solution (ODE)
Apply boundary conditions
d ⎛ dV ⎞
dV
V (0) = c2 = 0
∇V=
= c1
⎜⎝
⎟⎠ = 0;
dx dx
dx
VB
V (d) = c1d = VB ; c1 =
V (x) = c1x + c2
d
2
Lecture 14
Potential
x
V (x) = VB
d
Carl Bromberg - Prof. of Physics
9
Electric field
Parallel plates (cont’d)
⎛
E = −∇V = −∇ ⎜⎝ VB
En -> Surface charges
Electric field
x ⎞ VB
⎟⎠ − î
d
d
Surface 1r charge density
ε 0VB
σ 1r = −
d
σ 1r
VB
E1n =
=−
ε0
d
Surface 2 charge density
σ 2 VB
E2n =
=
ε0
d
σ 2
Surface charge
Q2 = −Q1r = σ A =
Lecture 14
ε 0VB A
d
ε 0VB
=+
d
Force on each plate
F2 = Q2 (E2n ) non-q
2
2
σ2
1
=
A
2 ε0
Carl Bromberg - Prof. of Physics
10
Parallel plates (cont’d)
Capacitance
Two ways to know the electric field
1
1
1r
VB
σ
E= ; E=
ε0
d
2
2r
E
Capacitance is a geometrical factor relating the
charge on a conductor and its potential.
ε0 A
C=
d
ε0 A
Q=σA=
VB = CVB
d
d
0
V = VB
V =0
–
x
VB
+
ground
Energy storage
2
ε0
ε 0 VB
2 3
1 ε0 A 2
U = ∫ E d x′ =
Ad =
VB
2
2
2 d
2 d
Lecture 14
2
1
2
U = CVB
2
Carl Bromberg - Prof. of Physics
11