Lecture 21 Carl Bromberg - Prof. of Physics

PHY481: Electromagnetism

HW4

Lecture 21 Carl Bromberg - Prof. of Physics 1

4.3Spread platesPlates charged

Potential changes to V1 Potential remains at VB

Q = CVB = ε0 AVB d0

Q = C1V1 = ε0 AV1 d1

Q1 = C1VB = ε0 AVB d1

V1 =VB d1 d0

Q1 = Qd0 d1

ε =

1

2CVB

2 =1

2

ε0 A

d0

VB

2

ε1 =

1

2C1V1

2 =d1

d0

ε ε1′ =

1

2C1VB

2 =d0

d1

ε

Stored energy

Charge remains at Q Charge changes to Q1

?

No battery With battery

Lecture 21 Carl Bromberg - Prof. of Physics 2

4.3

Force on +Q generated by fieldof the -Q charge (not both)

Field between the plates is thesum of the field of both plates

A

+Q −Q

ground+–

V0

V =V0 V = 0

E2

F = qE1

A

E2 =

σε0

E1 =

σ2ε0

F = qE1 = Q

σ2ε0

=Q

2

2ε0 A=ε0 AV

2

2x2

Work done to pull plates apart by Δx. Battery disconnected Battery connected

W = − F ⋅ i dxx0

x0 +Δx

∫ =

ε0 AV0

2

2

Δx

x0

2

ε0 AV0

2

2

Δx

x0 x0 + Δx( )Cap. energy storage

ε0 AV0

2

2

Δx

x0

2 −ε0 AV0

2

2

Δx

x0 x0 + Δx( ) U =

Q2

2C ΔU =

Q2

2C−

Q0

2

2C0

=

Energy stored by Battery

Cap. energy change

Δε = ΔQV0 = 0

ε0 AV0

2 Δx

x0 x0 + Δx( )

Q = CV

C =

ε0 A

x

negative

Lecture 21 Carl Bromberg - Prof. of Physics 3

4.6

V (ρ, z) =q

4πε0

1

ρ2 + z − z0( )2( )−

1

ρ2 + z + z0( )2( )⎡

⎣⎢⎢

⎤

⎦⎥⎥

Ez (ρ,0) = −∂V

∂z z=0= −

q

2πε0

z0

ρ2 + z0

2( )3 2

σ (ρ) = ε0En(ρ,0) = −q

2πz0

ρ2 + z0

2( )3 2

= −qz0

−1

ρ2 + z0

2( )1 2

⎡

⎣⎢⎢

⎤

⎦⎥⎥

0

∞

= −q asexpected( )

x = r z0 ; x = 3; r = 3z0

−q

2= 2π σ ρ( )

0

r

∫ ρdρ = qz0

1

ρ2 + z0

2( )1 2

⎡

⎣⎢⎢

⎤

⎦⎥⎥

0

r

1

2=

1

x2 +1( )1 2

; x = r z0

Potential of charge over conducting plane, via MoI is a discrete dipole potential

ρ2 = x

2 + y2Radius on

the mid-plane

Electric field on the mid-plane

Charge density on the conductor surface

Total charge on the on conductor surface Radius enclosing half the surface charge

Q = 2π σ ρ( )0

∞

∫ ρdρ = −qz0

ρdρ

ρ2 + z0

2( )3 20

∞

∫

Lecture 21 Carl Bromberg - Prof. of Physics 4

4.10a

Vd r,θ( ) = pcosθ

4πε0r2

V2d z,0( ) = p

4πε0

1

z − z0( )2+

1

z + z0( )2

⎡

⎣⎢⎢

⎤

⎦⎥⎥

Potential via M.o.I is two dipoles (SAME DIRECTION)

z0

−z0

z

z0

z

Conductor

+

–

+

–

+

–

Single dipole potential

Double dipole potential

V2d z,0( ) = p

2πε0z2

, z >> z0

E = −ΔV =2 p

2πε0z3

k

Find potential of an upward facing dipole above a grounded conductor

Lecture 21 Carl Bromberg - Prof. of Physics 5

4.15Potential via M.o.I uses

q2 at z2 Determine them using points a,b ⎫

⎬

⎪⎪⎪

⎭

⎪⎪⎪

⎫

⎬

⎪⎪⎪

⎭

⎪⎪⎪

z0

z

z0 − R

z1

R

q1

q0

R − z1

a

b

q0

z0 − R=

−q1

R − z1

q0

z0 + R=

−q1

z1 + R

Va = 0

Vb = 0

q0 R − z1( ) = −q1 z0 − R( )

q0 z1 + R( ) = −q1 z0 + R( )

q1 = −q0

R

z0

; z1 =R

2

z0

V (r,θ) =

1

4πε0

q0

r0

+q1

r1

⎡

⎣⎢

⎤

⎦⎥

Geometry

r0 = r2 + z0

2 − 2r z0 cosθ

r1 = z0

2r

2 + R4 − 2r z0R

2cosθ z0

F =−q0

2Rz0

4πε0 z0

2 − R2( )2

k

U = − Fz∞

z0

∫ dz = −q0

2R

4πε0

zdz

z2 − R

2( )2∞

z0

∫

= −q0

2R

8πε0

1

z0

2 − R2( )

ImageCross multiply

Er (r,θ) = −∂V

∂r=

q

4πε0

r − z0 cosθ

r0

3+

R z0

2r − z0R

2cosθ( )

r13

⎡

⎣⎢⎢

⎤

⎦⎥⎥

Eθ (r,θ) = −∂V

r∂θ=

qz0 sinθ4πε0

1

r0

3+

R3

z0

3

r13

⎡

⎣⎢⎢

⎤

⎦⎥⎥

Lecture 21 Carl Bromberg - Prof. of Physics 6

4.18

V0 =V b( ) −V a( ) = Q

4πε0b−

−Q

4πε0a

=Q a + b( )4πε0ab

C =Q

V0

= 4πε0

ab

a + b( )

Two spheres charged by a battery

Charged by a battery, each must have the same magnitude of charge

Because the two spheres are very far apart thepotentials of each sphere are independent.

Comparing the potentials of each spherewith respect to V = 0 at infinity

Reference each potential withrespect to V = 0 at infinity

V r( ) = q

4πε0r

Lecture 21 Carl Bromberg - Prof. of Physics 7

4.20

′V r,θ( ) = A

r+

E0a3cosθ

r2

− E0r cosθ

σ (θ) = 3ε0E0 cosθ

See lecture 18 for derivation of charge densityon a grounded sphere in an external field

′Er r,θ( ) = −∂V r,θ( )

∂r=

Q0

4πε0r2+

2E0a3cosθ

r3

+ E0 cosθ

Q = Q0 ⇒ A ≠ 0

Q = 0 ⇒ A = 0

V (a,θ) = 0; V (r,θ) r→∞ = −E0z

′Er a,θ( ) = Q0

4πε0a2+ 3E0 cosθ

′σ (θ) = ε0Er a,θ( ) = Q0

4πa2+ 3ε0E0 cosθ

V r,θ( ) = A

r+ B +

C cosθ

r2

+ Dr cosθ

′V a,θ( ) = A

a+ E0acosθ − E0acosθ =

Q0

4πε0a

′V a,θ( ) = Q0

4πε0a

A =

Q0

4πε0

′σ (θ) > 0; cosθ = −1

Q0

4πa2= 3ε0E0

Q0 = 12πa2ε0E0

General potential sphericalcoordinates in simple situations

Grounded sphere in external field

New boundary conditions

New term

Add positivecharge Q0 here

Lecture 21 Carl Bromberg - Prof. of Physics 8

4.22

r1

r

y y0

− y0

r2

θ

−λ line charge

λ line charge

x y0 − y

y0 + y

V (r) =

−λ ln r

2πε0

V (r) =

−λ ln r1 r2( )2πε0

Ey = −∇V =λ

2πε0

y − y0

r2 + y0

2 − 2yy0

−y + y0

r2 + y0

2 + 2yy0

⎡

⎣⎢⎢

⎤

⎦⎥⎥

Ey (x) =−λ y0

πε0 x2 + y0

2( )

σ x( ) = ε0Ey (x) =−λ y0

π x2 + y0

2( )

λz = σ x( )dx−∞

∞

∫ =−λ y0

πdx

x2 + y0

2( )−∞

∞

∫ =−λπ

arctanx

y0

⎡

⎣⎢

⎤

⎦⎥−∞

∞

λz =

−λπ

arctan ∞( ) − arctan −∞( )[ ] = −λ

Line charge above a grounded conducting plane Image with another line charge-λ below the mid-planeSingle line

charge potential

Electric field

Double line charge potential

Field onmid-plane

Charge densityon conductor

Linear charge densityon conductor

Lecture 21 Carl Bromberg - Prof. of Physics 9

5.3

V x, y( ) = Cn cos 2n +1( )π x

a⎡⎣⎢

⎤⎦⎥cosh 2n +1( )π y

a⎡⎣⎢

⎤⎦⎥

V x,± a 2( ) =V0 cosπ x

a

⎛⎝⎜

⎞⎠⎟

= Cn cos 2n +1( )π x

a⎡⎣⎢

⎤⎦⎥cosh 2n +1( )π y

a⎡⎣⎢

⎤⎦⎥

n = 0, Cn = C0

V =V0cos

π x

a

⎛⎝⎜

⎞⎠⎟ at y = ± a 2

V x,± a 2( ) = C0 cosπ x

a⎡⎣⎢

⎤⎦⎥cosh

π2

⎡⎣⎢

⎤⎦⎥=V0 cos

π x

a⎡⎣⎢

⎤⎦⎥

C0 =V0

coshπ2

⎡⎣⎢

⎤⎦⎥

V x, y( ) =V0 cos π x a[ ]cosh π y a[ ]

cosh π 2[ ]

Boundary condition

General solution (for even boundary conditions)

Only non-zero terms

Apply boundary conditionto determine coefficient

Potential inside pipe

Square pipe cross section

Lecture 21 Carl Bromberg - Prof. of Physics 10

5.11See lecture 18 for derivation of potential for ona conducting sphere in an external field

V r,θ( ) = +

E0a3cosθ

r2

− E0r cosθ

V r,θ( ) = pcosθ

4πε0r2

p = 4πε0E0a

3

p = αE0

α = 4πε0a3

σ θ( ) = ε0Er a,θ( ) = −ε0

∂V

∂r r=a

= 2ε0

E0a3cosθ

a3

+ ε0E0 cosθ = 3ε0E0 cosθ

σ θ( ) = 3ε0E0 cosθ

r = acosθ k + asinθ ρ

p = σ (θ)rdS∫= 2πa

33ε0E0 cosθ( )cosθ sinθdθ∫ k

pz = 6πa3ε0E0 cos

2θ sinθdθ0

π

∫

= 6πa3ε0E0 −u

2du

1

−1

∫

pz = 4πa

3ε0E0

Compare 1/r2 potentials

Polarizability

Dipole potential

Charge density

Dipole moment analogous to p = qd

Dipole moment

Lecture 21 Carl Bromberg - Prof. of Physics 11

5.13

V (r,θ) =q

4πε0

1

r2 + a

2 − 2ar cosθ+

2

r+

1

r2 + a

2 + 2ar cosθ

⎧⎨⎩⎪

⎫⎬⎭⎪

1

1+ ε= 1−

1

2ε + 3

8ε 2 +…

1

r2 + a

2 ± 2ar cosθ=

1

r1±

acosθr

+3a

2cos

2θ − a2

2r2

⎧⎨⎪

⎩⎪

⎫⎬⎪

⎭⎪

V (r,θ) =

qa2

4πε0

3cos2θ −1

2r3

⎧⎨⎪

⎩⎪

⎫⎬⎪

⎭⎪=

2qa2

4πε0

P3(cosθ)

r3

+q

+q

–2q

–a

+a r1

r2

r

Linear quadrupoleLinear quadrupole potential

Expansions

Monopole and dipole terms cancel

Lecture 21 Carl Bromberg - Prof. of Physics 12

5.16

V (r,φ) = Aln r + B + Anr

n + Bnr−n( ) Cn cos nφ + Dn sin nφ( )

n=1

∞

∑

Er (Rext ) − Er (Rint ) =

σε0

V (rint ,φ) = Ar cosφ En(rint ,φ) = −Acosφ

V (rext ,φ) =B

rcosφ En(rext ,φ) = +

Bcosφ

rext

2

Bcosφ

rext

2+ Acosφ =

σ0 cosφε0

B

R2+ A =

σ0

ε0

Vint (R,φ) =Vext (R,φ)

ARcosφ =B

Rcosφ

AR =B

R A =

σ0

2ε0

B =σ0R

2

2ε0

V (rint ,φ) =

σ0

2ε0

r cosφ; V (rext ,φ) =σ0R

2

2ε0rcosφ

General solution in cylindrical coordinates

Boundary conditioncrossing the boundary

Matching terms

Applyboundaryconditions

Solution insideand out

Lecture 21 Carl Bromberg - Prof. of Physics 13

5.32

V =pcosθ

4πε0r2

V (r,θ) = Ar

+B

r+1

⎛⎝⎜

⎞⎠⎟=0

∞

∑ P cosθ( )

V (r,θ) = Ar +

B

r2

⎛⎝⎜

⎞⎠⎟

P1 cosθ( )

B =

p

4πε0

V (R,θ) = AR +B

R2

⎛⎝⎜

⎞⎠⎟= 0

A =−B

R3=

− p

4πε0R3

V (r,θ) =

p

4πε0

−r

R3+

1

r2

⎛⎝⎜

⎞⎠⎟

cosθ

Dipole in groundedconducting sphere

General solution in spherical coordinates

Only one term = 1

Dipole potential

Apply boundary conditions

Solution