Lecture 3 Electrostatics And Conductors

Electrostatics and Conductors 1

Carl Gottfried Neumann (May 7th 1832 – March 27th 1925)

Electrostatic properties of conductors

Peter Gustav Lejeune Dirichlet (February 13th 1805 – May 5th 1859)

2

Conductors in Equilibrium with Static Electric Fields Property 1 Inside a conductor in electrostatic equilibrium, the electric field is zero: E(x) = 0. Property 2 Inside a conductor in electrostatic equilibrium, the electric potential is constant: V (x) = V0. The surface of the conductor, which is also at potential V0, is an Property 3 surface. equipotential Inside a conductor in electrostatic equilibrium, the charge density is zero: ρ (x) = ε 0∇ ⋅ E(x) = 0. Any excess charge on a conductor must therefore reside on its surface, described by a surface charge density σ .

3

Conductors in Equilibrium with Static Electric Fields(cont’) Property 4 At the surface of a conductor, E tangential ≡ Et = 0 E normal

σ ≡ En = ε0

There is an outward force per unit area on the surface of a conductor, or electrostatic pressure, given by 2     σ 2 F 1 1 1 σ lim = lim lim  ε 0   × Aδl   = A→ 0 A A→0 A δl →0 δl 2   ε0     2ε 0 The origin of this force is the Coulomb repulsion between like charges: 1 F σ  σ2 lim = lim  × ( σA) × =  A→ 0 A A→ 0 A 2ε 0  2ε 0  4

Conductors in Equilibrium with Static Electric Fields(cont’) Property 5 (Uniqueness theorem) The potential V (x) throughout space is determined (uniquely up to an additive constant) when (a) the external charge densities are specified, and (b) either the potentials of all conductors are specified: V (x) = V0 – Dirichlet boundary condition or the charge densities on all surfaces are specified: − nˆ ⋅ ∇V ( x ) = σ( x ) ε 0 Neumann boundary “The practical importance condition of the uniqueness theorem is that when a potential function satisfying the equations and boundary conditions has been found, by any method, then we need look no further – any other function satisfying these conditions can only differ by an additive constant.” 5

The Uniqueness Theorem for Laplace’s Equation Suppose V1(x) and V2(x) are both functions that satisfy Laplace’s equation in a volume V 2of space: ∇ V1 = 0 ∇ 2V2 = 0 and that satisfy the same boundary conditions on the set of boundary surfaces S of the volume. Consider the function

F( x ) = ( V1 − V2 ) ∇( V1 − V2 )

and the integral

3 ∇ ⋅ F ( x ) d x = ∫ F ⋅ nˆ dA = ∫ ( V1 − V2 )( nˆ ⋅ ∇V1 − nˆ ⋅ ∇V2 ) dA = 0 ∫ V

S

S

since both V1 and V2 satisfy the same boundary conditions on S: V1 = V2 Dirichlet boundary condition nˆ ⋅ ∇V1 = nˆ ⋅ ∇V2

Neumann boundary condition

6

The Uniqueness Theorem for Laplace’s Equation (cont’) 2 ∇ ⋅ F ( x ) = ∇ ∇ ( V ⋅ F − ( x V ) = ) ⋅ ∇ ∇ ( ( V V − − V V ) ) ⋅ + ∇ ( ( V V − − V V ) ) ∇ ( V1 − V2 ) Now, 1 2 11 22 11 22 Therefore,

2 3 3 [ ] ∇ ( V − V ) d x = ∇ ⋅ F ( x ) d x=0 ∫V 1 2 ∫V

The integrand is never negative, so the only way the integral can be 0 is if ∇( V1 − V2 ) = 0 V1 ( x ) − V2 ( x ) = constant for all x. Dirichlet boundary condition

V1 = V2 ⇒ V1 ( x ) = V2 ( x )

nˆ ⋅ ∇V1 = nˆ ⋅ ∇V2 ⇒ E1 ( x ) = E 2 ( x ) Neumann boundary condition 7

Conductors in Equilibrium with Static Electric Fields(cont’) Property 6 The electric field is zero in any empty cavity in a conductor in electrostatic equilibrium, regardless of the electric field outside the conductor. Property 7 If a charged conductor has within it an empty cavity, the field in the cavity is zero and the charge on the cavity wall is zero – the net charge of the conductor lies entirely on the outer surface.

8

Electrostatic problems with rectangular symmetry

9

Charged Plates Consider two plane conducting plates 1 and 2 separated by a vacuum. Laplace ’s V ( d ) = V0 > 0 equatio 2 ∇ V ( x) = 0 n V ( 0) = 0 Assume the plates each have a “large” area A - translation invariance in directions parallel V ( x ) = V ( z ) to the plates: V0 V0 ˆ d2 V ( z) = z E = −∇ V = − k V ( z) = 0 2 d d dz The surface charge densities εV σ1u = ε 0kˆ ⋅ E = − 0 0 d

εV σ 2l = ε 0 ( − kˆ ) ⋅ E = 0 0 d

10

Charged Plates (cont’) The total charges on the conductors are equal and opposite: ε A ε A Q1 = − 0 V0 Q2 = 0 V0 d d Recall two conductors that carry equal but opposite charges ± Q form a capacitor. The capacitance of the system Q C≡ ∆V where ∆ V is the potential difference between the conductors. It follows from the above that A C = ε0 d for a parallel plate capacitor.

Homework: Work through Example 1 @ Page 100

11

Charged Plates (cont’) Consider a capacitor with arbitrarily shaped conductors at potentials V1 and V2 with respective Assuming is no charges –Qthere and +Q. external field or other charges anywhere, the total electrostatic1 energy 1 1 1 U = ∫ ρVd 3 x = ∫ σVdA = ( − QV1 + QV2 ) = Q∆V 2 2 2 2 where the difference of potential between the conductors ∆ V ≡ V2 – V1.

12

External Point Charges. The Method of Images Consider a point charge at distance z0 from a grounded plane. (0, 0, z0)

For z < 0, V = 0 and E = 0; while for z ≥ 0, V satisfies the Poisson’s equation 1 ∇ 2V ( x ) = qδ( x ) δ( y ) δ( z − z0 ) ε0 with the boundary condition

V ( x , y ,0 ) = 0

13

External Point Charges. The Method of Images (cont’) Method of image charge  q  1 1 V ( x) = − 2  2 2 2 2 2 4π ε0  x + y + ( z − z0 ) x + y + ( z + z0 )   xˆi + yˆj + ( z − z ) kˆ  ˆ ˆ ˆ x i + y j + ( z + z ) k 0 0 −  2  ε0  x + y 2 + ( z − z ) 2 3 2 x 2 + y 2 + ( z + z ) 2 3 2  0 0 qz0 E( x, y,0 ) = − kˆ 3 2 2π ε0 ( x 2 + y 2 + z02 )

q E( x ) = −∇ V ( x ) = 4π

[

On the surface z = 0, It follows that the surface charge density:

The total charge induced on the plane:

]

σ( x, y,0 ) = − ∞

[

]

qz0

2 π( x + y + z 2

2

)

2 32 0

∞

∫ ∫ σ( x, y,0) dxdy = −q −∞ −∞

When q is placed at (0, 0, z0), an equal but opposite charge is drawn onto the conductor from ground.

14

Example 2 Consider a point-like electric dipole p located at the point (0, 0, z0) above the grounded plane z = 0. What is the asymptotic electric field on the z axis for z >> z0? Solution: rˆ ⋅ p V ( x ) = Recall the potential of a point-like dipole: 4π ε0 r 2 It follows that the potential of the above configuration is given by  p0  rˆ ⋅ ˆj rˆ ′ ⋅ ˆj V ( x) = − 2  2 2 2 2 4π ε0  x + y + ( z − z0 ) x + y 2 + ( z + z0 )  p0  3( rˆ ⋅ ˆj)rˆ − ˆj and E( x ) =  2 4π ε0  x + y 2 + ( z − z ) 2 0

[

]

32

−

[x

3( rˆ ′ ⋅ ˆj)rˆ ′ − ˆj

2

+ y 2 + ( z + z0 )

2

]

 32 

15

Example 2 (cont’) Solution (cont’):

For points on the z axis, p0  1 1 ˆ E( 0,0, z ) = − + j  3 3 4π ε0  ( z − z0 ) ( z + z0 )  For z >> z0, −3 −3  p0   z 0  z 3 p0 z 0 ˆ  0  ˆ E( 0,0, z ) = − 1 −  + 1 +   j ≈ − j 3   4 4π ε0 z   z z  2π ε0 z 

– the far field is that of a quadrupole. 16

Multiple Images Consider a point charge +q located at the point (0, y0, z0), with mutually perpendicular planes y = 0 and z = 0 both grounded.

Recall the potential of a point charge: q 1 V ( 0, y0 , z0 ) = 4π ε0 x 2 + ( y − y0 ) 2 + ( z − z0 ) 2

17

Multiple Images (cont’) It follows that the potential of the above configuration is given by q 1 V ( x) = 4π ε0 x 2 + ( y − y0 ) 2 + ( z − z0 ) 2 q 1 − 4π ε0 x 2 + ( y + y0 ) 2 + ( z − z0 ) 2 −

q 1 4π ε0 x 2 + ( y − y0 ) 2 + ( z + z0 ) 2

q 1 + 4π ε0 x 2 + ( y + y0 ) 2 + ( z + z0 ) 2

Homework: Work through Example 3 @ Page 106

18

Electrostatic problems with spherical symmetry

19

Charged Spheres Consider a conducting sphere of radius a that carries a charge Q0. The potential inside the conductor, i.e., for r < a, V = V0. For r ≥ a, V satisfies the Laplace’s equation 1d 2dV  22 dV  2 ∇ V ( x) = 0 +r ==00 2 2  r drdr  r dr  with boundary conditions V ( a ) = V0 It follows that

V ( ∞) = 0 A V ( r ) = + B with A = aV0 and B = 0. r

aV0 d  aV0  E( x ) = −∇ V ( x ) = −  rˆ = 2 rˆ dr  r  r Q0 aV0 = σ = ε 0rˆ ⋅ E( a ) = ε 0 2 2 4πa a

Q0 V0 = 4π ε0 a

20

Charged Spheres (cont’) Consider two concentric conducting spheres. For a ≤ r ≤ b, V satisfies the Laplace’s equation d 2V 2 dV + =0 2 dr r dr with boundary conditions V ( a ) = V0

V ( b) = 0

A It follows that V ( r ) = + B r A V ( a ) = + B = V0 a or

with A and B satisfying A V ( b) = + B = 0 b

A A A( b − r ) aV0 ( b − r ) V ( r) = − = = r b br ( b − a) r

21

Charged Spheres (cont’) d  aV0 ( b − r )  E( x ) = −∇ V ( x ) = −  rˆ  dr  ( b − a ) r  E( r ) =

abV0 rˆ 2 ( b − a) r

The surface charge densities bV0 σ a = ε 0rˆ ⋅ E( a ) = ε 0 ( b − a) a aV0 σb = ε 0 ( − rˆ ) ⋅ E( b ) = −ε 0 ( b − a)b The total charges on the conductors are equal and opposite: 2 4π ε0 ab 4π ε0 ab Qa = 4πa σ a = V0 Qb = 4πb 2 σb = − V0 b−a b−a The capacitance of the two concentric C ≡ Q = 4π ε0 ab 22 ∆V b−a spherical conductors

Spherical Problems with Dependence on θ Laplace’s equation

∇ 2V ( x ) = 0

∂  ∂ =0 reduce 1 ∂ r 2 ∂ V ( r , θ)  + 1 sin θ V ( r , θ )  r 2 sin θ ∂θ   r 2 ∂r  ∂r ∂θ s to or

∂ 2V 2 ∂V 1  ∂ 2V cos θ ∂V  + + 2 2 + =0 2 ∂r r ∂r r  ∂θ sin θ ∂θ 

It can be shown that the following is a solution: A C cos θ V ( r , θ) = + B + + Dr cos θ 2 r r The final term corresponds to a field uniform throughout space: ∂ 1 ∂ − ∇( Dr cos θ) = − ( Dr cos θ) rˆ − ( Dr cos θ) θˆ ∂r r ∂θ = − D( cos θrˆ − sin θθˆ ) = − Dkˆ

23

Example 4 Consider a grounded conducting sphere, of radius a and centered at the origin, in an externally applied field in the z direction: E appl ( x ) = E0kˆ The presence of the sphere changes the field. What are the potential V (r, θ ), field E(x), and surface charge density σ (θ )? Solution: A C cos θ V ( r , θ) = + B + + Dr cos θ 2 r r A = 0, B = 0

D = − E0

C = a 3 E0

24

Example 4 (cont’) Solution: (cont’) For r < a, V = 0; while for r ≥ a, a 3 E0 cos θ V ( r , θ) = − E0 r cos θ 2 r It follows that ∂V  2a 3  Er ( r , θ ) = − =  3 + 1 E0 cos θ ∂r  r  1 ∂V  a 3  Eθ ( r , θ ) = − =  3 − 1 E0 sin θ r ∂θ  r  σ( θ) = ε 0rˆ ⋅ E( a ) = 3ε 0 E0 cos θ The total surface charge π

Qtot = ∫ σ( θ)( 2πa sin θ) adθ = 0 0

25

External Charges Consider a grounded spherical conductor of radius R, centered at the origin; and an external point charge q located at (0, 0, z0):

For r ≥ R 1 V ( r , θ) = 4π

  ′ q q   + 2 2 2 2 ε0  r + z0 − 2rz0 cos θ r + z0′ − 2rz0′ cos θ 

26

External Charges (cont’)  1  q q′ =0 + At A, V ( R,0 ) = 2 2 2 2 4π ε0  R + z0 − 2 Rz0 R + z0′ − 2 Rz0′  1 At B, V ( R, π ) = 4π

  ′ q q  =0 + 2 2 2 2  ε0  R + z0 + 2 Rz0 R + z0′ + 2 Rz0′ 

It follows that q( R − z0′ ) = − q′( z0 − R )

conjugate R −points z0′ z0 − R = ⇒ z0 z0′ = R 2 R + z0′ z0 + R

q( R + z0′ ) = − q′( z0 + R )

 R2  R ′ ′ q R +  = − q ( z0 + R ) ⇒ q = − q z0  z0 

q And V ( r , θ) = 4π q = 4π

  1 R   − 2 2 2 2  ε0  r + z0 − 2rz0 cos θ z0 r + z0′ − 2rz0′ cos θ 

  1 R   − 2 2 2 2 4 2 ε0  r + z0 − 2rz0 cos θ z0 r + R − 2 z0 R r cos θ 

27

External Charges (cont’) 1 1 − −   2 4 2 2 2    q 1  z0 2 z0 R  R 2R 1 + 2 2 − V ( r , θ) = cos θ  − cos θ  1 + 2 − 4π ε0  r  r r z0 r  z0 r z0 r     

The first two leading terms in the potential for the far field 2  q 1  z0  R  R V ( r , θ) ≈ 1+ cos θ  1 + cos θ  −   4π ε0  r  r  z0 r  z0 r  1  R 1  R3  1 − q + 1 − 3 qz0 cos θ = 2  4π ε0 r  z0  4π ε0 r  z0  Electric field

R ( z02 r − z0 R 2 cos θ)

q Er ( r , θ ) = 4π

 r − z0 cos θ  − 3 2 2 ε0  r + z − 2rz cos θ 0 0 

  3 2 2 4 2 z0 r + R − 2 z0 R r cos θ 

q Eθ ( r , θ ) = 4π

 z0 sin θ  − 3 2 2 ε0  r + z − 2rz cos θ 0 0 

  283 2 2 4 2 z0 r + R − 2 z0 R r cos θ 

(

(

) (

) (

R 3 z0 sin θ

)

)

External Charges (cont’) The surface charge density: q   = 4π  

(

σ( θ) = ε 0rˆ ⋅ E( R, θ)

R − z0 cos θ R 2 + z02 − 2 Rz0 cos θ

q   = 4πR  

(

The total charge on the conductor: qR ( R − z = 2 2

2 0

)

−1

(R

− z0 R cos θ)

2 0

π

+ z − 2 Rz0u )

)

)

3

R σ( θ)( 2πR sin θ × Rdθ) = − q z0

q( R − z = 2 z0 2

3 2

)

  3 z02 + R 2 − 2 z0 R cos θ  2 0

R + z − 2 Rz0 cos θ 2

0

2 0

(z

R 2 − z02

du 2

−

) ( 3

Qtot = ∫

1

∫

  3 z02 R 2 + R 4 − 2 z0 R 3 cos θ 

−

R 2 + z02 − 2 Rz0 cos θ q = 4πR

R ( z02 R − z0 R 2 cos θ)

) ( 3

R 2 − z0 R cos θ

(

Pg 115 – 116

2 0

)

1

1

R 2 + z02 − 2 Rz0u29−1

Electrostatic problems with cylindrical symmetry

30

Charged Lines and Cylinders Consider a charged line with constant linear charge density λ .two symmetries: translation invariance in z, the There are coordinate along the line; and rotational invariance in φ , the azimuthal angle. 2 1 dV  1dV d dV ∇ 2V ( x ) = 0 r + =0 2  rdrdr  rdrdr Laplace’s whichequation has the general solution:

r V ( r ) = A ln  + B  r0  dV A E( x ) = −∇ V ( x ) = − rˆ = − rˆ dr r

A 1 λ ∫ E( x ) ⋅ dA = − r × 2πrl = ε0 λl ⇒ A = − 2π ε0 31

Charged Lines and Cylinders (cont’) Consider a conducting cylinder of radius a which carries the constant line charge density λ spread uniformly on its surface: λ λl = 2πalσ ⇒ σ = 2πa In the region r ≥ a, V satisfies d 2V 1 dV + =0 2 dr r dr

∇ V ( x) = 0 2

Laplace’s whichequation has the general solution:

r  V ( r ) = A ln  + B a dV A E( x ) = −∇ V ( x ) = − rˆ = − rˆ dr r

A 1 λ ∫ E( x ) ⋅ dA = − r × 2πrl = ε0 λl ⇒ A = − 2π ε0

32

Charged Lines and Cylinders (cont’) Consider two concentric conducting cylinders: In the region a ≤ r ≤ b, V satisfies Laplace’s equation d 2V 1 dV + =0 2 dr r dr with boundary conditions V ( a ) = V0 It follows that

V ( b) = 0 r  V ( r ) = A ln  + B b

a  V ( a ) = A ln  + B = V0 b or

ln( b r ) V ( r) = V0 ln( b a )

with A and B satisfying V ( b) = B = 0

33

Charged Lines and Cylinders (cont’) d  ln( b r )  E( x ) = −∇ V ( x ) = −  V0 rˆ dr  ln ( b a )  E( r ) =

V0 1 rˆ ln( b a ) r

The surface charge densities V0 σ a = ε 0rˆ ⋅ E( a ) = ε 0 ln( b a ) a V0 σb = ε 0 ( − rˆ ) ⋅ E( b ) = −ε 0 ln( b a ) b The total charges on the conductors are equal and opposite: 2π ε0l 2π ε0l Qa = 2πalσ a = V0 Qb = 2πblσb = − V0 ln( b a ) ln( b a ) The capacitance per unit length of the C ′ ≡ Q = 2π ε0 34 l∆V ln( b a ) two concentric cylindrical conductors

Two Infinitely Long Line Charges Consider two infinitely long line charges with equal but opposite linear charge densities +λ and –λ , parallel to the z axis, a distance 2d apart. The potential at (r, φ ): λ λ V ( r , φ) = − ln r+ + ln r− 2π ε0 2π ε0  r−  λ = ln  2π ε0  r+  λ r 2 + d 2 + 2dr cos φ V ( r , φ) = ln 2 4π ε0 r + d 2 − 2dr cos φ λ x + y + d + 2dx λ ( x + d ) + y2 V ( x, y ) = ln 2 = ln 2 2 4π ε0 x + y + d − 2dx 4π ε0 ( x − d ) 2 + y 2 2

2

2

2

35

Two Infinitely Long Line Charges (cont’) λ x + y + d + 2dx λ ( x + d ) + y2 V ( x, y ) = ln 2 = ln 2 2 4π ε0 x + y + d − 2dx 4π ε0 ( x − d ) 2 + y 2 2

2

2

2

The equation for an equipotential curve: λ ( x + d ) 2 + y2 ( x + d ) 2 + y2  4π ε0V0  ln = V0 = exp  2 2 2 2 4π ε0 ( x − d ) + y λ (x−d) + y  

[

]

[

2 2  2π ε0V0   − 2π ε0V0  2 exp ( x − d ) + y = exp ( x + d ) + y2    λ  λ   

]

 2π ε0V0 ( 2  2π ε0V0  2 2 ) sinh  x + d + y − 2 cosh   dx = 0  λ   λ  2

  2π ε0V0  2 2  2π ε 2 0V0  x − d coth − d csch + y =0        λ   λ 

( x − x0 ) 2 + y 2 = R 2 Homework: Work through Example 6 @ Page 122

36

Two Infinitely Long Line Charges (cont’) λ r 2 + d 2 + 2dr cos φ V ( r , φ) = ln 2 4π ε0 r + d 2 − 2dr cos φ In the dipole limit, where d → 0 and λ → ∞ with p′ = 2λdˆi 2λ d remaining finite, we have a line dipole described by the vector dipole moment per unit length: of a line dipole The potential λ r 2 + d 2 + 2dr cos φ V ( r , φ) = ln 2 4π ε0 r + d 2 − 2dr cos φ  2d cos φ d 2  λ   2d cos φ d 2  = ln1 + + 2  − ln1 − + 2   4π ε0   r r  r r   λ  2d cos φ d 2  2d cos φ d 2  ≈ + 2 −− + 2   4π ε0  r r  r r  2λd cos φ p′ ⋅ rˆ V ( r , φ) ≈ = 2π ε0 r 2π ε0 r

37

Cylindrical Problems with Dependence on φ Laplace’s equation reduce s to or

∇ 2V ( x ) = 0

1∂  ∂ 1 ∂2  r V ( r , φ)  + 2 2 V ( r , φ) = 0  r ∂r  ∂r  r ∂φ ∂ 2V 1 ∂V 1 ∂ 2V + + 2 2 =0 2 ∂r r ∂r r ∂φ

It can be shown that the following is a solution: C cos φ V ( r , φ) = A ln r + B + + Dr cos φ r The final term corresponds to a field uniform throughout space: ∂ 1 ∂ − ∇( Dr cos φ) = − ( Dr cos φ) rˆ − ( Dr cos φ) φˆ ∂r r ∂φ = − D( cos φrˆ − sin φφˆ ) = − Dˆi

38

Example 5 Consider a grounded conducting cylinder, of radius a and centered at the origin, in an externally applied field in the x direction: E appl ( x ) = E0 ˆi The presence of the sphere changes the field. What are the potential V (r, φ ), field E(x), and surface charge density σ (φ )? Solution: V ( r , φ) = A ln r + B + A = 0, B = 0

C cos φ + Dr cos φ r D = − E0

C = a 2 E0 39

Example 5 (cont’) Solution: (cont’) For r < a, V = 0; while for r ≥ a, a 2 E0 cos φ V ( r , φ) = − E0 r cos φ r

St. Elmo’s fire

It follows that ∂V  a 2  Er ( r , φ ) = − =  2 + 1 E0 cos φ ∂r  r  1 ∂V  a 2  Eφ ( r , φ ) = − =  2 − 1 E0 sin φ r ∂φ  r  σ( φ) = ε 0rˆ ⋅ E( a ) = 2ε 0 E0 cos φ The total surface charge 2π

λ tot = ∫ σ( φ) adφ = 0 0

40

External Line Charge Consider y = 0 to be a grounded conducting plane. There is an infinitely long line charge above it. The line charge is parallel to the z axis, intersects the xy plane at the point (0, y0) and has constant linear charge density +λ . x 2 + ( y − y0 ) x 2 + ( y + y0 ) λ λ V ( x, y ) = − ln + ln 2π ε0 r0 2π ε0 r0 2

λ E( x , y ) = 2π

2

 xˆi + ( y − y0 ) ˆj xˆi + ( y + y0 ) ˆj  − 2  2 2 2 ε0  x + ( y − y0 ) x + ( y + y0 ) 

y0 ˆj λ E( x , 0 ) = − π ε0 x 2 + y02

λ y0 ˆ σ ( x , z ) = ε 0 E( x , 0 ) ⋅ j = − π x 2 + y02

The induced charge per unit length on the plane ∞ λy0 ∞ dx ∫−∞ σ( x, z ) dx = − π ∫−∞ x 2 + y02 = −λ 41

External Line Charge (cont’) Consider an infinite line charge +λ in the presence of an infinite isolated charged conducting cylinder with linear charge density – λ .

(

1 V ( r , φ) = − λ ln r 2 + x02 − 2rx0 cos φ 2π ε0 + λ′ ln r 2 + x0′2 − 2rx0′ cos φ

)

1 [ λ ln( x0 − R ) + λ′ ln( R − x0′ ) ] V ( R,0 ) = − 2π ε0 1 [ λ ln( R + x0 ) + λ′ ln( R + x0′ ) ] = V ( R, π) =− 2π ε0

It follows if λ ’ = – λ , that x0 x0′ = R 2 conjugate points

42

External Line Charge (cont’)

R λ V ( R, φ) = ln  2π ε0  x0  r 2 + x0′2 − 2rx0′ cos φ λ V ( r , φ) = ln 2 4π ε0 r + x02 − 2rx0 cos φ r 2 x02 + R 4 − 2 R 2 rx0 cos φ λ = ln 2 2 4π ε0 x0 ( r + x02 − 2rx0 cos φ)

R 2 ( R 2 + r 2 x02 R 2 − 2rx0 cos φ) λ = ln 4π ε0 x02 ( r 2 + x02 − 2rx0 cos φ) 43

External Line Charge (cont’) Consider an isolated, uncharged infinite cylindrical conductor of radius R centered at the origin. An external line charge +λ passes through the point (x0, 0), parallel to the z axis, Since conductor uncharged, which the is the cylinderis axis. R 2 ( R 2 + r 2 x02 R 2 − 2rx0 cos φ) λ λ r V ( r , φ) = ln − ln 2 2 2 4π ε0 x0 ( r + x0 − 2rx0 cos φ) 2π ε0 α

44