time independent electromagnetic fields 611 spring... · symmetries. focus on the techniques,...

1Prof. Sergio B. MendesSpring 2018

Time Independent Electromagnetic Fields

Chapter 3 of “Modern Problems in Classical Electrodynamics” by Charles Brau

Chapter 2 and 3 of “Classical Electrodynamics” by John Jackson


𝑩𝑩 𝒓𝒓 = 𝜵𝜵 × 𝑨𝑨 𝒓𝒓

−𝛻𝛻𝟐𝟐𝑨𝑨 𝒓𝒓 = 𝜇𝜇𝑜𝑜 𝑱𝑱 𝒓𝒓−𝛻𝛻2Φ 𝒓𝒓 =𝜌𝜌 𝒓𝒓𝜖𝜖0

𝑬𝑬 𝒓𝒓 = −𝛻𝛻Φ 𝒓𝒓

Φ 𝒓𝒓 ≡1

4 𝜋𝜋 𝜖𝜖0�−∞

+∞𝜌𝜌 𝒓𝒓𝒓𝒓𝒓 − 𝒓𝒓′

𝑑𝑑𝑑𝑑𝒓 𝑨𝑨 𝒓𝒓 ≡𝜇𝜇𝑜𝑜

4 𝜋𝜋�−∞

+∞𝑱𝑱 𝒓𝒓𝒓𝒓𝒓 − 𝒓𝒓′

𝑑𝑑𝑑𝑑𝒓

Electrostatics Magnetostatics

𝜌𝜌 𝒓𝒓𝑱𝑱 𝒓𝒓

𝛁𝛁. 𝑱𝑱 𝒓𝒓, 𝑡𝑡 = 0−𝜕𝜕𝜌𝜌 𝒓𝒓, 𝑡𝑡𝜕𝜕𝑡𝑡

= 𝛁𝛁. 𝑱𝑱 𝒓𝒓, 𝑡𝑡𝜕𝜕𝜌𝜌 𝒓𝒓, 𝑡𝑡𝜕𝜕𝑡𝑡

= 0

charge conservation

𝑰𝑰


−𝛻𝛻𝟐𝟐𝑨𝑨 𝒓𝒓 = 𝜇𝜇𝑜𝑜 𝑱𝑱 𝒓𝒓

−𝛻𝛻2Φ 𝒓𝒓 =𝜌𝜌 𝒓𝒓𝜖𝜖0

Φ 𝒓𝒓 ≡1

4 𝜋𝜋 𝜖𝜖0�−∞



𝑨𝑨 𝒓𝒓 ≡𝜇𝜇𝑜𝑜

4 𝜋𝜋�−∞




Green’s Function

Symmetries

Focus on the techniques, rather than on the individual problems


Symmetries


Exploring Symmetries

𝒓𝒓 = (𝑥𝑥,𝑦𝑦, 𝑧𝑧)

𝑥𝑥𝑞𝑞

𝒅𝒅 = (𝑑𝑑, 0,0)

𝑦𝑦

Φ 𝒓𝒓 =1

4 𝜋𝜋 𝜖𝜖0𝑞𝑞

𝒓𝒓 − 𝒅𝒅Φ 𝑥𝑥 = 0, 𝑦𝑦, 𝑧𝑧 = 0 + ??

−𝑞𝑞−𝒅𝒅 = (−𝑑𝑑, 0,0)

=1


𝒓𝒓 − 𝒅𝒅+

14 𝜋𝜋 𝜖𝜖0

−𝑞𝑞𝒓𝒓 + 𝒅𝒅

Φ 𝑥𝑥 = 0, 𝑦𝑦, 𝑧𝑧 = 0

=1


𝑥𝑥 − 𝑑𝑑 2 + 𝑦𝑦2 +𝑧𝑧2−

𝑞𝑞𝑥𝑥 + 𝑑𝑑 2 + 𝑦𝑦2 +𝑧𝑧2


𝑬𝑬 𝒓𝒓 = −𝛻𝛻Φ 𝒓𝒓

Φ 𝑥𝑥,𝑦𝑦, 𝑧𝑧 =1


𝑥𝑥 − 𝑑𝑑 2 + 𝑦𝑦2 +𝑧𝑧2−

𝑞𝑞𝑥𝑥 + 𝑑𝑑 2 + 𝑦𝑦2 +𝑧𝑧2

𝐸𝐸𝑥𝑥 𝑥𝑥, 𝑦𝑦, 𝑧𝑧 = −1

4 𝜋𝜋 𝜖𝜖0−

𝑥𝑥 − 𝑑𝑑 𝑞𝑞𝑥𝑥 − 𝑑𝑑 2 + 𝑦𝑦2 +𝑧𝑧2 3/2 +

𝑥𝑥 + 𝑑𝑑 𝑞𝑞𝑥𝑥 + 𝑑𝑑 2 + 𝑦𝑦2 +𝑧𝑧2 3/2

𝐸𝐸𝑦𝑦 𝑥𝑥, 𝑦𝑦, 𝑧𝑧 = −1


𝑦𝑦 𝑞𝑞𝑥𝑥 − 𝑑𝑑 2 + 𝑦𝑦2 +𝑧𝑧2 3/2 +

𝑦𝑦 𝑞𝑞𝑥𝑥 + 𝑑𝑑 2 + 𝑦𝑦2 +𝑧𝑧2 3/2

𝐸𝐸𝑧𝑧 𝑥𝑥, 𝑦𝑦, 𝑧𝑧 = −1


𝑧𝑧 𝑞𝑞𝑥𝑥 − 𝑑𝑑 2 + 𝑦𝑦2 +𝑧𝑧2 3/2 +

𝑧𝑧 𝑞𝑞𝑥𝑥 + 𝑑𝑑 2 + 𝑦𝑦2 +𝑧𝑧2 3/2


𝑬𝑬 𝑥𝑥 = 0,𝑦𝑦, 𝑧𝑧 = −𝑞𝑞 𝑑𝑑

2 𝜋𝜋 𝜖𝜖0 𝑑𝑑2 + 𝑦𝑦2 +𝑧𝑧2 3/2 �𝒆𝒆𝑥𝑥

𝜎𝜎 𝑥𝑥 = 0,𝑦𝑦, 𝑧𝑧 = 𝜖𝜖0 𝑬𝑬.𝒏𝒏 = −𝑞𝑞 𝑑𝑑

2 𝜋𝜋 𝑑𝑑2 + 𝑦𝑦2 +𝑧𝑧2 3/2

𝑥𝑥

𝑦𝑦

𝑥𝑥 = 0


HW: What is the total induced charge on the grounded metallic surface ?

HW: What is the energy of attraction between a point charge and a grounded metallic surface ?


𝑥𝑥

+𝑞𝑞

𝑦𝑦

−𝑞𝑞

−𝑞𝑞+𝑞𝑞

𝑎𝑎

𝑏𝑏

Φ 𝑥𝑥, 𝑦𝑦, 𝑧𝑧 =1


𝑥𝑥 − 𝑎𝑎 2 + 𝑦𝑦 − 𝑏𝑏 2 +𝑧𝑧2

+1

4 𝜋𝜋 𝜖𝜖0−𝑞𝑞

𝑥𝑥 − 𝑎𝑎 2 + 𝑦𝑦 + 𝑏𝑏 2 +𝑧𝑧2

+1

4 𝜋𝜋 𝜖𝜖0−𝑞𝑞

𝑥𝑥 + 𝑎𝑎 2 + 𝑦𝑦 − 𝑏𝑏 2 +𝑧𝑧2

+1


𝑥𝑥 + 𝑎𝑎 2 + 𝑦𝑦 + 𝑏𝑏 2 +𝑧𝑧2


HW: What is the energy of attraction between a dipole and a grounded metallic surface ?

𝑦𝑦

𝜙𝜙 = 0


Grounded Conducting Sphere, Charge Outside the Sphere

𝑞𝑞

𝑎𝑎


Φ 𝒓𝒓 =1


𝒓𝒓 − 𝒓𝒓𝟎𝟎+ ??

𝑞𝑞𝒓𝒓𝟎𝟎 = 𝑟𝑟0 �𝒏𝒏𝟎𝟎

𝒓𝒓 = 𝑟𝑟 �𝒏𝒏

=1


𝒓𝒓 − 𝒓𝒓𝟎𝟎+


𝑄𝑄𝒓𝒓 − 𝑹𝑹𝟎𝟎

𝑎𝑎

𝑹𝑹𝟎𝟎 = 𝑅𝑅0 �𝒏𝒏𝟎𝟎

Φ 𝑎𝑎 �𝒏𝒏 =1


𝑎𝑎 �𝒏𝒏 − 𝑟𝑟0 �𝒏𝒏𝟎𝟎+


𝑄𝑄𝑎𝑎 �𝒏𝒏 − 𝑅𝑅0 �𝒏𝒏𝟎𝟎

= 0

𝑞𝑞 𝑎𝑎 �𝒏𝒏 − 𝑅𝑅0 �𝒏𝒏𝟎𝟎 + 𝑄𝑄 𝑎𝑎 �𝒏𝒏 − 𝑟𝑟0 �𝒏𝒏𝟎𝟎 = 0


𝒓𝒓 − 𝒓𝒓𝟎𝟎

for 𝑟𝑟 = 𝑎𝑎

𝑄𝑄

𝒓𝒓 − 𝑹𝑹𝟎𝟎

=1


𝑟𝑟 �𝒏𝒏 − 𝑟𝑟0 �𝒏𝒏𝟎𝟎+


𝑄𝑄𝑟𝑟 �𝒏𝒏 − 𝑅𝑅0 �𝒏𝒏𝟎𝟎


𝑞𝑞 𝑎𝑎 �𝒏𝒏 − 𝑅𝑅0 �𝒏𝒏𝟎𝟎 + 𝑄𝑄 𝑎𝑎 �𝒏𝒏 − 𝑟𝑟0 �𝒏𝒏𝟎𝟎 = 0

𝑞𝑞2 𝑎𝑎2 + 𝑅𝑅02 − 2 𝑎𝑎 𝑅𝑅0 𝑐𝑐𝑐𝑐𝑐𝑐 𝛾𝛾 = 𝑄𝑄2 𝑎𝑎2 + 𝑟𝑟02 − 2 𝑎𝑎 𝑟𝑟0 𝑐𝑐𝑐𝑐𝑐𝑐 𝛾𝛾

𝑄𝑄 = −𝑞𝑞𝑅𝑅0 = 𝑟𝑟0𝑞𝑞2 𝑎𝑎2 = 𝑄𝑄2 𝑟𝑟02

𝑞𝑞2 𝑅𝑅0 = 𝑄𝑄2 𝑟𝑟0

𝑅𝑅0 =𝑎𝑎2

𝑟𝑟0

Solution 1 Solution 2

𝑄𝑄 = −𝑞𝑞𝑎𝑎𝑟𝑟0

𝑞𝑞2 𝑅𝑅02 = 𝑄𝑄2 𝑎𝑎2

(1)

(2)

(3)

(2) ÷ (1) (2) or (3)

𝑐𝑐𝑐𝑐𝑐𝑐 𝛾𝛾 ≡ �𝒏𝒏 . �𝒏𝒏𝟎𝟎


Φ 𝒓𝒓 =1






𝑟𝑟0𝑄𝑄 = −𝑞𝑞

𝑎𝑎𝑟𝑟0

=1




−𝑞𝑞 𝑎𝑎𝑟𝑟0

𝑟𝑟 �𝒏𝒏 − 𝑎𝑎2𝑟𝑟0�𝒏𝒏𝟎𝟎

=𝑞𝑞


𝑟𝑟 �𝒏𝒏 − 𝑟𝑟0 �𝒏𝒏𝟎𝟎−

1𝑟𝑟0𝑎𝑎 𝑟𝑟 �𝒏𝒏 − 𝑎𝑎 �𝒏𝒏𝟎𝟎


𝑎𝑎 �𝒏𝒏 − 𝑟𝑟0 �𝒏𝒏𝟎𝟎

𝑟𝑟0 �𝒏𝒏𝟎𝟎

𝑎𝑎 �𝒏𝒏𝑟𝑟0 �𝒏𝒏 − 𝑎𝑎 �𝒏𝒏𝟎𝟎

𝑟𝑟0 �𝒏𝒏

𝑎𝑎 �𝒏𝒏𝟎𝟎

Φ 𝒓𝒓 = 𝑎𝑎 �𝒏𝒏 =𝑞𝑞


𝑎𝑎 �𝒏𝒏 − 𝑟𝑟0 �𝒏𝒏𝟎𝟎−

1𝑟𝑟0 �𝒏𝒏 − 𝑎𝑎 �𝒏𝒏𝟎𝟎

Checking the Potential on the Surface of

a Grounded Metallic Sphere

= 0


�𝒏𝒏 . �𝒏𝒏𝟎𝟎 = 𝑐𝑐𝑐𝑐𝑐𝑐 𝛾𝛾

Φ 𝒓𝒓 =𝑞𝑞


𝑟𝑟 �𝒏𝒏 − 𝑟𝑟0 �𝒏𝒏𝟎𝟎−

1𝑟𝑟0𝑎𝑎 𝑟𝑟 �𝒏𝒏 − 𝑎𝑎 �𝒏𝒏𝟎𝟎

=𝑞𝑞


𝑟𝑟2 + 𝑟𝑟02 − 2 𝑟𝑟 𝑟𝑟0 𝑐𝑐𝑐𝑐𝑐𝑐 𝛾𝛾−

1

𝑟𝑟0𝑎𝑎 𝑟𝑟

2+ 𝑎𝑎2 − 2 𝑟𝑟 𝑟𝑟0 𝑐𝑐𝑐𝑐𝑐𝑐 𝛾𝛾

𝑞𝑞

𝒓𝒓𝟎𝟎 = 𝑟𝑟0 �𝒏𝒏𝟎𝟎


𝑎𝑎

𝑹𝑹𝟎𝟎

𝑄𝑄𝛾𝛾

The Potential Outside the Grounded Metallic Sphere


𝜕𝜕Φ 𝒓𝒓𝜕𝜕𝑟𝑟 =

𝑞𝑞4 𝜋𝜋 𝜖𝜖0

−𝑟𝑟 + 𝑟𝑟0 𝑐𝑐𝑐𝑐𝑐𝑐 𝛾𝛾𝑟𝑟2 + 𝑟𝑟02 − 2 𝑟𝑟 𝑟𝑟0 𝑐𝑐𝑐𝑐𝑐𝑐 𝛾𝛾 3/2 +

𝑟𝑟0𝑎𝑎

2𝑟𝑟 − 𝑟𝑟0 𝑐𝑐𝑐𝑐𝑐𝑐 𝛾𝛾



3/2

= − 𝐸𝐸𝑟𝑟 𝑟𝑟, 𝛾𝛾




1



𝑟𝑟 ≥ 𝑎𝑎

𝛾𝛾

𝒓𝒓

�𝒆𝒆𝑟𝑟�𝒆𝒆𝛾𝛾


1𝑟𝑟𝜕𝜕Φ 𝒓𝒓𝜕𝜕𝛾𝛾

=1𝑟𝑟

𝑞𝑞4 𝜋𝜋 𝜖𝜖0

− 𝑟𝑟 𝑟𝑟0 𝑐𝑐𝑠𝑠𝑠𝑠 𝛾𝛾𝑟𝑟2 + 𝑟𝑟02 − 2 𝑟𝑟 𝑟𝑟0 𝑐𝑐𝑐𝑐𝑐𝑐 𝛾𝛾 3/2 +

𝑟𝑟 𝑟𝑟0 𝑐𝑐𝑠𝑠𝑠𝑠 𝛾𝛾



3/2




1



= − 𝐸𝐸𝛾𝛾 𝑟𝑟, 𝛾𝛾

𝑟𝑟 ≥ 𝑎𝑎

𝛾𝛾

𝒓𝒓

�𝒆𝒆𝑟𝑟�𝒆𝒆𝛾𝛾

𝐸𝐸𝛾𝛾 𝑟𝑟 = 𝑎𝑎, 𝛾𝛾 = 0


At the surface of the sphere, the electric field is perpendicular to the surface of the conducting sphere


𝐸𝐸𝑟𝑟 𝒓𝒓 = −𝑞𝑞

4 𝜋𝜋 𝜖𝜖0−𝑟𝑟 + 𝑟𝑟0 𝑐𝑐𝑐𝑐𝑐𝑐 𝛾𝛾

𝑟𝑟2 + 𝑟𝑟02 − 2 𝑟𝑟 𝑟𝑟0 𝑐𝑐𝑐𝑐𝑐𝑐 𝛾𝛾 3/2 +𝑟𝑟0𝑎𝑎

2𝑟𝑟 − 𝑟𝑟0 𝑐𝑐𝑐𝑐𝑐𝑐 𝛾𝛾



3/2

=𝜎𝜎𝜖𝜖0

𝐸𝐸𝑟𝑟 𝒓𝒓 = 𝑎𝑎 �𝒏𝒏 = −𝑞𝑞

4 𝜋𝜋 𝜖𝜖0 𝑎𝑎2𝑎𝑎𝑟𝑟0

1 − 𝑎𝑎𝑟𝑟0

2

1 + 𝑎𝑎𝑟𝑟0

2− 2 𝑎𝑎

𝑟𝑟0𝑐𝑐𝑐𝑐𝑐𝑐 𝛾𝛾

3/2


𝑄𝑄𝑖𝑖𝑖𝑖𝑖𝑖 = 𝑐𝑐𝑠𝑠𝑟𝑟𝑠𝑠𝑎𝑎𝑐𝑐𝑠𝑠 𝑐𝑐𝑐𝑎𝑎𝑟𝑟𝑐𝑐𝑠𝑠 = �0

2𝜋𝜋𝑑𝑑𝑑𝑑 �

0

𝜋𝜋𝜎𝜎 𝑎𝑎2 𝑐𝑐𝑠𝑠𝑠𝑠 𝛾𝛾 𝑑𝑑𝛾𝛾 = − 𝑞𝑞

𝑎𝑎𝑟𝑟0

= 𝑄𝑄

𝑄𝑄 < 𝑞𝑞

𝑟𝑟0 ≥ 𝑎𝑎because


𝑐𝑐𝑠𝑠𝑟𝑟𝑠𝑠𝑎𝑎𝑐𝑐𝑠𝑠 𝑐𝑐𝑐𝑎𝑎𝑟𝑟𝑐𝑐𝑠𝑠 = �0


0

𝜋𝜋𝜎𝜎 𝑎𝑎2 𝑐𝑐𝑠𝑠𝑠𝑠 𝛾𝛾 𝑑𝑑𝛾𝛾

= �0


0

𝜋𝜋−

𝑞𝑞4 𝜋𝜋

𝑎𝑎𝑟𝑟0


2


2− 2 𝑎𝑎

𝑟𝑟0𝑐𝑐𝑐𝑐𝑐𝑐 𝛾𝛾

3/2 𝑐𝑐𝑠𝑠𝑠𝑠 𝛾𝛾 𝑑𝑑𝛾𝛾

= −𝑞𝑞2

𝑎𝑎𝑟𝑟0

1 −𝑎𝑎𝑟𝑟0

2

�−1

1 1


2− 2 𝑎𝑎

𝑟𝑟0𝑥𝑥

3/2 𝑑𝑑𝑥𝑥

𝑥𝑥 ≡ 𝑐𝑐𝑐𝑐𝑐𝑐 𝛾𝛾

𝑑𝑑𝑥𝑥 = −𝑐𝑐𝑠𝑠𝑠𝑠 𝛾𝛾 𝑑𝑑 𝛾𝛾

= −𝑞𝑞2

𝑎𝑎𝑟𝑟0


2 𝑟𝑟0𝑎𝑎


2− 2 𝑎𝑎

𝑟𝑟0𝑥𝑥

1/2

𝑥𝑥=−1

𝑥𝑥=1

Proof:


= −𝑞𝑞2


2 1


2− 2 𝑎𝑎

𝑟𝑟0

1/2 −1


2+ 2 𝑎𝑎

𝑟𝑟0

1/2

= −𝑞𝑞2


2 1


−1


𝑟𝑟0 ≥ 𝑎𝑎

= −𝑞𝑞2


2 1


−1


= −𝑞𝑞2


2 2 𝑎𝑎𝑟𝑟0


2 = −𝑞𝑞𝑎𝑎𝑟𝑟0

because then:



𝑎𝑎

𝑹𝑹𝟎𝟎

𝑄𝑄


𝑟𝑟0𝑄𝑄 = − 𝑞𝑞

𝑎𝑎𝑟𝑟0

𝑭𝑭𝒒𝒒 𝒓𝒓𝟎𝟎 =𝑞𝑞 𝑄𝑄

4 𝜋𝜋 𝜖𝜖0𝒓𝒓𝟎𝟎 − 𝑹𝑹𝟎𝟎𝒓𝒓𝟎𝟎 − 𝑹𝑹𝟎𝟎 3

=𝑞𝑞 𝑄𝑄

4 𝜋𝜋 𝜖𝜖0𝑟𝑟0 − 𝑅𝑅0 �𝒏𝒏𝟎𝟎𝑟𝑟0 − 𝑅𝑅0 3

= −𝑞𝑞2

4 𝜋𝜋 𝜖𝜖0𝑎𝑎𝑟𝑟03

1


2 2 �𝒏𝒏𝟎𝟎

Force by the Grounded Metallic Sphere on Charge q



𝑎𝑎

𝑹𝑹𝟎𝟎

𝑄𝑄 𝑅𝑅0 =𝑎𝑎2

𝑟𝑟0

𝑄𝑄 = − 𝑞𝑞𝑎𝑎𝑟𝑟0

Insulated Conducting Sphere with Net Charge 𝑄𝑄𝑡𝑡

𝑄𝑄𝑡𝑡 − 𝑄𝑄

Φ 𝒓𝒓 =1





+1

4 𝜋𝜋 𝜖𝜖0𝑄𝑄𝑡𝑡 − 𝑄𝑄

𝑟𝑟



𝑎𝑎

𝑹𝑹𝟎𝟎

𝑄𝑄 𝑅𝑅0 =𝑎𝑎2


𝑎𝑎𝑟𝑟0

𝑭𝑭𝒒𝒒 𝒓𝒓𝟎𝟎 =𝑞𝑞

4 𝜋𝜋 𝜖𝜖0𝑄𝑄 𝒓𝒓𝟎𝟎 − 𝑹𝑹𝟎𝟎𝒓𝒓𝟎𝟎 − 𝑹𝑹𝟎𝟎 3 +

𝑄𝑄𝑡𝑡 − 𝑄𝑄 𝒓𝒓𝟎𝟎𝒓𝒓𝟎𝟎 3

=𝑞𝑞

4 𝜋𝜋 𝜖𝜖0 𝑟𝑟02𝑄𝑄𝑡𝑡 −

𝑞𝑞 𝑎𝑎3 2 𝑟𝑟02 − 𝑎𝑎2

𝑟𝑟0 𝑟𝑟02 − 𝑎𝑎2 2 �𝒏𝒏𝟎𝟎

Insulated Conducting Sphere with Net Charge 𝑄𝑄𝑡𝑡𝑄𝑄𝑡𝑡 − 𝑄𝑄


𝐹𝐹𝑞𝑞 4 𝜋𝜋 𝜖𝜖0 𝑟𝑟02

𝑞𝑞2𝑟𝑟0𝑎𝑎

𝑄𝑄𝑡𝑡𝑞𝑞



Grounded Conducting Sphere, Charge Inside the Sphere



𝑎𝑎

𝑹𝑹𝟎𝟎 = 𝑅𝑅0 �𝒏𝒏𝟎𝟎


Φ 𝒓𝒓 =1







𝑎𝑎𝑟𝑟0

for 𝑟𝑟 ≤ 𝑎𝑎

𝑄𝑄𝑖𝑖𝑖𝑖𝑖𝑖 = 𝑐𝑐𝑠𝑠𝑟𝑟𝑠𝑠𝑎𝑎𝑐𝑐𝑠𝑠 𝑐𝑐𝑐𝑎𝑎𝑟𝑟𝑐𝑐𝑠𝑠 = �0


0

𝜋𝜋𝜎𝜎 𝑎𝑎2 𝑐𝑐𝑠𝑠𝑠𝑠 𝛾𝛾 𝑑𝑑𝛾𝛾 = − 𝑞𝑞


Green’s Function


Φ 𝒓𝒓 ≡1

4 𝜋𝜋 𝜖𝜖0�−∞



𝑨𝑨 𝒓𝒓 ≡𝜇𝜇𝑜𝑜

4 𝜋𝜋�−∞



In principle, we need to take into account all charges and currents

present in the whole space

How can we simplify this ??


Φ 𝒓𝒓

Let’s consider two scalar functions:

𝜓𝜓 𝒓𝒓

and create a vector field from them:

𝑹𝑹 ≡ Φ 𝒓𝒓 𝛁𝛁𝜓𝜓 𝒓𝒓

&


�𝑆𝑆𝑹𝑹 . 𝒅𝒅𝒅𝒅 = �

𝑉𝑉𝛁𝛁.𝑹𝑹 𝑑𝑑𝑑𝑑

𝑹𝑹 ≡ Φ 𝒓𝒓 𝛁𝛁𝜓𝜓 𝒓𝒓

�𝑆𝑆Φ 𝒓𝒓 𝛁𝛁𝜓𝜓 𝒓𝒓 . 𝒅𝒅𝒅𝒅 = �

𝑉𝑉𝛁𝛁. Φ 𝒓𝒓 𝛁𝛁𝜓𝜓 𝒓𝒓 𝑑𝑑𝑑𝑑

= �𝑉𝑉

𝛁𝛁Φ 𝒓𝒓 .𝛁𝛁𝜓𝜓 𝒓𝒓 + Φ 𝒓𝒓 𝛻𝛻2𝜓𝜓 𝒓𝒓 𝑑𝑑𝑑𝑑


Φ 𝒓𝒓

Same two scalar functions:

𝜓𝜓 𝒓𝒓

but a different vector field:

𝒅𝒅 ≡ 𝜓𝜓 𝒓𝒓 𝛁𝛁Φ 𝒓𝒓

&


�𝑆𝑆𝑼𝑼 . 𝒅𝒅𝒅𝒅 = �

𝑉𝑉𝛁𝛁.𝑼𝑼 𝑑𝑑𝑑𝑑

𝑼𝑼 ≡ 𝜓𝜓 𝒓𝒓 𝛁𝛁Φ 𝒓𝒓

�𝑆𝑆𝜓𝜓 𝒓𝒓 𝛁𝛁Φ 𝒓𝒓 . 𝒅𝒅𝒅𝒅 = �

𝑉𝑉𝛁𝛁. 𝜓𝜓 𝒓𝒓 𝛁𝛁Φ 𝒓𝒓 𝑑𝑑𝑑𝑑

= �𝑉𝑉

𝛁𝛁𝜓𝜓 𝒓𝒓 .𝛁𝛁Φ 𝒓𝒓 + 𝜓𝜓 𝒓𝒓 𝛻𝛻2Φ 𝒓𝒓 𝑑𝑑𝑑𝑑


�𝑆𝑆Φ 𝒓𝒓 𝛁𝛁𝜓𝜓 𝒓𝒓 . 𝒅𝒅𝒅𝒅 = �

𝑉𝑉𝛁𝛁Φ 𝒓𝒓 .𝛁𝛁𝜓𝜓 𝒓𝒓 + Φ 𝒓𝒓 𝛻𝛻2𝜓𝜓 𝒓𝒓 𝑑𝑑𝑑𝑑

�𝑆𝑆𝜓𝜓 𝒓𝒓 𝛁𝛁Φ 𝒓𝒓 . 𝒅𝒅𝒅𝒅 = �

𝑉𝑉𝛁𝛁𝜓𝜓 𝒓𝒓 .𝛁𝛁Φ 𝒓𝒓 + 𝜓𝜓 𝒓𝒓 𝛻𝛻2Φ 𝒓𝒓 𝑑𝑑𝑑𝑑

�𝑆𝑆

Φ 𝒓𝒓 𝛁𝛁𝜓𝜓 𝒓𝒓 − 𝜓𝜓 𝒓𝒓 𝛁𝛁Φ 𝒓𝒓 . 𝒅𝒅𝒅𝒅 = �𝑉𝑉

Φ 𝒓𝒓 𝛻𝛻2𝜓𝜓 𝒓𝒓 − 𝜓𝜓 𝒓𝒓 𝛻𝛻2Φ 𝒓𝒓 𝑑𝑑𝑑𝑑

Green’s Theorem


Making Good Choices:

𝜓𝜓 𝒓𝒓, 𝒓𝒓𝒓 ≡1

4 𝜋𝜋 𝒓𝒓 − 𝒓𝒓𝒓

𝛻𝛻2𝜓𝜓 𝒓𝒓, 𝒓𝒓′ = 𝛻𝛻21

4 𝜋𝜋 𝒓𝒓 − 𝒓𝒓𝒓= − 𝛿𝛿3 𝒓𝒓 − 𝒓𝒓′

−𝛻𝛻2Φ 𝒓𝒓 =𝜌𝜌 𝒓𝒓𝜖𝜖0


= �𝑉𝑉

−Φ 𝒓𝒓 𝛿𝛿3 𝒓𝒓 − 𝒓𝒓′ +1

4 𝜋𝜋 𝒓𝒓 − 𝒓𝒓′𝜌𝜌 𝒓𝒓𝜖𝜖0

𝑑𝑑𝑑𝑑

�𝑆𝑆

Φ 𝒓𝒓 𝛁𝛁𝜓𝜓 𝒓𝒓, 𝒓𝒓𝒓 − 𝜓𝜓 𝒓𝒓, 𝒓𝒓′ 𝛁𝛁Φ 𝒓𝒓 . 𝒅𝒅𝒅𝒅

= �𝑉𝑉

Φ 𝒓𝒓 𝛻𝛻2𝜓𝜓 𝒓𝒓, 𝒓𝒓′ − 𝜓𝜓 𝒓𝒓, 𝒓𝒓′ 𝛻𝛻2Φ 𝒓𝒓 𝑑𝑑𝑑𝑑


= �𝑉𝑉

𝜌𝜌 𝒓𝒓4 𝜋𝜋 𝜖𝜖0 𝒓𝒓 − 𝒓𝒓𝒓

𝑑𝑑𝑑𝑑

+ �𝑆𝑆

𝜓𝜓 𝒓𝒓, 𝒓𝒓′ 𝛁𝛁Φ 𝒓𝒓 − Φ 𝒓𝒓 𝛁𝛁𝜓𝜓 𝒓𝒓, 𝒓𝒓𝒓 . 𝒅𝒅𝒅𝒅

Φ 𝒓𝒓𝒓

• One volume integral and one surface integral.

• Integration can be restricted to a finite volume and its surrounding surface.

• Surface integral accounts for what is left outside the volume.


4 𝜋𝜋 𝒓𝒓 − 𝒓𝒓𝒓


Recapping:

�𝑆𝑆

Φ 𝒓𝒓 𝛁𝛁𝜓𝜓 𝒓𝒓, 𝒓𝒓𝒓 − 𝜓𝜓 𝒓𝒓, 𝒓𝒓′ 𝛁𝛁Φ 𝒓𝒓 . 𝒅𝒅𝒅𝒅 = �𝑉𝑉

Φ 𝒓𝒓 𝛻𝛻2𝜓𝜓 𝒓𝒓, 𝒓𝒓′ − 𝜓𝜓 𝒓𝒓, 𝒓𝒓′ 𝛻𝛻2Φ 𝒓𝒓 𝑑𝑑𝑑𝑑


4 𝜋𝜋 𝒓𝒓 − 𝒓𝒓𝒓 −𝛻𝛻2Φ 𝒓𝒓 =𝜌𝜌 𝒓𝒓𝜖𝜖0

Φ 𝒓𝒓𝒓 = �𝑉𝑉

𝜌𝜌 𝒓𝒓4 𝜋𝜋 𝜖𝜖0 𝒓𝒓 − 𝒓𝒓𝒓

𝑑𝑑𝑑𝑑 + �𝑆𝑆

𝜓𝜓 𝒓𝒓, 𝒓𝒓′ 𝛁𝛁Φ 𝒓𝒓 − Φ 𝒓𝒓 𝛁𝛁𝜓𝜓 𝒓𝒓, 𝒓𝒓𝒓 . 𝒅𝒅𝒅𝒅

Green’s Theorem

and

then:


A small but crucial change:

�𝑆𝑆

Φ 𝒓𝒓 𝛁𝛁𝐺𝐺 𝒓𝒓, 𝒓𝒓𝒓 − 𝐺𝐺 𝒓𝒓, 𝒓𝒓′ 𝛁𝛁Φ 𝒓𝒓 . 𝒅𝒅𝒅𝒅 = �𝑉𝑉

Φ 𝒓𝒓 𝛻𝛻2𝐺𝐺 𝒓𝒓, 𝒓𝒓′ − 𝐺𝐺 𝒓𝒓, 𝒓𝒓′ 𝛻𝛻2Φ 𝒓𝒓 𝑑𝑑𝑑𝑑

𝜓𝜓 𝒓𝒓, 𝒓𝒓𝒓 −𝛻𝛻2Φ 𝒓𝒓 =𝜌𝜌 𝒓𝒓𝜖𝜖0

Φ 𝒓𝒓𝒓 = �𝑉𝑉𝐺𝐺 𝒓𝒓, 𝒓𝒓′

𝜌𝜌 𝒓𝒓𝜖𝜖0


𝐺𝐺 𝒓𝒓, 𝒓𝒓′ 𝛁𝛁Φ 𝒓𝒓 − Φ 𝒓𝒓 𝛁𝛁𝐺𝐺 𝒓𝒓, 𝒓𝒓𝒓 . 𝒅𝒅𝒅𝒅

If and only if 𝛻𝛻2𝐹𝐹 𝒓𝒓, 𝒓𝒓′ = 0

+ 𝐹𝐹 𝒓𝒓, 𝒓𝒓𝒓

𝛻𝛻2𝐺𝐺 𝒓𝒓, 𝒓𝒓′ = 𝛻𝛻21

4 𝜋𝜋 𝒓𝒓 − 𝒓𝒓′+ 𝐹𝐹 𝒓𝒓, 𝒓𝒓𝒓 = − 𝛿𝛿3 𝒓𝒓 − 𝒓𝒓′then

for all points inside V,

≡1

4 𝜋𝜋 𝒓𝒓 − 𝒓𝒓′𝐺𝐺 𝒓𝒓, 𝒓𝒓𝒓

which leads to:

,


We can use the freedom

𝐺𝐺 𝒓𝒓, 𝒓𝒓𝒓 ≡1

4 𝜋𝜋 𝒓𝒓 − 𝒓𝒓′+ 𝐹𝐹 𝒓𝒓, 𝒓𝒓𝒓

to find 𝐺𝐺 𝒓𝒓, 𝒓𝒓𝒓 such that:

𝐺𝐺 𝒓𝒓, 𝒓𝒓𝒓 = 0 on the surface SDirichlet’s boundary condition:





𝛁𝛁𝐺𝐺 𝒓𝒓, 𝒓𝒓𝒓 = 0 on the surface SNewmann’s boundary condition:

under the constraint that 𝛻𝛻2𝐹𝐹 𝒓𝒓, 𝒓𝒓′ = 0 inside V





or



4 𝜋𝜋 𝒓𝒓 − 𝒓𝒓′+ 𝐹𝐹 𝒓𝒓, 𝒓𝒓𝒓 = 0

on the closed surface S

Recapping the Dirichlet’s Boundary Condition:

then we can determine Φ 𝒓𝒓𝒓 anywhere inside 𝑑𝑑 by:

inside the closed surface S

Find 𝐹𝐹 𝒓𝒓, 𝒓𝒓𝒓 that satisfies simultaneously both conditions:

𝛻𝛻2𝐹𝐹 𝒓𝒓, 𝒓𝒓′ = 0

1)

2)



𝑑𝑑𝑑𝑑 −�𝑆𝑆Φ 𝒓𝒓 𝛁𝛁𝐺𝐺 𝒓𝒓, 𝒓𝒓𝒓 . 𝒅𝒅𝒅𝒅




𝑑𝑑𝑑𝑑 −�𝑆𝑆Φ 𝒓𝒓 𝛁𝛁𝐺𝐺 𝒓𝒓, 𝒓𝒓𝒓 . 𝒅𝒅𝒅𝒅

Φ 𝒓𝒓 = �𝑉𝑉𝐺𝐺 𝒓𝒓′, 𝒓𝒓

𝜌𝜌 𝒓𝒓𝒓𝜖𝜖0

𝑑𝑑𝑑𝑑𝒓 −�𝑆𝑆Φ 𝒓𝒓𝒓 𝛁𝛁𝒓𝐺𝐺 𝒓𝒓′, 𝒓𝒓 . 𝒅𝒅𝒅𝒅𝒓

Swapping 𝒓𝒓 and 𝒓𝒓′:


4 𝜋𝜋 𝒓𝒓 − 𝒓𝒓′+ 𝐹𝐹 𝒓𝒓, 𝒓𝒓𝒓 = 0

on the closed surface S

inside the closed surface S𝛻𝛻2𝐹𝐹 𝒓𝒓, 𝒓𝒓′ = 0

1)

2)


Uniqueness of a Solution to Poisson’s Equation:

Consider two possible solutions:

−𝛻𝛻2Φ1 𝒓𝒓 = −𝛻𝛻2Φ2 𝒓𝒓 =𝜌𝜌 𝒓𝒓𝜖𝜖0

Φ1 & Φ2


�𝑆𝑆

Φ1 −Φ2 𝛁𝛁 Φ1 − Φ2 .𝒅𝒅𝒅𝒅 = �𝑉𝑉

𝛁𝛁 Φ1 −Φ22 𝑑𝑑𝑑𝑑

Φ ≡ Φ1 − Φ2

𝜓𝜓 ≡ Φ1 − Φ2

�𝑆𝑆Φ 𝛁𝛁𝜓𝜓. 𝒅𝒅𝒅𝒅 = �

𝑉𝑉𝛁𝛁Φ .𝛁𝛁𝜓𝜓 + Φ 𝛻𝛻2𝜓𝜓 𝑑𝑑𝑑𝑑


Boundary conditions on the surface S:

Φ1 = Φ2

𝛁𝛁Φ1 = 𝛁𝛁Φ2

on the surface S:

on the surface S:

Dirichlet’s boundary conditions

Newmann’s boundary conditions

�𝑆𝑆

Φ1 − Φ2 𝛁𝛁 Φ1 − Φ2 . 𝒅𝒅𝒅𝒅 = �𝑉𝑉

𝛁𝛁 Φ1 −Φ22 𝑑𝑑𝑑𝑑

Imposing boundary conditions on the surrounding surface (either Φ or 𝛁𝛁Φ ) will guarantee the

uniqueness of the field −𝑬𝑬 = 𝛁𝛁Φ1 = 𝛁𝛁Φ2inside the volume.

= 0


Examples


Closed Surface = a Flat Surfaceand a Hemisphere

Example 1


𝒓𝒓

𝑥𝑥

𝑦𝑦

𝐺𝐺 𝒓𝒓′, 𝒓𝒓 ≡1

4 𝜋𝜋 𝒓𝒓𝒓 − 𝒓𝒓+ 𝐹𝐹 𝒓𝒓′, 𝒓𝒓

S1

S2S = S1 + S2

𝐺𝐺 𝒓𝒓2′ , 𝒓𝒓 → 0

when 𝑟𝑟2′ → ∞

𝑟𝑟2′ → ∞

over S2

over S1

𝐺𝐺 𝒓𝒓1′ , 𝒓𝒓 =1

4 𝜋𝜋 𝒓𝒓1′ − 𝒓𝒓+ 𝐹𝐹 𝒓𝒓1′ , 𝒓𝒓 = 0

𝒓𝒓1′ = (0, 𝑦𝑦′, 𝑧𝑧′)

𝒓𝒓1′ ∈

𝒓𝒓2′ ∈

Φ 𝒓𝒓1′

𝒓𝒓𝒓


𝐺𝐺 𝒓𝒓1′ , 𝒓𝒓 =1

4 𝜋𝜋 0 − 𝑥𝑥 2 + 𝑦𝑦𝒓 − 𝑦𝑦 2 + 𝑧𝑧𝒓 − 𝑧𝑧 2+ 𝐹𝐹 𝒓𝒓1′ ,𝒓𝒓

𝐹𝐹 𝒓𝒓1′ , 𝒓𝒓 =−1

4 𝜋𝜋 0 − 𝑥𝑥 2 + 𝑦𝑦𝒓 − 𝑦𝑦 2 + 𝑧𝑧𝒓 − 𝑧𝑧 2

𝐹𝐹 𝒓𝒓𝒓, 𝒓𝒓 =−1

4 𝜋𝜋 − 𝑥𝑥𝒓 − 𝑥𝑥 2 + 𝑦𝑦𝒓 − 𝑦𝑦 2 + 𝑧𝑧𝒓 − 𝑧𝑧 2

𝛻𝛻2𝐹𝐹 𝒓𝒓′, 𝒓𝒓 = 0 𝑥𝑥 > 0 𝑥𝑥𝒓 ≥ 0for and − 𝑥𝑥𝒓 − 𝑥𝑥 2

𝒓𝒓1′ = (0, 𝑦𝑦′, 𝑧𝑧′)

Two possible choices for 𝐹𝐹 𝒓𝒓𝒓, 𝒓𝒓 would be:

However we need:

𝐹𝐹 𝒓𝒓′, 𝒓𝒓 =−1

4 𝜋𝜋 ± 𝑥𝑥𝒓 − 𝑥𝑥 2 + 𝑦𝑦𝒓 − 𝑦𝑦 2 + 𝑧𝑧𝒓 − 𝑧𝑧 2

𝑥𝑥′ = 0over S1: or, in other words:

= 0



4 𝜋𝜋 𝒓𝒓𝒓 − 𝒓𝒓+ 𝐹𝐹 𝒓𝒓′, 𝒓𝒓

𝐹𝐹 𝒓𝒓𝒓, 𝒓𝒓 =−1

4 𝜋𝜋 −𝑥𝑥′ − 𝑥𝑥 2 + 𝑦𝑦𝒓 − 𝑦𝑦 2 + 𝑧𝑧𝒓 − 𝑧𝑧 2

𝐺𝐺 𝒓𝒓′, 𝒓𝒓 =1

4 𝜋𝜋 𝑥𝑥′ − 𝑥𝑥 2 + 𝑦𝑦𝒓 − 𝑦𝑦 2 + 𝑧𝑧𝒓 − 𝑧𝑧 2

+−1



Φ 𝒓𝒓 = �𝑉𝑉𝐺𝐺 𝒓𝒓𝒓, 𝒓𝒓


𝑑𝑑𝑑𝑑𝒓 −�𝑆𝑆Φ 𝒓𝒓𝒓 𝛁𝛁𝒓𝐺𝐺 𝒓𝒓𝒓, 𝒓𝒓 . 𝒅𝒅𝒅𝒅𝒓



+−1



𝛁𝛁′𝐺𝐺 𝒓𝒓′, 𝒓𝒓 =

+− −𝑥𝑥′ − 𝑥𝑥 �𝒆𝒆𝑥𝑥 + 𝑦𝑦′ − 𝑦𝑦 �𝒆𝒆𝑦𝑦 + 𝑧𝑧′ − 𝑧𝑧 �𝒆𝒆𝑧𝑧

4 𝜋𝜋 −𝑥𝑥′ − 𝑥𝑥 2 + 𝑦𝑦𝒓 − 𝑦𝑦 2 + 𝑧𝑧𝒓 − 𝑧𝑧 2 3/2



+−1


−𝑥𝑥′ − 𝑥𝑥 �𝒆𝒆𝑥𝑥 + 𝑦𝑦′ − 𝑦𝑦 �𝒆𝒆𝑦𝑦 + 𝑧𝑧′ − 𝑧𝑧 �𝒆𝒆𝑧𝑧

4 𝜋𝜋 𝑥𝑥′ − 𝑥𝑥 2 + 𝑦𝑦𝒓 − 𝑦𝑦 2 + 𝑧𝑧𝒓 − 𝑧𝑧 2 3/2


𝛁𝛁′𝐺𝐺 𝒓𝒓′, 𝒓𝒓 = −𝑥𝑥′ − 𝑥𝑥 �𝒆𝒆𝑥𝑥 + 𝑦𝑦′ − 𝑦𝑦 �𝒆𝒆𝑦𝑦 + 𝑧𝑧′ − 𝑧𝑧 �𝒆𝒆𝑧𝑧

4 𝜋𝜋 𝑥𝑥′ − 𝑥𝑥 2 + 𝑦𝑦𝒓 − 𝑦𝑦 2 + 𝑧𝑧𝒓 − 𝑧𝑧 2 3/2

𝒓𝒓1′ = (0,𝑦𝑦′, 𝑧𝑧′)

+− −𝑥𝑥′ − 𝑥𝑥 �𝒆𝒆𝑥𝑥 + 𝑦𝑦′ − 𝑦𝑦 �𝒆𝒆𝑦𝑦 + 𝑧𝑧′ − 𝑧𝑧 �𝒆𝒆𝑧𝑧

4 𝜋𝜋 −𝑥𝑥′ − 𝑥𝑥 2 + 𝑦𝑦𝒓 − 𝑦𝑦 2 + 𝑧𝑧𝒓 − 𝑧𝑧 2 3/2

𝛁𝛁′𝐺𝐺 𝒓𝒓1′ , 𝒓𝒓 =2 𝑥𝑥 �𝒆𝒆𝑥𝑥

4 𝜋𝜋 𝑥𝑥2 + 𝑦𝑦𝒓 − 𝑦𝑦 2 + 𝑧𝑧𝒓 − 𝑧𝑧 2 3/2







+−1



4 𝜋𝜋 𝑥𝑥2 + 𝑦𝑦𝒓 − 𝑦𝑦 2 + 𝑧𝑧𝒓 − 𝑧𝑧 2 3/2



𝑥𝑥𝑞𝑞

𝒅𝒅 = (𝑑𝑑, 0,0)

𝑦𝑦

Φ 0,𝑦𝑦𝒓, 𝑧𝑧𝒓 = Φ0 𝜌𝜌 𝒓𝒓𝒓 = 𝑞𝑞 𝛿𝛿3 𝒓𝒓′ − 𝒅𝒅

A Specific Case: A Point Charge Close to

a Flat Surface at a Fixed Potential


�𝑉𝑉𝐺𝐺 𝒓𝒓𝒓, 𝒓𝒓



=𝑞𝑞

4 𝜋𝜋 𝜖𝜖0 𝑑𝑑 − 𝑥𝑥 2 + 𝑦𝑦2 + 𝑧𝑧2+

−𝑞𝑞4 𝜋𝜋 𝜖𝜖0 −𝑑𝑑 − 𝑥𝑥 2 + 𝑦𝑦2 + 𝑧𝑧2



+−1


𝜌𝜌 𝒓𝒓𝒓 = 𝑞𝑞 𝛿𝛿3 𝒓𝒓′ − 𝒅𝒅𝒅𝒅 = (𝑑𝑑, 0,0)

= �𝑉𝑉𝐺𝐺 𝒓𝒓𝒓, 𝒓𝒓

𝑞𝑞 𝛿𝛿3 𝒓𝒓′ − 𝒅𝒅𝜖𝜖0



−�𝑆𝑆Φ 𝒓𝒓′ 𝛁𝛁′𝐺𝐺 𝒓𝒓′, 𝒓𝒓 .𝒅𝒅𝒅𝒅′ =


4 𝜋𝜋 𝑥𝑥2 + 𝑦𝑦𝒓 − 𝑦𝑦 2 + 𝑧𝑧𝒓 − 𝑧𝑧 2 3/2

Φ 0,𝑦𝑦𝒓, 𝑧𝑧𝒓 = Φ0

= Φ0�−∞

∞𝑑𝑑𝑦𝑦𝒓 𝑑𝑑𝑧𝑧𝒓

2 𝑥𝑥4 𝜋𝜋 𝑥𝑥2 + 𝑦𝑦𝒓 − 𝑦𝑦 2 + 𝑧𝑧𝒓 − 𝑧𝑧 2 3/2

𝒅𝒅𝒅𝒅′ = −𝑑𝑑𝑆𝑆′ �𝒆𝒆𝑥𝑥 = − 𝑑𝑑𝑦𝑦′ 𝑑𝑑𝑧𝑧′ �𝒆𝒆𝑥𝑥

= Φ0





=𝑞𝑞

4 𝜋𝜋 𝜖𝜖0 𝑑𝑑 − 𝑥𝑥 2 + 𝑦𝑦2 + 𝑧𝑧2+

−𝑞𝑞4 𝜋𝜋 𝜖𝜖0 −𝑑𝑑 − 𝑥𝑥 2 + 𝑦𝑦2 + 𝑧𝑧2

+ Φ0


𝑥𝑥𝑞𝑞

𝒅𝒅 = (𝑑𝑑, 0,0)

𝑦𝑦

Φ 0,𝑦𝑦𝒓, 𝑧𝑧𝒓 = Φ0

Consider the electric potential problem in the half-space defined by 𝑧𝑧 ≥ 0 with theDirichlet boundary conditions on the plane 𝑧𝑧 = 0 (and closed with a hemisphere atinfinity).

a) Write down an appropriate Green function 𝐺𝐺(𝒓𝒓, 𝒓𝒓𝒓) for this configuration.

Now, for questions (b) and (c) below, assume the absence of any electric charge. Alsoconsider that the potential on the plane 𝑧𝑧 = 0 is specified to be: Φ = 𝑑𝑑 inside a circle ofradius 𝑅𝑅 centered at the origin and Φ = 0 outside the circle (see figure below).

b) Find an integral expression for the electric potential at the point 𝑷𝑷 specified in termsof cylindrical coordinates by (𝜌𝜌,𝜑𝜑, 𝑧𝑧) in the half-space defined by 𝑧𝑧 ≥ 0.

c) Show that along the z-axis, which corresponds to the axis of the circle (𝜌𝜌 = 0), the potential is given by:


Φ 𝜌𝜌 = 0,𝜑𝜑, 𝑧𝑧 = 𝑑𝑑 1 −𝑧𝑧

𝑅𝑅2 + 𝑧𝑧2


Example 2

Spherical Surface:

Points Outside the Sphere


𝑞𝑞

𝒓𝒓𝟎𝟎 = 𝑟𝑟0 �𝒏𝒏𝟎𝟎


𝑎𝑎

𝑹𝑹𝟎𝟎



𝑄𝑄

Φ 𝒓𝒓 =1


𝒓𝒓 − 𝒓𝒓𝟎𝟎+



=𝑞𝑞



1




𝑟𝑟0

𝑄𝑄 = −𝑞𝑞𝑎𝑎𝑟𝑟0

𝛾𝛾

For a grounded metallic sphere, we previously got:


Green’s Function requires a closed surface, so we take two concentric spheres, and push the radius of the outer sphere to infinity

𝒓𝒓

𝒓𝒓𝒓𝑎𝑎


𝑎𝑎

𝑏𝑏 → ∞

𝒓𝒓

𝒓𝒓𝒓

S1

S2

S = S1 + S2

−�𝒆𝒆𝒓𝒓

𝒅𝒅𝒅𝒅𝒓 = 𝑑𝑑𝑆𝑆𝒓 �𝒆𝒆𝒓𝒓

𝒅𝒅𝒅𝒅′ = −𝑑𝑑𝑆𝑆𝒓 �𝒆𝒆𝒓𝒓

�𝒆𝒆𝒓𝒓


𝑎𝑎


4 𝜋𝜋 𝒓𝒓 − 𝒓𝒓′+ 𝐹𝐹 𝒓𝒓𝒓, 𝒓𝒓

𝒓𝒓

𝒓𝒓𝒓

𝐹𝐹 𝒓𝒓𝒓, 𝒓𝒓 =𝑄𝑄

4 𝜋𝜋 𝒓𝒓 − 𝑹𝑹𝒓

𝑅𝑅𝒓 =𝑎𝑎2

𝑟𝑟′𝑄𝑄 = −𝑞𝑞

𝑎𝑎𝑟𝑟′

= −𝑎𝑎𝑟𝑟′

𝛾𝛾𝐺𝐺 𝒓𝒓𝒓, 𝑟𝑟 = 𝑎𝑎 = 0

By imposing the Green’s function to be zero on the surface of the sphere (Dirichlet’s boundary condition)

we saw that we need:


𝑅𝑅𝒓 =𝑎𝑎2

𝑟𝑟′𝑄𝑄 = −

𝑎𝑎𝑟𝑟′


4 𝜋𝜋 𝒓𝒓 − 𝒓𝒓𝒓+

𝑄𝑄4 𝜋𝜋 𝒓𝒓 − 𝑹𝑹𝒓

=1

4 𝜋𝜋1

𝑟𝑟2 + 𝑟𝑟′2 − 2 𝑟𝑟 𝑟𝑟′𝑐𝑐𝑐𝑐𝑐𝑐 𝛾𝛾−

1

𝑟𝑟′𝑎𝑎 𝑟𝑟

2+ 𝑎𝑎2 − 2 𝑟𝑟 𝑟𝑟′𝑐𝑐𝑐𝑐𝑐𝑐 𝛾𝛾

𝑐𝑐𝑐𝑐𝑐𝑐 𝛾𝛾 = 𝑐𝑐𝑠𝑠𝑠𝑠 𝜃𝜃 𝑐𝑐𝑠𝑠𝑠𝑠 𝜃𝜃′ 𝑐𝑐𝑐𝑐𝑐𝑐 𝜑𝜑 − 𝜑𝜑′ + 𝑐𝑐𝑐𝑐𝑐𝑐 𝜃𝜃 𝑐𝑐𝑐𝑐𝑐𝑐 𝜃𝜃′


𝜕𝜕𝐺𝐺 𝒓𝒓′, 𝒓𝒓𝜕𝜕𝑟𝑟′

=1

4 𝜋𝜋− 1

2 2𝑟𝑟′ − 2 𝑟𝑟 𝑐𝑐𝑐𝑐𝑐𝑐 𝛾𝛾

𝑟𝑟2 + 𝑟𝑟′2 − 2 𝑟𝑟 𝑟𝑟′𝑐𝑐𝑐𝑐𝑐𝑐 𝛾𝛾3/2 −

− 12 2 𝑟𝑟′ 𝑟𝑟

𝑎𝑎2− 2 𝑟𝑟 𝑐𝑐𝑐𝑐𝑐𝑐 𝛾𝛾



3/2


4 𝜋𝜋1


1



𝜕𝜕𝐺𝐺 𝒓𝒓′ = 𝑎𝑎 �𝒆𝒆𝒓𝒓′ , 𝒓𝒓𝜕𝜕𝑟𝑟′

=1

4 𝜋𝜋 𝑎𝑎𝑟𝑟2 − 𝑎𝑎2

𝑟𝑟2 + 𝑎𝑎2 − 2 𝑟𝑟 𝑎𝑎 𝑐𝑐𝑐𝑐𝑐𝑐 𝛾𝛾 3/2

𝜕𝜕𝐺𝐺 𝒓𝒓′, 𝒓𝒓𝜕𝜕𝛾𝛾

=1

4 𝜋𝜋− 1

2 2 𝑟𝑟 𝑟𝑟′𝑐𝑐𝑠𝑠𝑠𝑠 𝛾𝛾

𝑟𝑟2 + 𝑟𝑟′2 − 2 𝑟𝑟 𝑟𝑟′𝑐𝑐𝑐𝑐𝑐𝑐 𝛾𝛾3/2 −

− 12 2 𝑟𝑟 𝑟𝑟′𝑐𝑐𝑠𝑠𝑠𝑠 𝛾𝛾



3/2

𝜕𝜕𝐺𝐺 𝒓𝒓′ = 𝑎𝑎 �𝒆𝒆𝒓𝒓′ , 𝒓𝒓𝜕𝜕𝛾𝛾

= 0


Φ 𝒓𝒓 = �𝑉𝑉1𝐺𝐺 𝒓𝒓𝒓, 𝒓𝒓


𝑑𝑑𝑑𝑑𝒓 −�𝑆𝑆1Φ 𝒓𝒓𝒓 𝛁𝛁𝒓𝐺𝐺 𝒓𝒓𝒓, 𝒓𝒓 .𝒅𝒅𝒅𝒅𝒓


4 𝜋𝜋1


1



𝛁𝛁′𝐺𝐺 𝒓𝒓1′ = 𝑎𝑎 �𝒆𝒆𝒓𝒓′ , 𝒓𝒓 =1


𝑟𝑟2 + 𝑎𝑎2 − 2 𝑟𝑟 𝑎𝑎 𝑐𝑐𝑐𝑐𝑐𝑐 𝛾𝛾 3/2 �𝒆𝒆𝒓𝒓′

𝑑𝑑1 = space outside sphere 𝑆𝑆1


Φ 𝑟𝑟,𝜑𝜑,𝜃𝜃 = −�𝑆𝑆1Φ 𝒓𝒓′ 𝛁𝛁′𝐺𝐺 𝒓𝒓′, 𝒓𝒓 . 𝒅𝒅𝒅𝒅′ =

𝜌𝜌 𝒓𝒓𝒓 = 0

=𝑎𝑎

4 𝜋𝜋�𝑆𝑆1Φ 𝑎𝑎,𝜑𝜑′,𝜃𝜃𝒓

𝑟𝑟2 − 𝑎𝑎2

𝑟𝑟2 + 𝑎𝑎2 − 2 𝑟𝑟 𝑎𝑎 𝑐𝑐𝑐𝑐𝑐𝑐 𝛾𝛾 3/2 𝑑𝑑Ω𝒓

𝑎𝑎

𝒓𝒓

𝒓𝒓𝒓𝛾𝛾

Φ 𝑎𝑎,𝜑𝜑′, 𝜃𝜃𝒓

outside sphere 𝑆𝑆1 𝑆𝑆1

Special Case 1: no charges outside

the sphere

=1

4 𝜋𝜋 𝑎𝑎�𝑆𝑆1Φ 𝑎𝑎,𝜑𝜑′,𝜃𝜃𝒓


𝑟𝑟2 + 𝑎𝑎2 − 2 𝑟𝑟 𝑎𝑎 𝑐𝑐𝑐𝑐𝑐𝑐 𝛾𝛾 3/2 𝑑𝑑𝑆𝑆𝒓

S1

− �𝒆𝒆𝒓𝒓′𝒅𝒅𝒅𝒅′ = − 𝑑𝑑𝑆𝑆′ �𝒆𝒆𝒓𝒓′


HW: Consider two metallic hemispheric shells of radius 𝑎𝑎, where the top one is

held at a constant potential Φ0 and the bottom one is grounded (Φ = 0).

Consider a scenario with no charges outside the sphere.

a) Use Green’s function to find an integral expression for the potential at any

point (x, y, z) outside the hemisphere ( 𝑥𝑥2 + 𝑦𝑦2 + 𝑧𝑧2 > 𝑎𝑎 ) and for 𝑧𝑧 > 0.

b) Find an analytical expression for the potential along the z-axis: (0, 0, 𝑧𝑧) for

𝑧𝑧 > 𝑎𝑎.

Φ = 0

Φ0


Φ 𝑟𝑟,𝜑𝜑,𝜃𝜃 = �𝑉𝑉1𝐺𝐺 𝒓𝒓𝒓, 𝒓𝒓


𝑑𝑑𝑑𝑑𝒓 =

𝜌𝜌 𝒓𝒓𝒓 = 𝑞𝑞 𝛿𝛿3 𝒓𝒓′ − 𝒓𝒓𝟎𝟎 𝑎𝑎

𝒓𝒓

𝒓𝒓𝒓𝛾𝛾

outside sphere 𝑆𝑆1 𝑆𝑆1

Special Case 2: single charge outside a grounded sphere

S1

−�𝒆𝒆𝒓𝒓𝒅𝒅𝒅𝒅′ = − 𝑑𝑑𝑆𝑆𝒓 �𝒆𝒆𝒓𝒓

=𝑞𝑞


𝑟𝑟2 + 𝑟𝑟′2 − 2 𝑟𝑟 𝑟𝑟0 𝑐𝑐𝑐𝑐𝑐𝑐 𝛾𝛾−

1



Φ 𝑟𝑟 = 𝑎𝑎,𝜑𝜑′, 𝜃𝜃𝒓 = 0


Spherical Surface:

Points Inside the Sphere

Example 3


𝑞𝑞

𝒓𝒓𝟎𝟎 = 𝑟𝑟0 �𝒏𝒏𝟎𝟎


𝑎𝑎

𝑹𝑹𝟎𝟎 = 𝑅𝑅0 �𝒏𝒏𝟎𝟎

𝑄𝑄

𝛾𝛾


𝑟𝑟0

𝑄𝑄 = − 𝑞𝑞𝑎𝑎𝑟𝑟0

Φ 𝒓𝒓 =1





For a grounded metallic sphere, we previously got:



4 𝜋𝜋1


1






𝑎𝑎

𝒓𝒓

𝒓𝒓𝒓

𝛾𝛾

�𝒆𝒆𝒓𝒓

𝒅𝒅𝒅𝒅𝒓 = 𝑑𝑑𝑆𝑆𝒓 �𝒆𝒆𝒓𝒓

S




𝑑𝑑𝑑𝑑𝒓 −�𝑆𝑆Φ 𝒓𝒓𝒓 𝛁𝛁𝒓𝐺𝐺 𝒓𝒓𝒓, 𝒓𝒓 .𝒅𝒅𝒅𝒅𝒓


4 𝜋𝜋1


1






𝑑𝑑 = inside the sphere


𝜌𝜌 𝒓𝒓𝒓 = 0inside the sphere

𝑎𝑎

𝒓𝒓

𝒓𝒓𝒓

𝛾𝛾

�𝒆𝒆𝒓𝒓′

𝒅𝒅𝒅𝒅′ = 𝑑𝑑𝑆𝑆′ �𝒆𝒆𝒓𝒓′

S

Φ 𝒓𝒓 = −�𝑆𝑆Φ 𝒓𝒓𝒓 𝛁𝛁𝒓𝐺𝐺 𝒓𝒓𝒓, 𝒓𝒓 .𝒅𝒅𝒅𝒅𝒓

= −𝑎𝑎

4 𝜋𝜋�𝑆𝑆1Φ 𝑎𝑎,𝜑𝜑′,𝜃𝜃𝒓


𝑟𝑟2 + 𝑎𝑎2 − 2 𝑟𝑟 𝑎𝑎 𝑐𝑐𝑐𝑐𝑐𝑐 𝛾𝛾 3/2 𝑑𝑑Ω𝒓

Φ 𝑎𝑎,𝜑𝜑′, 𝜃𝜃𝒓

Special Case: no charges inside the sphere

time independent electromagnetic fields 611 spring... · symmetries. focus on the techniques,...

Documents