Semester II, 2017-18
Department of Physics, IIT Kanpur
PHY103A: Lecture # 6
(Text Book: Intro to Electrodynamics by Griffiths, 3rd Ed.)
Anand Kumar Jha
15-Jan-2018
1
Summary of Lecture # 5:
• Gauss’s Law from Coulomb’s Law: 𝛁𝛁 ⋅ 𝐄𝐄 =
• Curl of the electric field : 𝛁𝛁 × 𝐄𝐄 = 0
• Electric Potential: 𝐄𝐄 = −𝛁𝛁V
𝜌𝜌
𝜖𝜖0
𝒃𝒃
V 𝐛𝐛 − V(𝐚𝐚) = − � 𝐄𝐄 ⋅ 𝑑𝑑𝐥𝐥
𝐫𝐫
V 𝐫𝐫 = − � 𝐄𝐄 ⋅ 𝑑𝑑𝐥𝐥
𝒂𝒂
∞
• Electric Potential due to a localized charge distribution
𝑞𝑞 1
V 𝐫𝐫 =
4𝜋𝜋𝜖𝜖0 r
Point charge
1
𝑑𝑑𝑑𝑑
V(𝐫𝐫) =
�
4𝜋𝜋𝜖𝜖0 r
Charge distribution
2
Charge distribution in terms of electric potential:
Gauss’s Law 𝛁𝛁 ⋅ 𝐄𝐄 =
𝜌𝜌
𝜖𝜖0
But 𝛁𝛁 × 𝐄𝐄 = 0 ⇒ 𝐄𝐄 = −𝛁𝛁V
Therefore, 𝛁𝛁 ⋅ 𝐄𝐄 = 𝛁𝛁 ⋅ −𝛁𝛁V = −𝛁𝛁 ⋅ 𝛁𝛁V = −𝛻𝛻 2 V =
𝛻𝛻 2 V
𝜌𝜌
=−
𝜖𝜖0
Poisson’s Equation
𝜌𝜌
𝜖𝜖0
In the region of space where there is no charge, 𝜌𝜌=0
𝛻𝛻 2 V = 0
Laplace’s Equation
3
Summary:
𝜌𝜌
V
𝐄𝐄 = −𝛁𝛁V
𝐫𝐫
V 𝐫𝐫 = − � 𝐄𝐄 ⋅ 𝑑𝑑𝐥𝐥
∞
Electrostatics
𝐄𝐄
4
Work and Energy in Electrostatics
There is a charge 𝑄𝑄 in an electrostatic field 𝐄𝐄. How
much work needs to be done in order to move the
charge from point 𝒂𝒂 to 𝒃𝒃?
𝒃𝒃
𝒃𝒃
𝑊𝑊 = � 𝐅𝐅 ⋅ 𝑑𝑑𝐥𝐥 = −𝑄𝑄 � 𝐄𝐄 ⋅ 𝑑𝑑𝐥𝐥 = Q V 𝐛𝐛 − V(𝐚𝐚)
𝒂𝒂
𝒂𝒂
• 𝐅𝐅 = −𝑄𝑄𝐄𝐄 is the force one has to exert in order to
counteract the electrostatic force 𝐅𝐅 = 𝑄𝑄𝐄𝐄.
• Work done to move a unit charge from point 𝒂𝒂 to 𝒃𝒃
is the potential difference between points 𝒃𝒃 and 𝒂𝒂
• Work is independent of the path.
Take V(𝐚𝐚)=V(∞)=0 and V 𝐛𝐛 = V 𝐫𝐫
𝑊𝑊 = QV 𝐫𝐫
If Q = 1,
𝑊𝑊 = V 𝐫𝐫
• Work done to construct a system of unit charge (to
bring a unit charge from ∞ to 𝐫𝐫 is the electric potential.
• Thus, electric potential is the potential energy per unit
5
charge
Work required to assemble 𝒏𝒏 point charges:
The work required to construct a system of
one point charge 𝑄𝑄 is: 𝑊𝑊 = QV 𝐫𝐫 .
Work required to bring in the charge 𝑞𝑞1 from ∞
to 𝐫𝐫𝟏𝟏 : 𝑊𝑊1 = 𝑞𝑞1 V0 = 𝑞𝑞1 × 0 = 0
Work required to bring in the charge 𝑞𝑞2 from ∞
to 𝐫𝐫𝟐𝟐 : 𝑊𝑊2 = 𝑞𝑞2 V1 = 𝑞𝑞2 1 𝑞𝑞1
4𝜋𝜋𝜖𝜖0 r12
Work required to bring in the charge 𝑞𝑞3 from ∞
to 𝐫𝐫𝟑𝟑 : 𝑊𝑊3 = 𝑞𝑞3 V2 = 𝑞𝑞 1 𝑞𝑞1 + 𝑞𝑞2
3 4𝜋𝜋𝜖𝜖 r
r23
0
13
Total work required to bring in the first three charges:
𝑊𝑊 = 𝑊𝑊1 + 𝑊𝑊2 + 𝑊𝑊3 =
1
𝑞𝑞1 𝑞𝑞2
4𝜋𝜋𝜖𝜖0 r12
+
𝑞𝑞1 𝑞𝑞3
r13
+
𝑞𝑞2 𝑞𝑞3
r23
Total work required to bring in the first four charges:
𝑊𝑊 = 𝑊𝑊1 + 𝑊𝑊2 + 𝑊𝑊3 + W4 =
1
𝑞𝑞1 𝑞𝑞2
4𝜋𝜋𝜖𝜖0 r12
+
𝑞𝑞1 𝑞𝑞3
r13
+
𝑞𝑞2 𝑞𝑞3
r23
+
𝑞𝑞1 𝑞𝑞4
r14
+
𝑞𝑞2 𝑞𝑞4
r24
+
𝑞𝑞3 𝑞𝑞4
r34
6
Work required to assemble 𝒏𝒏 point charges:
Total work required to bring in the first four charges:
𝑊𝑊 = 𝑊𝑊1 + 𝑊𝑊2 + 𝑊𝑊3 + W4 =
1
𝑞𝑞1 𝑞𝑞2
4𝜋𝜋𝜖𝜖0 r12
+
𝑞𝑞1 𝑞𝑞3
r13
Total work required to bring in 𝑛𝑛 point charges,
with charge 𝑞𝑞1 , 𝑞𝑞2 , 𝑞𝑞3 ⋯ 𝑞𝑞𝑛𝑛 , respectively:
𝑛𝑛
𝑛𝑛
𝑛𝑛
+
𝑞𝑞2 𝑞𝑞3
r23
+
𝑞𝑞1 𝑞𝑞4
r14
+
𝑞𝑞2 𝑞𝑞4
r24
+
𝑞𝑞3 𝑞𝑞4
r34
𝑛𝑛
𝑞𝑞𝑖𝑖 𝑞𝑞𝑗𝑗
𝑞𝑞𝑖𝑖 𝑞𝑞𝑗𝑗
1
1
1
𝑊𝑊 =
��
= ×
��
r𝑖𝑖𝑖𝑖
4𝜋𝜋𝜖𝜖0
r𝑖𝑖𝑖𝑖
2 4𝜋𝜋𝜖𝜖0
𝑖𝑖=1 𝑗𝑗>𝑖𝑖
𝑛𝑛
𝑛𝑛
𝑖𝑖=1
𝑗𝑗=1
𝑗𝑗≠𝑖𝑖
𝑞𝑞𝑗𝑗
1
1
= ×
� 𝑞𝑞𝑖𝑖 �
r𝑖𝑖𝑖𝑖
2 4𝜋𝜋𝜖𝜖0
𝑛𝑛
𝑖𝑖=1 𝑗𝑗=1
𝑗𝑗≠𝑖𝑖
• This is the total work required to assemble 𝑛𝑛 point charges
1
W = � 𝑞𝑞𝑖𝑖 𝑉𝑉(𝐫𝐫𝐢𝐢 ) • The potential 𝑉𝑉(𝐫𝐫 ) is the potential at 𝐫𝐫 due to all
𝐢𝐢
𝐢𝐢
2
𝑖𝑖=1
charges, except the charge at 𝐫𝐫𝐢𝐢.
7
The Work required to assemble a continuous charge Distribution:
𝑛𝑛
• This is the total worked required to assemble 𝑛𝑛 point charges
1
W = � 𝑞𝑞𝑖𝑖 𝑉𝑉(𝐫𝐫𝐢𝐢 ) • The potential 𝑉𝑉(𝐫𝐫 ) is the potential at 𝐫𝐫 due to all the other
𝐢𝐢
𝐢𝐢
2
𝑖𝑖=1
charges, except the charge at 𝐫𝐫𝐢𝐢.
What would be the required work if it is a continuous distribution of charge ?
𝑛𝑛
1
W = � 𝑞𝑞𝑖𝑖 𝑉𝑉 𝐫𝐫𝐢𝐢
2
𝑖𝑖=1
1
→
� 𝑑𝑑𝑑𝑑 𝑉𝑉(𝐫𝐫)
2 𝑣𝑣𝑜𝑜𝑜𝑜
1
→
� 𝜌𝜌 𝑉𝑉(𝐫𝐫)𝑑𝑑𝑑𝑑
2 𝑣𝑣𝑜𝑜𝑜𝑜
Is this correct? Not really !
The potential 𝑉𝑉(𝐫𝐫) inside the integral is the potential
at point 𝐫𝐫. However, the potential 𝑉𝑉 𝐫𝐫𝐢𝐢 inside the
summation in the potential at 𝐫𝐫𝐢𝐢 due to all the charges
except the charge at 𝐫𝐫𝐢𝐢. Because of this difference in
the definition of the potentials, the integral formula
turns out to be different.
8
The Work required to assemble a continuous charge Distribution:
𝜖𝜖0
1
=
� (𝛁𝛁 ⋅ 𝐄𝐄) 𝑉𝑉𝑉𝑉𝑉𝑉
W = � 𝜌𝜌 𝑉𝑉𝑉𝑉𝑉𝑉
2 𝑣𝑣𝑜𝑜𝑜𝑜
2 𝑣𝑣𝑜𝑜𝑜𝑜
W=−
Using 𝜌𝜌 = 𝜖𝜖0 (𝛁𝛁 ⋅ 𝐄𝐄)
𝜖𝜖0
𝜖𝜖0
� 𝐄𝐄 ⋅ 𝛁𝛁𝑉𝑉𝑉𝑉𝑉𝑉 + � 𝛁𝛁 ⋅ 𝑉𝑉𝐄𝐄𝑑𝑑𝑑𝑑
2 𝑣𝑣𝑜𝑜𝑜𝑜
2 𝑣𝑣𝑜𝑜𝑜𝑜
𝜖𝜖0
𝜖𝜖0
W = − � 𝐄𝐄 ⋅ 𝛁𝛁𝑉𝑉𝑉𝑉𝑉𝑉 + � 𝑉𝑉𝐄𝐄 ⋅ 𝑑𝑑𝐚𝐚
2 𝑣𝑣𝑜𝑜𝑜𝑜
2 𝑠𝑠𝑢𝑢𝑢𝑢𝑢𝑢
𝜖𝜖0
𝜖𝜖0
2
W = � 𝐸𝐸 𝑑𝑑𝑑𝑑 + � 𝑉𝑉𝐄𝐄 ⋅ 𝑑𝑑𝐚𝐚
2 𝑣𝑣𝑜𝑜𝑜𝑜
2 𝑠𝑠𝑢𝑢𝑢𝑢𝑢𝑢
𝜖𝜖0
W= �
𝐸𝐸 2 𝑑𝑑𝑑𝑑
2 𝑎𝑎𝑙𝑙𝑙𝑙 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠
Using the product rule
𝛁𝛁 ⋅ 𝑓𝑓𝐀𝐀 = 𝑓𝑓 𝛁𝛁 ⋅ 𝐀𝐀 + 𝐀𝐀 ⋅ (𝛁𝛁𝑓𝑓)
Using the divergence theorem
∫𝑉𝑉𝑜𝑜𝑜𝑜 𝛁𝛁 ⋅ 𝐀𝐀 𝑑𝑑𝑑𝑑 = ∮𝑆𝑆𝑢𝑢𝑢𝑢𝑢𝑢 𝐀𝐀 ⋅ 𝑑𝑑𝐚𝐚
Using -𝛁𝛁𝑉𝑉 = 𝐄𝐄
When the volume we are integrating over is
very large, the contribution due to the surface
integral is negligibly small.
9
The Energy of a Continuous Charge Distribution:
W=
1
� 𝜌𝜌 𝑉𝑉𝑉𝑉𝑉𝑉
2 𝑣𝑣𝑜𝑜𝑜𝑜
𝜖𝜖0
W= �
𝐸𝐸 2 𝑑𝑑𝑑𝑑
2 𝑎𝑎𝑙𝑙𝑙𝑙 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠
So, what is the energy of a point charge using the above formula?
1 𝑞𝑞
Electric field of a point charge is 𝐄𝐄 =
r̂
2
4𝜋𝜋𝜖𝜖0 r
The energy of a point charge is therefore,
2
𝑞𝑞2
1
1
𝑞𝑞 2 2
𝜖𝜖0
�
𝑑𝑑𝑟𝑟
W= �
𝑟𝑟 sin𝜃𝜃𝜃𝜃𝜃𝜃𝜃𝜃𝜃𝜃𝜃𝜃𝜃𝜃 =
2
2
8𝜋𝜋𝜖𝜖0 𝑎𝑎𝑙𝑙𝑙𝑙 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 𝑟𝑟
r
2 𝑎𝑎𝑙𝑙𝑙𝑙 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 4𝜋𝜋𝜖𝜖0
𝑛𝑛
= ∞ ?? due to incorrect conversion
1
1
W
=
�
𝑞𝑞
𝑉𝑉
𝐫𝐫
→
� 𝜌𝜌 𝑉𝑉𝑉𝑉𝑉𝑉
of the sum into an integral
𝑖𝑖
𝐢𝐢
2
2 𝑣𝑣𝑜𝑜𝑜𝑜
This is the total work done
to assemble a set of point
charges. This does not
include the self energy of
assembling a point charge.
𝑖𝑖=1
This is the total energy of a charge distribution
including the self energy of assembling the charge
distribution. Assembling a point charge requires
infinite energy. This is why this expression gives
10
infinity for the energy of a point charge.
The Electrostatic Energy (Summary):
The work required to construct a system of a point charge 𝑄𝑄 is: 𝑊𝑊 = QV 𝐫𝐫
Total work required to put together 𝑛𝑛 point charges is: 𝑊𝑊 =
𝑛𝑛
1
� 𝑞𝑞𝑖𝑖 𝑉𝑉(𝐫𝐫𝐢𝐢 )
2
𝑖𝑖=1
𝑉𝑉(𝐫𝐫𝐢𝐢 ) is the potential at 𝐫𝐫𝐢𝐢 due to all
charges, except the charge at 𝐫𝐫𝐢𝐢.
1
The Energy of a Continuous Charge Distribution: W = � 𝜌𝜌 𝑉𝑉𝑉𝑉𝑉𝑉
2 𝑣𝑣𝑜𝑜𝑜𝑜
𝜖𝜖0
= �
𝐸𝐸 2 𝑑𝑑𝑑𝑑
2 𝑎𝑎𝑙𝑙𝑙𝑙 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠
Note # 1: The total work required to assemble a continuous charge distribution
= the total energy of a continuous charge distribution
Note # 2: The self energy of assembling a point charge is infinite. Therefore, the
total energy of a point charge is infinite.
Note # 3: For systems consisting of point charges, we do not talk about the total
energy. We only discuss the total work required to put together the system.
11