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Semester II, 2017-18 Department of Physics, IIT Kanpur
PHY103A: Lecture # 7
(Text Book: Intro to Electrodynamics by Griffiths, 3rd Ed.)
Anand Kumar Jha 17-Jan-2018
Notes β’ HW # 3 is uploaded on the course webpage. β’ Solutions to HW#2 have also been uploaded.
β’ Class this Saturday (Jan 20th).
β’ Are lecture notes sufficient for exams? β’ Office Hours??
2
Summary of Lecture # 6: β’ Poissonβs Equation: ππ2V = β ππ
ππ0
β’ The work required to construct a system of one point charge ππ:
ππ = QV π«π«
ππ =12οΏ½ππππ
ππ
ππ=1
ππ(π«π«π’π’)
β’ Total work required to put together ππ point charges:
β’ Energy of a continuous charge distribution: W =
12οΏ½ ππ ππππππ
π£π£ππππ=ππ02οΏ½ πΈπΈ2ππππ
ππππππ π π π π πππ π π π
ππ
V ππ ππ = βππV
V π«π« = β οΏ½ππ β πππ₯π₯π«π«
β
3
Where is the electrostatic energy stored ?
W =12οΏ½ ππ ππππππ
π£π£ππππ
The Energy of a Continuous Charge Distribution:
β’ The first expression has a volume integral of ππ ππ over the localized space whereas the second one has the volume integral of πΈπΈ2over all space.
β’ Just as both the integrals are mathematically correct, both the interpretations are also correct. The electrostatic energy can be interpreted as stored locally within the charge distribution or globally over all space.
β’ Again, at this point, as regarding fields, we know how to calculate different physical quantities but we donβt really know what exactly the field is. 4
So, where is the electrostatic energy stored? Within the localized charge distribution or over all space?
=ππ02οΏ½ πΈπΈ2ππππ
ππππππ π π π π πππ π π π
Where is the electrostatic energy stored ?
Ex. 2.8 (Griffiths, 3rd Ed. ): Find the energy of a uniformly charged spherical shell of total charge ππ and radius π π .
ππshell =12οΏ½ ππ ππππππ
π£π£ππππ
ππshell =ππ02οΏ½ πΈπΈ2ππππ
ππππππ π π π π πππ π π π
=12οΏ½
ππ4ππ π π 2
Γππ
4ππππ0π π π π 2sinΞΈ ππππ ππππ
=12
ππ4ππ π π 2
Γππ
4ππππ0π π π π 2 4ππ =
ππ2
8ππππ0π π
=12οΏ½ππ ππππππ
Alternatively:
What is πΈπΈ? ππ = 0 inside and ππ = ππ4ππππ0ππ2
πποΏ½ outside
ππshell =ππ02οΏ½
ππ4ππππ0ππ2
2
ππ2sinΞΈ ππππππππ ππππ
ππππππ π π π π πππ π π π =
ππ02
ππ4ππππ0
2
4πποΏ½1ππ2ππππ
β
ππ=π π
=ππ2 8ππππ0
1π π
The two expressions for energy indeed give us the same information
Where is the electrostatic energy stored ?
Ex. 2.8 (Griffiths, 3rd Ed. ): Find the energy of a uniformly charged solid sphere of total charge ππ and radius π π .
Wsphere =ππ02οΏ½ πΈπΈ2ππππ
ππππππ π π π π πππ π π π
HW Prob 2.5(a): ππ = ππππ3ππ0
πποΏ½
Wsphere
ππ = ππ4ππππ0ππ2
πποΏ½, outside
For the same charge and radius, a solid sphere has more total energy than a spherical shell.
ππshell =ππ2 8ππππ0
1π π
=ππ2
4ππππ012π π
Recall:
=ππ
4πππ π 3/3ππ
3ππ0 πποΏ½ =
ππππ 4ππππ0π π 3
πποΏ½ , inside
=ππ02
14ππππ0 2 ππ
2 οΏ½1ππ44ππππ2ππππ
π π
0+ οΏ½
ππ2
π π 64ππππ2ππππ
β
π π =
ππ2
4ππππ035π π
Electrostatic Boundary Conditions (Consequences of the fundamental laws):
1. Normal component of E is Discontinuous
ππ β ππ =ππ ππ0
οΏ½ ππ β ππππ
π π π’π’πππ’π’=ππencππ0
ππaboveπ΄π΄ β ππbelowπ΄π΄ + 0 + 0 + 0 + 0 =ππ π΄π΄ππ0
EaboveβEbelow =ππ ππ0
2. Parallel component of E is Continuous ππ Γ ππ = 0 οΏ½ ππ β πππ₯π₯
π π πππππ= ππ
ππabove ππ β ππbelowππ + 0 + 0 = 0
Eabove β Ebelow = 0
ππaboveβ ππbelow=ππ ππ0π§π§οΏ½
7
How does electric field (ππ) change across a boundary containing surface charge ππ?
The electrostatic boundary condition
How does electric potential (V) change across a boundary containing surface charge ππ?
V ππ β V(ππ) = βοΏ½ππ β πππ₯π₯ππ
ππ
Vabove β Vbelow = βοΏ½ππ β πππ₯π₯ππ
ππ
For ππ β 0,
Vabove β Vbelow = 0
3. Potential ππ is continuous across a boundary
βοΏ½ππ β πππ₯π₯ππ
ππ
= 0
8
Electrostatic Boundary Conditions (Consequences of the fundamental laws):
Uniformly charged spherical shell
9
Ex. 2.6 (Griffiths, 3rd Ed. ): Find the electric field and electric potential inside and outside a uniformly charged sphere of radius π π and total charge ππ.
The electric field outside the shell: ππ(π«π«) =1
4ππππ0ππr2
rΜ
The electric field inside the shell: The electric potential at a point outside the shell (r > π π ):
V π«π« = β οΏ½ππ β πππ₯π₯ππ
β
The electric potential for a point inside the shell (r < π π ):
V π«π« = β οΏ½ππ β πππ₯π₯π π
β
β οΏ½ππ β πππ₯π₯ππ
π π
Check the boundary condition on Electric field at r = π π
EaboveβEbelow =ππ ππ0
Check the boundary condition on Electric potential
Vabove β Vbelow = 0
OK Βͺ
OK Βͺ
Eabove β 0 =ππ
ππ0 4πππ π 2 Eabove =
1 4ππππ0
ππ π π 2
1
4ππππ0ππR β
14ππππ0
ππR = 0
ππ π«π« = 0
=1
4ππππ0ππr = β οΏ½
14ππππ0
ππrβ²2ππππβ²
ππ
β
=1
4ππππ0ππR = β οΏ½
14ππππ0
ππrβ²2ππππβ² β 0
π π
β
Conductors (Materials containing unlimited supply of electrons)
10
This is true even when the conductor is placed in an external electric field ππππππππ.
(2) The charge density ππ = 0 inside a conductor.
(3) Any net charge resides on the surface.
This is because ππ = 0.
(5) ππ is perpendicular to the surface, just outside the conductor.
> Wsphere =ππ2
4ππππ035π π
ππshell =ππ2
4ππππ012π π
To minimize the energy Why?
This is because ππ = 0 inside a conductor
(1) The electric field ππ = 0 inside a conductor
(4) A conductor is an equipotential.
and therefore ππ = ππ0ππ β ππ = 0.
V ππ β V ππ
So, for any two points ππ and ππ,
This means V ππ = V ππ . = 0. = βοΏ½ ππ β πππ₯π₯ππ
ππ
Induced Charges
11
Charge inside the cavity
No charge inside the cavity
No charge inside the cavity, Conductor in an external field
Induced Charges
12
Prob. 2.36 (Griffiths, 3rd Ed. ):
- Surface charge ππππ? ππππ = βππππ4ππππ2
- Surface charge ππππ? ππππ = βππππ4ππππ2
- Surface charge πππ π ? πππ π =ππππ + ππππ4πππ π 2
- Force on ππππ ? 0
- Force on ππππ ? 0
- ππout(π«π«) ? ππout =1
4ππππ0ππππ + ππππππ2
π«π«οΏ½
- ππ(π«π«ππ) ? ππout =1
4ππππ0ππππππππ2
π«π«πποΏ½
- ππ(π«π«ππ) ? ππout =1
4ππππ0ππππππππ2
π«π«πποΏ½
Induced Charges
13
Prob. 2.36 (Griffiths, 3rd Ed. ):
- Surface charge ππππ? ππππ = βππππ4ππππ2
- Surface charge ππππ? ππππ = βππππ4ππππ2
- Surface charge πππ π ? πππ π =ππππ + ππππ4πππ π 2
- Force on ππππ ? 0
- Force on ππππ ? 0
Same Βͺ
Same Βͺ
Changes οΏ½
Same Βͺ
Same Βͺ
Same Βͺ
Same Βͺ
Changes οΏ½ - ππout(π«π«) ? ππout =
14ππππ0
ππππ + ππππππ2
π«π«οΏ½
- ππ(π«π«ππ) ? ππout =1
4ππππ0ππππππππ2
π«π«πποΏ½
- ππ(π«π«ππ) ? ππout =1
4ππππ0ππππππππ2
π«π«πποΏ½
Bring in a third charge πππ π