PHY103A: Lecture # 5akjha/PHY103_Notes_HW_Solutions/...Semester II, 2017-18 Department of Physics,...
Transcript of PHY103A: Lecture # 5akjha/PHY103_Notes_HW_Solutions/...Semester II, 2017-18 Department of Physics,...
Semester II, 2017-18 Department of Physics, IIT Kanpur
PHY103A: Lecture # 5
(Text Book: Intro to Electrodynamics by Griffiths, 3rd Ed.)
Anand Kumar Jha 12-Jan-2018
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Summary of Lecture # 4:
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โข Gaussโs Law
๐๐(๐ซ๐ซ) =1
4๐๐๐๐0๏ฟฝ๐๐๐๐r2
rฬ
ฮฆ๐ธ๐ธ = ๏ฟฝ ๐๐ โ ๐๐๐๐
๐ ๐ ๐ข๐ข๐ข๐ข๐ข๐ข
๐๐ โ ๐๐ =๐๐ ๐๐0
โข Scalar Potential : if ๐๐ ร ๐ ๐ = 0 everywhere, ๐ ๐ = โ๐๐V
โข Vector Potential : if ๐๐ โ ๐ ๐ = 0 everywhere, ๐ ๐ = ๐๐ ร ๐๐
โข Coulombโs Law:
โข Electric Flux
๏ฟฝ ๐๐ โ ๐๐๐๐
๐ ๐ ๐ข๐ข๐ข๐ข๐ข๐ข=๐๐enc๐๐0
(in integral form)
(in differential form)
๏ฟฝ ๐๐ โ ๐๐๐๐
๐ ๐ ๐ข๐ข๐ข๐ข๐ข๐ข=๐๐enc๐๐0
This is the Gaussโs law in integral form.
Q: (Griffiths: Ex 2.10): What is the flux through the shaded face of the cube due to the charge ๐๐ at the corner
Answer:
๏ฟฝ ๐๐ โ ๐๐๐๐
๐ ๐ ๐ข๐ข๐ข๐ข๐ข๐ข=
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๐๐๐๐0
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Correction in Lecture # 4:
๏ฟฝ ๐๐ โ ๐๐๐๐ ? ?
๐ ๐ ๐ข๐ข๐ข๐ข๐ข๐ข
24 ๏ฟฝ ๐๐ โ ๐๐๐๐
๐ ๐ ๐ข๐ข๐ข๐ข๐ข๐ข=๐๐๐๐0
Gaussโs Law from Coulombโs Law:
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๐๐(๐ซ๐ซ) =1
4๐๐๐๐0๏ฟฝ
rฬr2
๐๐ ๐ซ๐ซโฒ ๐๐๐๐
Coulombโs law gives the electric field due to a volume charge ๐๐ ๐ซ๐ซโฒ
If Coulombโs Law and Gaussโs law have the same information content, can we derive Gaussโs law from Coulombโs law?
Take the divergence of both sides of the equation
๐๐ โ ๐๐(๐ซ๐ซ) =1
4๐๐๐๐0๏ฟฝ๐๐ โ
rฬr2
๐๐ ๐ซ๐ซโฒ ๐๐๐๐
We have: ๐๐ โ rฬr2
= 4๐๐ ๐ฟ๐ฟ(r) = 4๐๐ ๐ฟ๐ฟ(๐ซ๐ซ โ ๐ซ๐ซ๐ซ)
๐๐ โ ๐๐ ๐ซ๐ซ =1
4๐๐๐๐0๏ฟฝ 4๐๐ ๐ฟ๐ฟ ๐ซ๐ซ โ ๐ซ๐ซโฒ ๐๐ ๐ซ๐ซโฒ ๐๐๐๐ Therefore,
๐๐ โ ๐๐ =๐๐ ๐๐0
The divergence of electric field is equal to the charge density divided by ๐๐0
=๐๐(๐ซ๐ซ) ๐๐0
Curl of the Electric Field:
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๐๐(๐ซ๐ซ) =1
4๐๐๐๐0๐๐r2
rฬ
Electric field due to a single point charge ๐๐ is:
Letโs take the simplest electric field:
We need to find the curl of it
๐๐ ร ๐๐ ๐ซ๐ซ =๐๐
4๐๐๐๐0๐๐ ร
1r2
rฬ
Take the area integral
๏ฟฝ ๐๐ ร ๐๐ ๐ซ๐ซ โ ๐๐๐๐
๐ ๐ ๐ข๐ข๐ข๐ข๐ข๐ข=
๐๐4๐๐๐๐0
๏ฟฝ ๐๐ ร1r2
rฬ โ ๐๐๐๐
๐ ๐ ๐ข๐ข๐ข๐ข๐ข๐ข
๏ฟฝ ๐๐ ร ๐๐ ๐ซ๐ซ โ ๐๐๐๐
๐ ๐ ๐ข๐ข๐ข๐ข๐ข๐ข=
๐๐4๐๐๐๐0
๏ฟฝ1r2
rฬ โ ๐๐๐ฅ๐ฅ
Use Stokesโs theorem
๏ฟฝ ๐๐ ร ๐๐ ๐ซ๐ซ โ ๐๐๐๐
๐ ๐ ๐ข๐ข๐ข๐ข๐ข๐ข=
=๐๐
4๐๐๐๐0๏ฟฝ
1r2๐๐๐๐
since ๐๐๐ฅ๐ฅ = ๐๐๐๐๐๐๏ฟฝ + ๐๐๐๐๐๐๐ฝ๐ฝ๏ฟฝ
+๐๐sin๐๐๐๐๐๐๐๐๏ฟฝ
implies ๐๐ ร ๐๐ = ๐๐ = 0 ๐๐
4๐๐๐๐0ร
1๐๐๏ฟฝ๐ข๐ข๐๐
๐ข๐ข๐๐
Curl of the Electric Field (Digression):
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๐๐ ร ๐๐ = ๐๐ Curl of an electric field is zero. We have shown this for the simplest field, which is the field of a point charge. But it can be shown to be true for any electric field, as long as the field is static.
What if the field is dynamic, that is, what if the field changes as a function of time?
๐๐ ร ๐๐ = โ๐๐๐๐๐๐๐๐
Faradayโs Law in differential form.
Integrate over a surface
๏ฟฝ ๐๐ ร ๐๐ โ ๐๐๐๐
๐๐๐๐๐๐๐๐= ๏ฟฝ โ
๐๐๐๐๐๐๐๐
โ ๐๐๐๐
๐๐๐๐๐๐๐๐
Apply Stokesโ theorem
๏ฟฝ ๐๐ โ ๐๐๐ฅ๐ฅ
๐๐๐๐๐๐๐= โ
๐๐๐๐๐๐๏ฟฝ ๐๐ โ ๐๐๐๐
๐๐๐๐๐๐๐๐
E = โ๐๐ฮฆ๐๐๐๐
EMF
Magnetic flux Faradayโs Law in integral form.
Maxwellโs Equations (Digression 2):
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๐๐ ร ๐๐ = โ๐๐๐๐๐๐๐๐
๏ฟฝ ๐๐ โ ๐๐๐ฅ๐ฅ
๐๐๐๐๐๐๐= โ
๐๐๐๐๐๐๏ฟฝ ๐๐ โ ๐๐๐๐
๐๐๐๐๐๐๐๐
๐๐ โ ๐๐ =๐๐ ๐๐0
๏ฟฝ ๐๐ โ ๐๐๐๐
๐ ๐ ๐ข๐ข๐ข๐ข๐ข๐ข=๐๐enc๐๐0
๐๐ โ ๐๐ = 0
๐๐ ร ๐๐ = ๐๐0๐๐ โ ๐๐0๐๐0๐๐๐๐๐๐๐๐
Gaussโs Law
Faradayโs Law
Amperesโs Law with Maxwellโs correction
No name; Magnetic Monopole does not exist
When fields do not vary as a function of time, it is called Electrostatics / Magnetostatics. (before mid-sem) When fields do vary as a function of time, then the two fields have to be studied together as an electromagnetic field, and one consequence of a changing electric and magnetic field is the electromagnetic radiation. (after mid-sem) When the energy of the field is quantized (photons) then it is called quantum electrodynamics. (Not for this course). Applications: Quantum computers, Quantum cryptography, Quantum teleportation
Electric Potential:
(1) โซ ๐ ๐ โ d๐ฅ๐ฅb
a is independent of path. (2) โฎ๐ ๐ โ ๐๐๐ฅ๐ฅ = 0 for any closed loop.
This is because of Stokesโ theorem
๏ฟฝ ๐๐ ร ๐ ๐ โ ๐๐๐๐ = ๏ฟฝ ๐ ๐ โ ๐๐๐ฅ๐ฅ
๐๐๐๐๐๐๐
๐๐๐ข๐ข๐ข๐ข๐ข๐ข
โข This is because Curl of a gradient is zero ๐๐ ร ๐๐V = ๐๐ (3) ๐ ๐ is the gradient of a scalar function: ๐ ๐ = โ๐๐V
If the curl of a vector field ๐ ๐ is zero, that is, if ๐๐ ร ๐ ๐ = 0 everywhere, then:
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Recall:
(1) โซ ๐๐ โ d๐ฅ๐ฅb
a is independent of path. (2) โฎ๐๐ โ ๐๐๐ฅ๐ฅ = 0 for any closed loop.
This is because of Stokesโ theorem
๏ฟฝ ๐๐ ร ๐๐ โ ๐๐๐๐ = ๏ฟฝ ๐๐ โ ๐๐๐ฅ๐ฅ
๐๐๐๐๐๐๐
๐๐๐ข๐ข๐ข๐ข๐ข๐ข
(3) ๐๐ is the gradient of a scalar function:
The curl of Electric field ๐๐ is zero, that is,๐๐ ร ๐๐ = 0 everywhere. Therefore:
V is called the electric potential. It is a scalar quantity, the gradient of which is equal to the electric field
๐๐ = โ๐๐V
Electric Potential:
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Use the fundamental Theorem for Gradient:
How to write electric potential in terms of the electric field?
โ๐๐V = ๐๐
Take the line integral of the above equation over a path
๏ฟฝ๐๐V โ ๐๐๐ฅ๐ฅ๐๐
๐๐
= โ๏ฟฝ๐๐ โ ๐๐๐ฅ๐ฅ๐๐
๐๐
V ๐๐ โ V(๐๐) = โ๏ฟฝ๐๐ โ ๐๐๐ฅ๐ฅ๐๐
๐๐
๏ฟฝ ๐๐V โ ๐๐๐ฅ๐ฅ = V ๐๐ โ V(๐๐) ๐๐
๐๐ ๐๐๐๐๐๐๐
โข Absolute potential cannot be defined. โข Only potential differences can be defined.
Since ๐๐ ร ๐๐ = 0 everywhere, ๐๐ = โ๐๐V
Electric Potential:
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(5) V ๐๐ โ V(๐๐) = โโซ ๐๐ โ ๐๐๐ฅ๐ฅ๐๐๐๐ . Absolute potential cannot be defined. In
electrostatics, usually one takes the reference point to infinity and set the potential at infinity to zero, that is, take V ๐๐ = V โ = 0. Also if V ๐๐ = V(๐ซ๐ซ),
V ๐ซ๐ซ = โ ๏ฟฝ๐๐ โ ๐๐๐ฅ๐ฅ๐ซ๐ซ
โ
(1) Electric potential is different from electric potential energy. Unit of electric potential is Newton-meter per Coulomb ( Nโ m
C ) or Volt.
(2) The potential obeys superposition principle, i.e., the potential due to several charges is equal to the sum of the potentials due to individual ones: V = V1 + V2 + โฏ
(4) The electric field is a vector quantity, but we still get all the information from the potential (a scalar quantity). This is because different components are interrelated: ๐๐ ร ๐๐ = 0, i.e., ๐๐Ex
๐๐๐= ๐๐Ey
๐๐๐; ๐๐Ez
๐๐๐= ๐๐Ey
๐๐๐; ๐๐Ex
๐๐๐= ๐๐Ez
๐๐๐;
(3) If one knows the electrical potential (a scalar quantity), the electric field (a vector quantity) can be calculated
Electric Potential due to a point charge at origin:
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๐๐(๐ซ๐ซ๐๐) =1
4๐๐๐๐0๐๐๐๐12
๐ซ๐ซ๐๐๏ฟฝ
Electric field ๐๐(๐ซ๐ซ๐๐) at ๐ซ๐ซ๐๐ due to a single point charge ๐๐ at origin:
V ๐ซ๐ซ = โ ๏ฟฝ๐๐ โ ๐๐๐ฅ๐ฅ๐ข๐ข
โ
Electric potential V ๐ซ๐ซ at ๐ซ๐ซ due to a single point charge ๐๐ at origin:
The line element is: ๐๐๐ฅ๐ฅ๐๐ = ๐๐๐๐1 r๐๐๏ฟฝ + ๐๐1๐๐๐๐1 ๐๐๐๐๏ฟฝ + ๐๐1sin๐๐1๐๐๐๐1 ๐๐๐๐๏ฟฝ
V ๐ซ๐ซ = โ ๏ฟฝ1
4๐๐๐๐0๐๐๐๐12
๐๐
โ
๐๐๐๐1
V ๐ซ๐ซ =๐๐
4๐๐๐๐01๐๐
= โ ๏ฟฝ๐๐(๐ซ๐ซ๐๐) โ ๐๐๐ฅ๐ฅ๐๐
๐๐
โ
=๐๐
4๐๐๐๐01๐๐
= โ๐๐
4๐๐๐๐0๏ฟฝ
1๐๐12
๐๐
โ
๐๐๐๐1
Electric Potential due to localized charge distribution:
V ๐ซ๐ซ =๐๐
4๐๐๐๐01r
V ๐ซ๐ซ =1
4๐๐๐๐0๏ฟฝ
๐๐๐๐ri
๐๐
๐๐=1
V ๐ซ๐ซ =๐๐
4๐๐๐๐01๐๐
Potential due to a point charge ๐๐ at origin:
Potential due to a point charge ๐๐ at ๐ซ๐ซ๐ซ:
Potential due to a collection of point charges
V(๐ซ๐ซ) =1
4๐๐๐๐0๏ฟฝ๐๐๐๐r
For a line charge ๐๐๐๐ = ๐๐ ๐ซ๐ซโฒ ๐๐๐๐๐ซ For a surface charge ๐๐๐๐ = ๐๐ ๐ซ๐ซโฒ ๐๐๐๐๐ซ For a volume charge ๐๐๐๐ = ๐๐ ๐ซ๐ซโฒ ๐๐๐๐๐ซ
Potential due to a a continuous charge distribution is
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Ease of calculating the Electric Field
โข If the above two is not applicable, one has to go back to the Coulombโs law and then calculate the electric field.
โข The easiest way to calculate the electric field is using Gaussโs law. But this is possible only when there is some symmetry in the problem.
โข The next best thing: if the electric potential is known, one can calculate the electric field by just taking the gradient of the potential ๐๐ = โ๐๐V. Sometimes, it is very effective to calculate the electric potential first and then the electric field from there.
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