PHY103A: Lecture # 5 (Text Book: Intro to Electrodynamics by Griffiths, 3Rd Ed.)
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
1 Summary of Lecture # 4:
โข Scalar Potential : if ร = 0 everywhere, = V
๐๐ ๐ ๐ ๐ ๐ โ๐๐ โข Vector Potential : if = 0 everywhere, = ร
๐๐ โ ๐ ๐ ๐ ๐ ๐๐ ๐๐ 1 โข Coulombโs Law: ( ) = r r ๐๐๐๐ 2 ๐๐ ๐ซ๐ซ 0 ๏ฟฝ ฬ 4๐๐ ๐๐ โข Electric Flux =
๐ธ๐ธ ฮฆ ๏ฟฝ๐ ๐ ๐ข๐ข๐ข๐ข๐ข๐ข๐๐ โ ๐๐๐๐
โข Gaussโs Law = (in integral form) ๐๐enc ๏ฟฝ๐ ๐ ๐ข๐ข๐ข๐ข๐ข๐ข๐๐ โ ๐๐๐๐ 0 ๐๐ = (in differential form) ๐๐ ๐๐ โ ๐๐ 2 ๐๐0 Correction in Lecture # 4:
This is the Gaussโs law in integral form. = ๐๐enc ๏ฟฝ ๐๐ โ ๐๐๐๐ Q๐ ๐ : ๐ข๐ข๐ข๐ข๐ข๐ข(Griffiths: Ex 2.10):๐๐0 What is the flux through the shaded face of the cube due to the charge at the corner ? ? ๐๐ ๏ฟฝ๐ ๐ ๐ข๐ข๐ข๐ข๐ข๐ข๐๐ โ ๐๐๐๐
Answer: 24 = ๐๐ ๏ฟฝ ๐๐ โ ๐๐๐๐ ๐ ๐ ๐ข๐ข๐ข๐ข๐ข๐ข ๐๐10 = 24 ๐๐ 3 ๏ฟฝ ๐๐ โ ๐๐๐๐ ๐ ๐ ๐ข๐ข๐ข๐ข๐ข๐ข ๐๐0 Gaussโs Law from Coulombโs Law: If Coulombโs Law and Gaussโs law have the same information content, can we derive Gaussโs law from Coulombโs law?
Coulombโs law gives the electric field due to a volume charge โฒ 1 r ( ) = ๐๐ ๐ซ๐ซ r ฬ โฒ Take the divergence๐๐ ๐ซ๐ซ of both๏ฟฝ 2 sides๐๐ ๐ซ๐ซ of๐๐๐๐ the equation 4๐๐๐๐0 1 r ( ) = r ฬ โฒ ๐๐ โ ๐๐ ๐ซ๐ซr ๏ฟฝ ๐๐ โ 2 ๐๐ ๐ซ๐ซ ๐๐๐๐ We have: =4๐๐4 ๐๐ 0 (r) = 4 ( ) r ฬ ๐๐ โ 2 ๐๐1๐ฟ๐ฟ ๐๐ ๐ฟ๐ฟ ๐ซ๐ซ โ ๐ซ๐ซํซ ( ) Therefore, = 4 = โฒ โฒ ๐๐ ๐ซ๐ซ ๐๐ โ ๐๐ ๐ซ๐ซ 0 ๏ฟฝ ๐๐ ๐ฟ๐ฟ ๐ซ๐ซ โ ๐ซ๐ซ ๐๐ ๐ซ๐ซ ๐๐๐๐ 0 4๐๐๐๐The divergence of electric field is๐๐ equal = to the charge density divided by ๐๐ 4 ๐๐ โ ๐๐ ๐๐0 ๐๐0 Curl of the Electric Field: Letโs take the simplest electric field: Electric field due to a single point charge is: 1 ( ) = r r ๐๐ ๐๐ We need to find the2 curl of it ๐๐ ๐ซ๐ซ 0 ฬ 4๐๐๐๐ 1 ร = ร r r ๐๐ Take๐๐ the ๐๐area๐ซ๐ซ integral ๐๐ 2 ฬ 4๐๐๐๐0 1 ร = ร r r ๐๐ Use๏ฟฝ Stokesโs๐๐ ๐๐theorem๐ซ๐ซ โ ๐๐๐๐ ๏ฟฝ ๐๐ 2 ฬ โ ๐๐๐๐ ๐ ๐ ๐ข๐ข๐ข๐ข๐ข๐ข 4๐๐๐๐0 ๐ ๐ ๐ข๐ข๐ข๐ข๐ข๐ข 1 1 since ร = r = = + r r + sin ๐๐ ๐๐ 2 2 ๐๐๐ฅ๐ฅ ๐๐๐๐๐๐๏ฟฝ ๐๐๐๐๐๐๐ฝ๐ฝ๏ฟฝ ๏ฟฝ ๐๐ ๐๐ ๐ซ๐ซ โ ๐๐๐๐ ๏ฟฝ ฬ โ ๐๐๐ฅ๐ฅ 0 ๏ฟฝ ๐๐๐๐ ๐ ๐ ๐ข๐ข๐ข๐ข๐ข๐ข 4๐๐๐๐0 4๐๐๐๐ ๐๐ ๐๐๐๐๐๐๐๐๏ฟฝ 1 ร = ร ๐๐๐๐ = 0 implies ร = ๐๐ ๏ฟฝ ๐๐ ๐๐ ๐ซ๐ซ โ ๐๐๐๐ ๏ฟฝ ๐ ๐ ๐ข๐ข๐ข๐ข๐ข๐ข 4๐๐๐๐0 ๐๐ ๐๐๐๐ ๐๐ ๐๐ ๐๐ 5 Curl of the Electric Field (Digression):
Curl of an electric field is zero. We have shown this for the ร = 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. ๐๐ ๐๐ ๐๐ What if the field is dynamic, that is, what if the field changes as a function of time? ร = Faradayโs Law in differential ๐๐๐๐ form. ๐๐ ๐๐ โ Integrate over a surface๐๐๐๐
ร = ๐๐๐๐ ๏ฟฝ ๐๐ ๐๐ โ ๐๐๐๐ ๏ฟฝ โ โ ๐๐๐๐ Apply Stokesโ๐๐๐๐ ๐๐theorem๐๐ ๐๐๐๐๐๐๐๐ ๐๐๐ก๐ก
= ๐๐ ๏ฟฝ ๐๐ โ ๐๐๐ฅ๐ฅ โ ๏ฟฝ ๐๐ โ ๐๐๐๐ ๐๐๐๐๐๐ํ ๐๐๐ก๐ก ๐๐๐๐Magnetic๐๐๐๐ flux Faradayโs Law in integral form. = E 6 EMF ๐๐ฮฆ โ ๐๐๐ก๐ก Maxwellโs Equations (Digression 2):
Gaussโs Law = = ๐๐ ๐๐enc ๐๐ โ ๐๐ 0 ๏ฟฝ ๐๐ โ ๐๐๐๐ ๐๐ ๐ ๐ ๐ข๐ข๐ข๐ข๐ข๐ข ๐๐0 ร = = Faradayโs Law ๐๐๐๐ ๐๐ ๐๐ ๐๐ โ ๏ฟฝ๐๐๐๐๐๐ํ๐๐ โ ๐๐๐ฅ๐ฅ โ ๏ฟฝ๐๐๐๐๐๐๐๐๐๐ โ ๐๐๐๐ ๐๐๐๐ ๐๐๐ก๐ก No name; Magnetic = 0 Monopole does not exist ๐๐ โ ๐๐ Amperesโs Law with ร = Maxwellโs correction ๐๐๐๐ When๐๐ fields๐๐ do๐๐0 not๐๐ โ vary๐๐0๐๐0 as a function of time, it is called Electrostatics / ๐๐๐๐ 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 7
Electric Potential:
Recall: If the curl of a vector field is zero, that is, if ร = 0 everywhere, then: (1) d is independent of path. ๐ ๐ ๐๐This๐ ๐ is because of Stokesโ theorem b a (2) โซ ๐ ๐ โ =๐ฅ๐ฅ 0 for any closed loop. ร =
(3) โฎ is๐ ๐ theโ ๐๐๐ฅ๐ฅ gradient of a scalar function: = V๏ฟฝ๐๐ ๐ข๐ข๐ข๐ข๐ข๐ข ๐๐ ๐ ๐ โ ๐๐๐๐ ๏ฟฝ๐๐๐๐๐๐ํ๐ ๐ โ ๐๐๐ฅ๐ฅ โข This is because Curl of a gradient is zero ร V = ๐ ๐ ๐ ๐ โ๐๐ ๐๐ ๐๐ ๐๐ The curl of Electric field is zero, that is, ร = 0 everywhere. Therefore: (1) d is independent of path. ๐๐ ๐๐ ๐๐ This is because of Stokesโ theorem b a (2) โซ ๐๐ โ =๐ฅ๐ฅ 0 for any closed loop. ร =
โฎ ๐๐ โ ๐๐๐ฅ๐ฅ ๏ฟฝ๐๐๐ข๐ข๐ข๐ข๐ข๐ข ๐๐ ๐๐ โ ๐๐๐๐ ๏ฟฝ๐๐๐๐๐๐ํ๐๐ โ ๐๐๐ฅ๐ฅ (3) is the gradient of a scalar function: = V
๐๐ ๐๐ โ๐๐ V is called the electric potential. It is a scalar quantity, the gradient of which is equal to the electric field
8 Electric Potential:
Since ร = 0 everywhere, = V
๐๐ ๐๐ ๐๐ โ๐๐ How to write electric potential in terms of the electric field? V =
Take theโ๐๐ line integral๐๐ of the above equation over a path
๐๐ V = ๐๐
๏ฟฝ ๐๐ โ ๐๐๐ฅ๐ฅ โ ๏ฟฝ ๐๐ โ ๐๐๐ฅ๐ฅ ๐๐ ๐๐ Use the fundamental Theorem for Gradient: ๐๐ V = V V( )
๏ฟฝ ๐๐ โ ๐๐๐ฅ๐ฅ ๐๐ โ ๐๐ ๐๐ ๐๐๐๐๐๐๏ฟฝ โข Absolute potential cannot be defined. V V( ) = ๐๐ โข Only potential differences can be defined. ๐๐ โ ๐๐ โ ๏ฟฝ ๐๐ โ ๐๐๐ฅ๐ฅ ๐๐
9 Electric Potential: (1) Electric potential is different from electric potential energy. Unit of electric potential is Newton-meter per Coulomb ( ) or Volt. Nโ m (2) The potential obeys superposition principle,C i.e., the potential due to several charges is equal to the sum of the potentials due to individual ones: V = V + V +
(3) If one knows the electrical potential (a scalar quantity), the electric field1 (a 2 โฏ vector quantity) can be calculated (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: ร = 0, i.e., = ; = ; = ; ๐๐Ex ๐๐Ey ๐๐Ez ๐๐Ey ๐๐Ex ๐๐Ez (5) V ๐๐V( ๐๐) = ๐๐ํ . Absolute๐๐ํ ๐๐ํ potential๐๐ํ cannot๐๐ํ be๐๐ํ defined. In electrostatics, usually one๐๐ takes the reference point to infinity and set the ๐๐ potential๐๐ at โinfinity๐๐ to โzero,โซ ๐๐thatโ ๐๐ ๐ฅ๐ฅis, take V = V = 0. Also if V = V( ),
๐๐ โ ๐๐ ๐ซ๐ซ V = ๐ซ๐ซ 10 ๐ซ๐ซ โ ๏ฟฝ ๐๐ โ ๐๐๐ฅ๐ฅ โ Electric Potential due to a point charge at origin:
Electric field ( ) at due to a single point charge at origin: ๐๐ ๐๐ ๐๐ ๐ซ๐ซ 1 ๐ซ๐ซ ( ) = ๐๐ ๐๐ ๐๐ ๐ซ๐ซ๐๐ 2 ๐ซ๐ซ๏ฟฝ๐๐ Electric potential4๐๐ V ๐๐0 ๐๐at1 due to a single point charge at origin: ๐ซ๐ซ ๐ซ๐ซ ๐๐ V = ๐๐ = ๐๐ ( )
๐ซ๐ซ โ ๏ฟฝ ๐๐ โ ๐๐๐ฅ๐ฅ โ ๏ฟฝ ๐๐ ๐ซ๐ซ๐๐ โ ๐๐๐ฅ๐ฅ๐๐ The line elementโ is: = โ r + + sin
๐๐1๐ฅ๐ฅ๐๐ ๐๐๐๐1 ๏ฟฝ๐๐ ๐๐1๐๐๐๐1 ๐๐๏ฟฝ๐๐ 1๐๐1 ๐๐1๐๐๐๐1 ๐๐๏ฟฝ๐๐1 V = ๐๐ = ๐๐ = ๐๐ ๐๐ ๐๐ 2 1 2 1 ๐ซ๐ซ โ ๏ฟฝ 0 1 ๐๐๐๐ โ 0 ๏ฟฝ 1 ๐๐๐๐ 0 โ 4๐๐๐๐ ๐๐ 4๐๐๐๐ โ ๐๐ 4๐๐๐๐ ๐๐ 1 V = 11 ๐๐ ๐ซ๐ซ 4๐๐๐๐0 ๐๐ Electric Potential due to localized charge distribution: Potential due to a point 1 charge at origin: V = ๐๐ ๐๐ ๐ซ๐ซ 4๐๐๐๐0 ๐๐ Potential due to a point 1 V = charge at : r ๐๐ ๐ซ๐ซ ๐๐ ๐ซ๐ซโฒ 4๐๐๐๐0 Potential due to a collection 1 of point charges V = ๐๐ r ๐๐i๐๐ ๐ซ๐ซ 0 ๏ฟฝ Potential due to a a continuous 4๐๐๐๐ ๐๐=1 charge distribution is 1 V( ) = r ๐๐๐๐ ๐ซ๐ซ ๏ฟฝ For a line charge = 4๐๐๐๐ 0 For a surface charge = โฒ For a volume charge๐๐๐๐ ๐๐=๐ซ๐ซ ๐๐๐๐โฒ๏ฟฝ 12 ๐๐๐๐ ๐๐ ๐ซ๐ซโฒ ๐๐๐๐๏ฟฝ ๐๐๐๐ ๐๐ ๐ซ๐ซ ๐๐๐๐๏ฟฝ Ease of calculating 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 = V. Sometimes, it is very effective to calculate the electric potential first and then the electric field from there. ๐๐ โ๐๐
โข If the above two is not applicable, one has to go back to the Coulombโs law and then calculate the electric field.
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