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PHY103A: Lecture # 5 (Text Book: Intro to Electrodynamics by Griffiths, 3Rd Ed.)

Semester II, 2017-18 Department of , 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:

โข : if ร = 0 everywhere, = V

๐๐ ๐๐ ๐๐ โ๐๐ โข : if = 0 everywhere, = ร

๐๐ โ ๐๐ ๐๐ ๐๐ ๐๐ 1 โข โ Law: ( ) = r ๐๐๐๐ 2 ๐๐ ๐ซ๐ซ 0 ๏ฟฝ ฬ 4๐๐ ๐๐ โข Electric =

๐ธ๐ธ ฮฆ ๏ฟฝ๐ ๐ ๐ข๐ข๐ข๐ข๐ข๐ข๐๐ โ ๐๐๐๐

โข โs Law = (in form) ๐๐enc ๏ฟฝ๐ ๐ ๐ข๐ข๐ข๐ข๐ข๐ข๐๐ โ ๐๐๐๐ 0 ๐๐ = (in differential form) ๐๐ ๐๐ โ ๐๐ 2 ๐๐0 Correction in Lecture # 4:

This is the Gaussโs law in integral form. = ๐๐enc ๏ฟฝ ๐๐ โ ๐๐๐๐ ๐ ๐ : ๐ข๐ข๐ข๐ข๐ข๐ข(Griffiths: Ex 2.10):๐๐0 What is the flux through the shaded face of the cube due to the at the corner ? ? ๐๐ ๏ฟฝ๐ ๐ ๐ข๐ข๐ข๐ข๐ข๐ข๐๐ โ ๐๐๐๐

Answer: 24 = ๐๐ ๏ฟฝ ๐๐ โ ๐๐๐๐ ๐ ๐ ๐ข๐ข๐ข๐ข๐ข๐ข ๐๐10 = 24 ๐๐ 3 ๏ฟฝ ๐๐ โ ๐๐๐๐ ๐ ๐ ๐ข๐ข๐ข๐ข๐ข๐ข ๐๐0 Gaussโs Law from Coulombโs Law: If Coulombโs Law and Gaussโs law have the same information content, can we derive Gaussโs law from Coulombโs law?

Coulombโs law gives the electric due to a volume charge โฒ 1 r ( ) = ๐๐ ๐ซ๐ซ r ฬ โฒ Take the ๐๐ ๐ซ๐ซ of both๏ฟฝ 2 sides๐๐ ๐ซ๐ซ of๐๐๐๐ the equation 4๐๐๐๐0 1 r ( ) = r ฬ โฒ ๐๐ โ ๐๐ ๐ซ๐ซr ๏ฟฝ ๐๐ โ 2 ๐๐ ๐ซ๐ซ ๐๐๐๐ We have: =4๐๐4 ๐๐ 0 (r) = 4 ( ) r ฬ ๐๐ โ 2 ๐๐1๐ฟ๐ฟ ๐๐ ๐ฟ๐ฟ ๐ซ๐ซ โ ๐ซ๐ซํซ ( ) Therefore, = 4 = โฒ โฒ ๐๐ ๐ซ๐ซ ๐๐ โ ๐๐ ๐ซ๐ซ 0 ๏ฟฝ ๐๐ ๐ฟ๐ฟ ๐ซ๐ซ โ ๐ซ๐ซ ๐๐ ๐ซ๐ซ ๐๐๐๐ 0 4๐๐๐๐The divergence of is๐๐ equal = to the divided by ๐๐ 4 ๐๐ โ ๐๐ ๐๐0 ๐๐0 of the Electric Field: Letโs take the simplest electric field: Electric field due to a single point charge is: 1 ( ) = r r ๐๐ ๐๐ We need to find the2 curl of it ๐๐ ๐ซ๐ซ 0 ฬ 4๐๐๐๐ 1 ร = ร r r ๐๐ Take๐๐ the ๐๐area๐ซ๐ซ integral ๐๐ 2 ฬ 4๐๐๐๐0 1 ร = ร r r ๐๐ Use๏ฟฝ Stokesโs๐๐ ๐๐theorem๐ซ๐ซ โ ๐๐๐๐ ๏ฟฝ ๐๐ 2 ฬ โ ๐๐๐๐ ๐ ๐ ๐ข๐ข๐ข๐ข๐ข๐ข 4๐๐๐๐0 ๐ ๐ ๐ข๐ข๐ข๐ข๐ข๐ข 1 1 since ร = r = = + r r + sin ๐๐ ๐๐ 2 2 ๐๐๐ฅ๐ฅ ๐๐๐๐๐๐๏ฟฝ ๐๐๐๐๐๐๐ฝ๐ฝ๏ฟฝ ๏ฟฝ ๐๐ ๐๐ ๐ซ๐ซ โ ๐๐๐๐ ๏ฟฝ ฬ โ ๐๐๐ฅ๐ฅ 0 ๏ฟฝ ๐๐๐๐ ๐ ๐ ๐ข๐ข๐ข๐ข๐ข๐ข 4๐๐๐๐0 4๐๐๐๐ ๐๐ ๐๐๐๐๐๐๐๐๏ฟฝ 1 ร = ร ๐๐๐๐ = 0 implies ร = ๐๐ ๏ฟฝ ๐๐ ๐๐ ๐ซ๐ซ โ ๐๐๐๐ ๏ฟฝ ๐ ๐ ๐ข๐ข๐ข๐ข๐ข๐ข 4๐๐๐๐0 ๐๐ ๐๐๐๐ ๐๐ ๐๐ ๐๐ 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. ๐๐ ๐๐ ๐๐ What if the field is dynamic, that is, what if the field changes as a of time? ร = Faradayโs Law in differential ๐๐๐๐ form. ๐๐ ๐๐ โ Integrate over a surface๐๐๐๐

ร = ๐๐๐๐ ๏ฟฝ ๐๐ ๐๐ โ ๐๐๐๐ ๏ฟฝ โ โ ๐๐๐๐ Apply Stokesโ๐๐๐๐ ๐๐theorem๐๐ ๐๐๐๐๐๐๐๐ ๐๐๐ก๐ก

= ๐๐ ๏ฟฝ ๐๐ โ ๐๐๐ฅ๐ฅ โ ๏ฟฝ ๐๐ โ ๐๐๐๐ ๐๐๐๐๐๐ํ ๐๐๐ก๐ก ๐๐๐๐Magnetic๐๐๐๐ flux Faradayโs Law in integral form. = 6 EMF ๐๐ฮฆ โ ๐๐๐ก๐ก โs Equations (Digression 2):

Gaussโs Law = = ๐๐ ๐๐enc ๐๐ โ ๐๐ 0 ๏ฟฝ ๐๐ โ ๐๐๐๐ ๐๐ ๐ ๐ ๐ข๐ข๐ข๐ข๐ข๐ข ๐๐0 ร = = Faradayโs Law ๐๐๐๐ ๐๐ ๐๐ ๐๐ โ ๏ฟฝ๐๐๐๐๐๐ํ๐๐ โ ๐๐๐ฅ๐ฅ โ ๏ฟฝ๐๐๐๐๐๐๐๐๐๐ โ ๐๐๐๐ ๐๐๐๐ ๐๐๐ก๐ก No name; Magnetic = 0 Monopole does not exist ๐๐ โ ๐๐ โs Law with ร = Maxwellโs correction ๐๐๐๐ When๐๐ ๐๐ do๐๐0 not๐๐ โ vary๐๐0๐๐0 as a function of time, it is called / ๐๐๐๐ . (before mid-sem) When fields do vary as a function of time, then the two fields have to be studied together as an , and one consequence of a changing electric and is the electromagnetic . (after mid-sem) When the of the field is quantized () then it is called electrodynamics. (Not for this course). Applications: Quantum computers, Quantum cryptography, Quantum teleportation 7

Electric Potential:

Recall: If the curl of a is zero, that is, if ร = 0 everywhere, then: (1) is independent of path. ๐๐ ๐๐This๐๐ is because of Stokesโ theorem a (2) โซ ๐๐ โ =๐ฅ๐ฅ 0 for any closed loop. ร =

(3) โฎ is๐๐ theโ ๐๐๐ฅ๐ฅ of a scalar function: = V๏ฟฝ๐๐ ๐ข๐ข๐ข๐ข๐ข๐ข ๐๐ ๐๐ โ ๐๐๐๐ ๏ฟฝ๐๐๐๐๐๐ํ๐๐ โ ๐๐๐ฅ๐ฅ โข This is because Curl of a gradient is zero ร V = ๐๐ ๐๐ โ๐๐ ๐๐ ๐๐ ๐๐ The curl of Electric field is zero, that is, ร = 0 everywhere. Therefore: (1) d is independent of path. ๐๐ ๐๐ ๐๐ This is because of Stokesโ theorem b a (2) โซ ๐๐ โ =๐ฅ๐ฅ 0 for any closed loop. ร =

โฎ ๐๐ โ ๐๐๐ฅ๐ฅ ๏ฟฝ๐๐๐ข๐ข๐ข๐ข๐ข๐ข ๐๐ ๐๐ โ ๐๐๐๐ ๏ฟฝ๐๐๐๐๐๐ํ๐๐ โ ๐๐๐ฅ๐ฅ (3) is the gradient of a scalar function: = V

๐๐ ๐๐ โ๐๐ V is called the . It is a scalar quantity, the gradient of which is equal to the electric field

8 Electric Potential:

Since ร = 0 everywhere, = V

๐๐ ๐๐ ๐๐ โ๐๐ How to write electric potential in terms of the electric field? V =

Take theโ๐๐ ๐๐ of the above equation over a path

๐๐ V = ๐๐

๏ฟฝ ๐๐ โ ๐๐๐ฅ๐ฅ โ ๏ฟฝ ๐๐ โ ๐๐๐ฅ๐ฅ ๐๐ ๐๐ Use the fundamental Theorem for Gradient: ๐๐ V = V V( )

๏ฟฝ ๐๐ โ ๐๐๐ฅ๐ฅ ๐๐ โ ๐๐ ๐๐ ๐๐๐๐๐๐๏ฟฝ โข Absolute potential cannot be defined. V V( ) = ๐๐ โข Only potential differences can be defined. ๐๐ โ ๐๐ โ ๏ฟฝ ๐๐ โ ๐๐๐ฅ๐ฅ ๐๐

9 Electric Potential: (1) Electric potential is different from electric . Unit of electric potential is -meter per Coulomb ( ) or . โ (2) The potential obeys , i.e., the potential due to several charges is equal to the sum of the due to individual ones: V = V + V +

(3) If one knows the electrical potential (a scalar quantity), the electric field1 (a 2 โฏ 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., = ; = ; = ; ๐๐Ex ๐๐Ey ๐๐Ez ๐๐Ey ๐๐Ex ๐๐Ez (5) V ๐๐V( ๐๐) = ๐๐ํ . Absolute๐๐ํ ๐๐ํ potential๐๐ํ cannot๐๐ํ be๐๐ํ defined. In electrostatics, usually one๐๐ takes the reference point to infinity and set the ๐๐ potential๐๐ at โinfinity๐๐ to โzero,โซ ๐๐thatโ ๐๐ ๐ฅ๐ฅis, take V = V = 0. Also if V = V( ),

๐๐ โ ๐๐ ๐ซ๐ซ V = ๐ซ๐ซ 10 ๐ซ๐ซ โ ๏ฟฝ ๐๐ โ ๐๐๐ฅ๐ฅ โ Electric Potential due to a point charge at origin:

Electric field ( ) at due to a single point charge at origin: ๐๐ ๐๐ ๐๐ ๐ซ๐ซ 1 ๐ซ๐ซ ( ) = ๐๐ ๐๐ ๐๐ ๐ซ๐ซ๐๐ 2 ๐ซ๐ซ๏ฟฝ๐๐ Electric potential4๐๐ V ๐๐0 ๐๐at1 due to a single point charge at origin: ๐ซ๐ซ ๐ซ๐ซ ๐๐ V = ๐๐ = ๐๐ ( )

๐ซ๐ซ โ ๏ฟฝ ๐๐ โ ๐๐๐ฅ๐ฅ โ ๏ฟฝ ๐๐ ๐ซ๐ซ๐๐ โ ๐๐๐ฅ๐ฅ๐๐ The line elementโ is: = โ r + + sin

๐๐1๐ฅ๐ฅ๐๐ ๐๐๐๐1 ๏ฟฝ๐๐ ๐๐1๐๐๐๐1 ๐๐๏ฟฝ๐๐ 1๐๐1 ๐๐1๐๐๐๐1 ๐๐๏ฟฝ๐๐1 V = ๐๐ = ๐๐ = ๐๐ ๐๐ ๐๐ 2 1 2 1 ๐ซ๐ซ โ ๏ฟฝ 0 1 ๐๐๐๐ โ 0 ๏ฟฝ 1 ๐๐๐๐ 0 โ 4๐๐๐๐ ๐๐ 4๐๐๐๐ โ ๐๐ 4๐๐๐๐ ๐๐ 1 V = 11 ๐๐ ๐ซ๐ซ 4๐๐๐๐0 ๐๐ Electric Potential due to localized charge distribution: Potential due to a point 1 charge at origin: V = ๐๐ ๐๐ ๐ซ๐ซ 4๐๐๐๐0 ๐๐ Potential due to a point 1 V = charge at : r ๐๐ ๐ซ๐ซ ๐๐ ๐ซ๐ซโฒ 4๐๐๐๐0 Potential due to a collection 1 of point charges V = ๐๐ r ๐๐i๐๐ ๐ซ๐ซ 0 ๏ฟฝ Potential due to a a continuous 4๐๐๐๐ ๐๐=1 charge distribution is 1 V( ) = r ๐๐๐๐ ๐ซ๐ซ ๏ฟฝ For a line charge = 4๐๐๐๐ 0 For a = โฒ For a volume charge๐๐๐๐ ๐๐=๐ซ๐ซ ๐๐๐๐โฒ๏ฟฝ 12 ๐๐๐๐ ๐๐ ๐ซ๐ซโฒ ๐๐๐๐๏ฟฝ ๐๐๐๐ ๐๐ ๐ซ๐ซ ๐๐๐๐๏ฟฝ Ease of calculating the Electric Field

โข The easiest way to calculate the electric field is using Gaussโs law. But this is possible only when there is some symmetry in the problem.

โข 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. ๐๐ โ๐๐

โข If the above two is not applicable, one has to go back to the Coulombโs law and then calculate the electric field.

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