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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:

: if × = 0 everywhere, = V

𝛁𝛁 𝐅𝐅 𝐅𝐅 −𝛁𝛁 • : if = 0 everywhere, = ×

𝛁𝛁 ⋅ 𝐅𝐅 𝐅𝐅 𝛁𝛁 𝐀𝐀 1 • Law: ( ) = r 𝑑𝑑𝑑𝑑 2 𝐄𝐄 𝐫𝐫 0 � ̂ 4𝜋𝜋 𝜖𝜖 • Electric =

𝐸𝐸 Φ �𝑠𝑠𝑢𝑢𝑢𝑢𝑢𝑢𝐄𝐄 ⋅ 𝑑𝑑𝐚𝐚

’s Law = (in form) 𝑄𝑄enc �𝑠𝑠𝑢𝑢𝑢𝑢𝑢𝑢𝐄𝐄 ⋅ 𝑑𝑑𝐚𝐚 0 𝜖𝜖 = (in differential form) 𝜌𝜌 𝛁𝛁 ⋅ 𝐄𝐄 2 𝜖𝜖0 Correction in Lecture # 4:

This is the Gauss’s law in integral form. = 𝑄𝑄enc � 𝐄𝐄 ⋅ 𝑑𝑑𝐚𝐚 𝑠𝑠: 𝑢𝑢𝑢𝑢𝑢𝑢(Griffiths: Ex 2.10):𝜖𝜖0 What is the flux through the shaded face of the cube due to the at the corner ? ? 𝑞𝑞 �𝑠𝑠𝑢𝑢𝑢𝑢𝑢𝑢𝐄𝐄 ⋅ 𝑑𝑑𝐚𝐚

Answer: 24 = 𝑞𝑞 � 𝐄𝐄 ⋅ 𝑑𝑑𝐚𝐚 𝑠𝑠𝑢𝑢𝑢𝑢𝑢𝑢 𝜖𝜖10 = 24 𝑞𝑞 3 � 𝐄𝐄 ⋅ 𝑑𝑑𝐚𝐚 𝑠𝑠𝑢𝑢𝑢𝑢𝑢𝑢 𝜖𝜖0 Gauss’s Law from Coulomb’s Law: If Coulomb’s Law and Gauss’s law have the same information content, can we derive Gauss’s law from Coulomb’s law?

Coulomb’s law gives the electric due to a volume charge ′ 1 r ( ) = 𝜌𝜌 𝐫𝐫 r ̂ ′ Take the 𝐄𝐄 𝐫𝐫 of both� 2 sides𝜌𝜌 𝐫𝐫 of𝑑𝑑𝑑𝑑 the equation 4𝜋𝜋𝜖𝜖0 1 r ( ) = r ̂ ′ 𝛁𝛁 ⋅ 𝐄𝐄 𝐫𝐫r � 𝛁𝛁 ⋅ 2 𝜌𝜌 𝐫𝐫 𝑑𝑑𝑑𝑑 We have: =4𝜋𝜋4 𝜖𝜖 0 (r) = 4 ( ) r ̂ 𝛁𝛁 ⋅ 2 𝜋𝜋1𝛿𝛿 𝜋𝜋 𝛿𝛿 𝐫𝐫 − 𝐫𝐫퐫 ( ) Therefore, = 4 = ′ ′ 𝜌𝜌 𝐫𝐫 𝛁𝛁 ⋅ 𝐄𝐄 𝐫𝐫 0 � 𝜋𝜋 𝛿𝛿 𝐫𝐫 − 𝐫𝐫 𝜌𝜌 𝐫𝐫 𝑑𝑑𝑑𝑑 0 4𝜋𝜋𝜖𝜖The divergence of is𝜖𝜖 equal = to the divided by 𝜌𝜌 4 𝛁𝛁 ⋅ 𝐄𝐄 𝜖𝜖0 𝜖𝜖0 of the Electric Field: Let’s take the simplest electric field: Electric field due to a single point charge is: 1 ( ) = r r 𝑞𝑞 𝑞𝑞 We need to find the2 curl of it 𝐄𝐄 𝐫𝐫 0 ̂ 4𝜋𝜋𝜖𝜖 1 × = × r r 𝑞𝑞 Take𝛁𝛁 the 𝐄𝐄area𝐫𝐫 integral 𝛁𝛁 2 ̂ 4𝜋𝜋𝜖𝜖0 1 × = × r r 𝑞𝑞 Use� Stokes’s𝛁𝛁 𝐄𝐄theorem𝐫𝐫 ⋅ 𝑑𝑑𝐚𝐚 � 𝛁𝛁 2 ̂ ⋅ 𝑑𝑑𝐚𝐚 𝑠𝑠𝑢𝑢𝑢𝑢𝑢𝑢 4𝜋𝜋𝜖𝜖0 𝑠𝑠𝑢𝑢𝑢𝑢𝑢𝑢 1 1 since × = r = = + r r + sin 𝑞𝑞 𝑞𝑞 2 2 𝑑𝑑𝐥𝐥 𝑑𝑑𝑑𝑑𝒓𝒓� 𝑟𝑟𝑟𝑟𝑟𝑟𝜽𝜽� � 𝛁𝛁 𝐄𝐄 𝐫𝐫 ⋅ 𝑑𝑑𝐚𝐚 � ̂ ⋅ 𝑑𝑑𝐥𝐥 0 � 𝑑𝑑𝑑𝑑 𝑠𝑠𝑢𝑢𝑢𝑢𝑢𝑢 4𝜋𝜋𝜖𝜖0 4𝜋𝜋𝜖𝜖 𝑟𝑟 𝜃𝜃𝜃𝜃𝜃𝜃𝝓𝝓� 1 × = × 𝑟𝑟𝑎𝑎 = 0 implies × = 𝑞𝑞 � 𝛁𝛁 𝐄𝐄 𝐫𝐫 ⋅ 𝑑𝑑𝐚𝐚 � 𝑠𝑠𝑢𝑢𝑢𝑢𝑢𝑢 4𝜋𝜋𝜖𝜖0 𝑟𝑟 𝑟𝑟𝑎𝑎 𝛁𝛁 𝐄𝐄 𝟎𝟎 5 Curl of the Electric Field (Digression):

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. 𝛁𝛁 𝐄𝐄 𝟎𝟎 What if the field is dynamic, that is, what if the field changes as a of time? × = Faraday’s Law in differential 𝑑𝑑𝐁𝐁 form. 𝛁𝛁 𝐄𝐄 − Integrate over a surface𝑑𝑑𝑑𝑑

× = 𝑑𝑑𝐁𝐁 � 𝛁𝛁 𝐄𝐄 ⋅ 𝑑𝑑𝐚𝐚 � − ⋅ 𝑑𝑑𝐚𝐚 Apply Stokes’𝒔𝒔𝒖𝒖 𝒖𝒖theorem𝒖𝒖 𝒔𝒔𝒖𝒖𝒖𝒖𝒖𝒖 𝑑𝑑𝑡𝑡

= 𝑑𝑑 � 𝐄𝐄 ⋅ 𝑑𝑑𝐥𝐥 − � 𝐁𝐁 ⋅ 𝑑𝑑𝐚𝐚 𝑝𝑝𝑎𝑎𝑎𝑎푎 𝑑𝑑𝑡𝑡 𝒔𝒔𝒖𝒖Magnetic𝒖𝒖𝒖𝒖 flux Faraday’s Law in integral form. = 6 EMF 𝑑𝑑Φ − 𝑑𝑑𝑡𝑡 ’s Equations (Digression 2):

Gauss’s Law = = 𝜌𝜌 𝑄𝑄enc 𝛁𝛁 ⋅ 𝐄𝐄 0 � 𝐄𝐄 ⋅ 𝑑𝑑𝐚𝐚 𝜖𝜖 𝑠𝑠𝑢𝑢𝑢𝑢𝑢𝑢 𝜖𝜖0 × = = Faraday’s Law 𝑑𝑑𝐁𝐁 𝑑𝑑 𝛁𝛁 𝐄𝐄 − �𝑝𝑝𝑎𝑎𝑎𝑎푎𝐄𝐄 ⋅ 𝑑𝑑𝐥𝐥 − �𝒔𝒔𝒖𝒖𝒖𝒖𝒖𝒖𝐁𝐁 ⋅ 𝑑𝑑𝐚𝐚 𝑑𝑑𝑑𝑑 𝑑𝑑𝑡𝑡 No name; Magnetic = 0 Monopole does not exist 𝛁𝛁 ⋅ 𝐁𝐁 ’s Law with × = Maxwell’s correction 𝑑𝑑𝐄𝐄 When𝛁𝛁 𝐁𝐁 do𝜇𝜇0 not𝐉𝐉 − vary𝜇𝜇0𝜖𝜖0 as a function of time, it is called / 𝑑𝑑𝑑𝑑 . (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 × = 0 everywhere, then: (1) is independent of path. 𝐅𝐅 𝛁𝛁This𝐅𝐅 is because of Stokes’ theorem a (2) ∫ 𝐅𝐅 ⋅ =𝐥𝐥 0 for any closed loop. × =

(3) ∮ is𝐅𝐅 the⋅ 𝑑𝑑𝐥𝐥 of a scalar function: = V�𝑆𝑆 𝑢𝑢𝑢𝑢𝑢𝑢 𝛁𝛁 𝐅𝐅 ⋅ 𝑑𝑑𝐚𝐚 �𝑃𝑃𝑎𝑎𝑎𝑎푎𝐅𝐅 ⋅ 𝑑𝑑𝐥𝐥 • This is because Curl of a gradient is zero × V = 𝐅𝐅 𝐅𝐅 −𝛁𝛁 𝛁𝛁 𝛁𝛁 𝟎𝟎 The curl of Electric field is zero, that is, × = 0 everywhere. Therefore: (1) d is independent of path. 𝐄𝐄 𝛁𝛁 𝐄𝐄 This is because of Stokes’ theorem b a (2) ∫ 𝐄𝐄 ⋅ =𝐥𝐥 0 for any closed loop. × =

∮ 𝐄𝐄 ⋅ 𝑑𝑑𝐥𝐥 �𝑆𝑆𝑢𝑢𝑢𝑢𝑢𝑢 𝛁𝛁 𝐄𝐄 ⋅ 𝑑𝑑𝐚𝐚 �𝑃𝑃𝑎𝑎𝑎𝑎푎𝐄𝐄 ⋅ 𝑑𝑑𝐥𝐥 (3) is the gradient of a scalar function: = V

𝐄𝐄 𝐄𝐄 −𝛁𝛁 V is called the . It is a scalar quantity, the gradient of which is equal to the electric field

8 Electric Potential:

Since × = 0 everywhere, = V

𝛁𝛁 𝐄𝐄 𝐄𝐄 −𝛁𝛁 How to write electric potential in terms of the electric field? V =

Take the−𝛁𝛁 𝐄𝐄 of the above equation over a path

𝒃𝒃 V = 𝒃𝒃

� 𝛁𝛁 ⋅ 𝑑𝑑𝐥𝐥 − � 𝐄𝐄 ⋅ 𝑑𝑑𝐥𝐥 𝒂𝒂 𝒂𝒂 Use the fundamental Theorem for Gradient: 𝒃𝒃 V = V V( )

� 𝛁𝛁 ⋅ 𝑑𝑑𝐥𝐥 𝑏𝑏 − 𝑎𝑎 𝒂𝒂 𝑃𝑃𝑃𝑃𝑃𝑃� • Absolute potential cannot be defined. V V( ) = 𝒃𝒃 • Only potential differences can be defined. 𝐛𝐛 − 𝐚𝐚 − � 𝐄𝐄 ⋅ 𝑑𝑑𝐥𝐥 𝒂𝒂

9 Electric Potential: (1) Electric potential is different from electric . Unit of electric potential is -meter per Coulomb ( ) or . (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 ⋯ 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: × = 0, i.e., = ; = ; = ; 𝜕𝜕Ex 𝜕𝜕Ey 𝜕𝜕Ez 𝜕𝜕Ey 𝜕𝜕Ex 𝜕𝜕Ez (5) V 𝛁𝛁V( 𝐄𝐄) = 𝜕𝜕휕 . Absolute𝜕𝜕휕 𝜕𝜕휕 potential𝜕𝜕휕 cannot𝜕𝜕휕 be𝜕𝜕휕 defined. In electrostatics, usually one𝒃𝒃 takes the reference point to infinity and set the 𝒂𝒂 potential𝐛𝐛 at −infinity𝐚𝐚 to −zero,∫ 𝐄𝐄that⋅ 𝑑𝑑 𝐥𝐥is, take V = V = 0. Also if V = V( ),

𝐚𝐚 ∞ 𝐛𝐛 𝐫𝐫 V = 𝐫𝐫 10 𝐫𝐫 − � 𝐄𝐄 ⋅ 𝑑𝑑𝐥𝐥 ∞ Electric Potential due to a point charge at origin:

Electric field ( ) at due to a single point charge at origin: 𝟏𝟏 𝟏𝟏 𝐄𝐄 𝐫𝐫 1 𝐫𝐫 ( ) = 𝑞𝑞 𝑞𝑞 𝐄𝐄 𝐫𝐫𝟏𝟏 2 𝐫𝐫�𝟏𝟏 Electric potential4𝜋𝜋 V 𝜖𝜖0 𝑟𝑟at1 due to a single point charge at origin: 𝐫𝐫 𝐫𝐫 𝑞𝑞 V = 𝑟𝑟 = 𝒓𝒓 ( )

𝐫𝐫 − � 𝐄𝐄 ⋅ 𝑑𝑑𝐥𝐥 − � 𝐄𝐄 𝐫𝐫𝟏𝟏 ⋅ 𝑑𝑑𝐥𝐥𝟏𝟏 The line element∞ is: = ∞ r + + sin

𝑑𝑑1𝐥𝐥𝟏𝟏 𝑑𝑑𝑟𝑟1 �𝟏𝟏 𝑟𝑟1𝑑𝑑𝜃𝜃1 𝛉𝛉�𝟏𝟏 1𝑟𝑟1 𝜃𝜃1𝑑𝑑𝜙𝜙1 𝝓𝝓�𝟏𝟏1 V = 𝒓𝒓 = 𝒓𝒓 = 𝑞𝑞 𝑞𝑞 𝑞𝑞 2 1 2 1 𝐫𝐫 − � 0 1 𝑑𝑑𝑟𝑟 − 0 � 1 𝑑𝑑𝑟𝑟 0 ∞ 4𝜋𝜋𝜖𝜖 𝑟𝑟 4𝜋𝜋𝜖𝜖 ∞ 𝑟𝑟 4𝜋𝜋𝜖𝜖 𝑟𝑟 1 V = 11 𝑞𝑞 𝐫𝐫 4𝜋𝜋𝜖𝜖0 𝑟𝑟 Electric Potential due to localized charge distribution: Potential due to a point 1 charge at origin: V = 𝑞𝑞 𝑞𝑞 𝐫𝐫 4𝜋𝜋𝜖𝜖0 𝑟𝑟 Potential due to a point 1 V = charge at : r 𝑞𝑞 𝐫𝐫 𝑞𝑞 𝐫𝐫′ 4𝜋𝜋𝜖𝜖0 Potential due to a collection 1 of point charges V = 𝑛𝑛 r 𝑞𝑞i𝑖𝑖 𝐫𝐫 0 � Potential due to a a continuous 4𝜋𝜋𝜖𝜖 𝑖𝑖=1 charge distribution is 1 V( ) = r 𝑑𝑑𝑑𝑑 𝐫𝐫 � For a line charge = 4𝜋𝜋𝜖𝜖 0 For a = ′ For a volume charge𝑑𝑑𝑑𝑑 𝜆𝜆=𝐫𝐫 𝑑𝑑𝑑𝑑′� 12 𝑑𝑑𝑑𝑑 𝜎𝜎 𝐫𝐫′ 𝑑𝑑𝑑𝑑� 𝑑𝑑𝑑𝑑 𝜌𝜌 𝐫𝐫 𝑑𝑑𝑑𝑑� Ease of calculating the Electric Field

• The easiest way to calculate the electric field is using Gauss’s law. But this is possible only when there is some symmetry in the problem.

• 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. 𝐄𝐄 −𝛁𝛁

• If the above two is not applicable, one has to go back to the Coulomb’s law and then calculate the electric field.

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