1. Coulombs Law Lecture-3

2. Coulombs Law Like charges repel, unlike charges attract. The electric force acting on a point charge q 1 as a result of the presence of a second point charge q 2 is given by Coulomb's Law: where  0 = permittivity of space

3. Quiz A charge Q 1 = 1nC is located at the origin in free space and another charge Q at (2,0,0). If the X – component of the electric field at (3,1,1) is zero, calculate the value of Q. Is the Y component zero at (3,1,1)?Ans: -3(3/11) 1.5 Q 1 Calculation of E: due to 1.Dipole, 2.Rod (line charge), Ring (Line charge), 3.Circular plate (surface charge), Square sheet, 4.Sphere or Cylinder (Volume charge density)

5. Gauss's Law The total of the electric flux out of a closed surface is equal to the charge enclosed divided by the permittivity. Electric Flux over a Closed Surface = Charge enclosed by the Surface divided by  o.   o = the permittivity of free space 8.854x10 -12 C 2 /(N m 2 ) = 1/4  k) Φ=Φ= E da +q

6. Case(i) Charge is placed inside the body: E da +q Where dΩ is solid angle * Flux passing through an area da: Total flux over a closed surface: E da da cosθ +q dΩdΩ

7. Case(ii) Charge is placed outside the surface E da Where dΩ is solid angle +q da E A B C D Note: dΩ will remain same for but direction of the flux will be opposite. Flux passing through AB: Net flux = 0 Flux passing through CD: where

8. . A more intuitive statement: the total number of electric field lines entering or leaving a closed volume of space is directly proportional to the charge enclosed by the volume If there is no net charge inside some volume of space then the electric flux over the surface of that volume is always equal to zero. Electric Flux: The Electric Flux  E is the product of Component of the Electric Field Perpendicular to a surface times the surface area.

9. -ve flux +ve Flux

11. Differential form of Gauss Law: Proof: Gauss Law Φ=Φ= Gauss divergence theorem: or Where D=ε 0 E, called Electric field displacement vector. Note: Gauss law is also known as Maxwell’s first equation.

12. Quiz: 1.How electric flux is related to number of electric field lines? 2.How will you determine flux for nonuniform surfaces? 3.How will you explain negative flux or positive flux? 4.A positive charge of 1 μC is placed at the centre of a hollow cube. Calculate the electric flux diverging 1.through the cube. Ans: 1.12 x 10 5 V/m 2.through each face.Ans: 1.86 x 10 4 V/m 5.If E=3i+4j-5k, calculate electric flux through the surface S=2 x 10 -5 k m 2. Ans: -10 -4 V/m

13. Applications of Gauss law Lecture-4

14. (Spherical systems: Conducting Sphere) 1)Conducting Sphere of charge ‘q’ and radius ‘R’: 1)E at an external point: E o r>R 2)E at the surface: E s r=R 3)E at an internal point: E i r<R Case-I: E at an external point; Net electric fux through ‘P’: Gaussia n surface R r P S1 The Electric field strength at any point outside a spherical charge distribution is the same as through the whole charge were concentrated at the centre.

15. Case-II: E at the Surface; Gaussian surface r=R Case-III: E at an internal point; Gaussia n surface R r (Spherical systems: Conducting Sphere)

16. 1)Conducting Sphere of charge ‘q’ and radius ‘R’: 1)E at an external point: E o r>R 2)E at the surface: E s r=R 3)E at an internal point: E i r<R (Spherical systems: Conducting Sphere) R P r EsEs EoEo r=R r E i =0 E r=0

17. Nonconducting sphere (Volume charge density) E at an external point: E o E at the surface: E s E at an internal point: E i (Spherical systems: Nonconducting Sphere)

18. 1)Nonconducting Sphere of charge ‘q’ and radius ‘R’: 1)E at an external point: E o r>R 2)E at the surface: E s r=R 3)E at an internal point: E i r<R Case-I: E at an external point; Net electric flux through ‘P’: Gaussia n surface R r P S1 Volume charge density

19. Case-II: E at the Surface; Gaussian surface r=R Case-II: E at an internal point; Gaussia n surface R r (Spherical systems: Nonconducting Sphere)

20. 1)Nonconducting Sphere of charge ‘q’ and radius ‘R’: 1)E at an external point: E o r>R 2)E at the surface: E s r=R 3)E at an internal point: E i r<R (Spherical systems: Nonconducting Sphere) R P r EsEs EoEo r=R r EiEi E r=0

21. Applications of Gauss law (Cylindrical distribution systems)

22. 1)Conducting long Cylinder of charge ‘q’ and radius ‘R’: 1)E at an external point: E o 2)E at the surface: E s 3)E at an internal point: E i 2)Nonconducting long Cylinder 1)E at an external point: E o 2)E at the surface: E s 3)E at an internal point: E i

23. Cylindrical distribution systems: Conducting Cylinder 1)Conducting long Cylinder of charge ‘q’ and radius ‘R’ : 1)E at an external point: E o r>R 2)E at the surface: E s r=R 3)E at an internal point: E i r<R Gaussian surface E R r l PO Case-I: E at an external point; Net electric flux through ‘P’:

24. Case-II: E at the Surface;Case-III: E at an internal point; EsEs EoEo r=R r E i =0 E r=0 E R l PO

25. For infinite long line charge density ‘λ’

26. Applications of Gauss law (Infinitely long sheet of Charge)

27. (1) +σ+σ +σ+σ (2) -σ-σ +σ+σ (3) -σ-σ -σ-σ E

28. Quiz For a conducting sphere: with surface charge density ‘σ’ and radius R, determine E o, E s and E i. For a spherical shell volume charge density is ρ=k/r 2 for a≤r≤b otherwise zero. determine E for each region. For a cylinder of radius ‘R’ and height ‘h’ volume charge density is ρ=kr. Determine E o, E s and E i

29. Electric field in inside and just outside (very close) of the surface of a charged conductor Ans: Inside E=0, Outside E= σ/ε 0