Coulombs Law Lecture3 Coulombs Law Like charges repel

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Coulombs Law Lecture-3

Coulombs Law Lecture-3

Coulombs Law Like charges repel, unlike charges attract. The electric force acting on a

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 e 0 = permittivity of space

Quiz • A charge Q 1= 1 n. C is located at the origin

Quiz • A charge Q 1= 1 n. C 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)

Gauss's Law • The total of the electric flux out of a closed surface

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 eo. da Φ= • +q eo = the permittivity of free space 8. 854 x 10 -12 C 2/(N m 2) = 1/4 pk) E

Case(i) Charge is placed inside the body: da * Flux passing through an area

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

Case(ii) Charge is placed outside the surface • Flux passing through AB: da +q

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

 • A more intuitive statement: the total number of electric field lines entering

• 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 FE is the product of Component of the Electric Field Perpendicular to a surface times the surface area.

-ve flux +ve Flux

-ve flux +ve Flux

Differential form of Gauss Law: Proof: Gauss Law Φ= Gauss divergence theorem: or Where

Differential form of Gauss Law: Proof: Gauss Law 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.

Quiz: 1. 2. 3. 4. How electric flux is related to number of electric

Quiz: 1. 2. 3. 4. How electric flux is related to number of electric field lines? How will you determine flux for nonuniform surfaces? How will you explain negative flux or positive flux? A positive charge of 1 μC is placed at the centre of a hollow cube. Calculate the electric flux diverging 1. 2. 5. through the cube. through each face. Ans: 1. 12 x 105 V/m Ans: 1. 86 x 104 V/m If E=3 i+4 j-5 k, calculate electric flux through the surface S=2 x 10 -5 k m 2. Ans: -10 -4 V/m

Applications of Gauss law Lecture-4

Applications of Gauss law Lecture-4

(Spherical systems: Conducting Sphere) 1)Conducting Sphere of charge ‘q’ and radius ‘R’: 1) E

(Spherical systems: Conducting Sphere) 1)Conducting Sphere of charge ‘q’ and radius ‘R’: 1) E at an external point: Eo r>R 2) E at the surface: Es r=R 3) E at an internal point: Ei r<R Gaussian surface Case-I: E at an external point; Net electric fux through ‘P’: R P r S 1 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.

(Spherical systems: Conducting Sphere) Gaussian surface Case-II: E at the Surface; r=R Gaussian surface

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

(Spherical systems: Conducting Sphere) 1)Conducting Sphere of charge ‘q’ and radius ‘R’: 1) E

(Spherical systems: Conducting Sphere) 1)Conducting Sphere of charge ‘q’ and radius ‘R’: 1) E at an external point: Eo r>R 2) E at the surface: Es r=R 3) E at an internal point: Ei r<R R r E Es Ei=0 r=0 P r=R Eo r

(Spherical systems: Nonconducting Sphere) Nonconducting sphere (Volume charge density) • E at an external

(Spherical systems: Nonconducting Sphere) Nonconducting sphere (Volume charge density) • E at an external point: Eo • E at the surface: Es • E at an internal point: Ei

(Spherical systems: Nonconducting Sphere) 1)Nonconducting Sphere of charge ‘q’ and radius ‘R’: 1) E

(Spherical systems: Nonconducting Sphere) 1)Nonconducting Sphere of charge ‘q’ and radius ‘R’: 1) E at an external point: Eo r>R Volume charge density 2) E at the surface: Es r=R 3) E at an internal point: Ei r<R Case-I: E at an external point; Net electric flux through ‘P’: Gaussian surface R P r S 1

(Spherical systems: Nonconducting Sphere) Gaussian surface Case-II: E at the Surface; r=R Case-II: E

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

(Spherical systems: Nonconducting Sphere) 1)Nonconducting Sphere of charge ‘q’ and radius ‘R’: 1) E

(Spherical systems: Nonconducting Sphere) 1)Nonconducting Sphere of charge ‘q’ and radius ‘R’: 1) E at an external point: Eo r>R 2) E at the surface: Es r=R 3) E at an internal point: Ei r<R R r E Es Eo Ei r=0 P r=R r

Applications of Gauss law (Cylindrical distribution systems)

Applications of Gauss law (Cylindrical distribution systems)

Applications of Gauss law (Cylindrical distribution systems) 1)Conducting long Cylinder of charge ‘q’ and

Applications of Gauss law (Cylindrical distribution systems) 1)Conducting long Cylinder of charge ‘q’ and radius ‘R’: 1) E at an external point: Eo 2) E at the surface: Es 3) E at an internal point: Ei 2)Nonconducting long Cylinder 1) E at an external point: Eo 2) E at the surface: Es 3) E at an internal point: Ei

Cylindrical distribution systems: Conducting Cylinder 1)Conducting long Cylinder of charge ‘q’ and radius ‘R’

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

Case-II: E at the Surface; Case-III: E at an internal point; l E R

Case-II: E at the Surface; Case-III: E at an internal point; l E R O P E Es Ei=0 r=R Eo r

For infinite long line charge density ‘λ’

For infinite long line charge density ‘λ’

Applications of Gauss law (Infinitely long sheet of Charge)

Applications of Gauss law (Infinitely long sheet of Charge)

Quiz • For a conducting sphere: with surface charge density ‘σ’ and radius R,

Quiz • For a conducting sphere: with surface charge density ‘σ’ and radius R, determine Eo, Es and Ei. • 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 Eo, Es and Ei

Electric field in inside and just outside (very close) of the surface of a

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