Application of Gauss’ Law to calculate Electric field:

Application of Gauss’ Law to calculate Electric field: • draw a figure with location of all charges and direction of E ; • draw a Gaussian surface, so that it contains the field point; • find qencl& dA and substitute in Gauss’ law equation; Solve for E.

Gauss’ Law is always true: BUT not always useful ! Symmetry is crucial Spherical Symmetry → a concentric sphere. 2. Cylindrical Symmetry → a coaxial cylinder. 3. Plane Symmetry → a pill-box straddling the surface.

+ Application of Gauss’ Law A point charge A spherical Gaussian surface of radius r centered on the charge +q.

The Electric Field Due to a Point Charge Eis parallel todAat each point. => E . dA = E dA

(Surface Area of sphere)

+Q a P +Q a A solid sphere An insulating solid sphere of radius a has a uniform volume charge density  carrying a total positive charge Q. To calculate E: P (i) outside the sphere (ii) inside the sphere.

P (i) For r > a +Q Gaussian surface : A concentric sphere of radius r a r

(ii) For r < a P r a Gaussian surface : A concentric sphere of radius r +Q

For r < a For r > a

Pr: 2.9 : Griffiths In spherical co-ordinates, the electric field in some region is given by: ( k being an arbitrary constant) (a) Find the charge density . (b) Find the total charge Q in a sphere of radius R, centered at the origin. (Do it using Gauss’ law and by direct integration).

Soln. Pr. 2.9 (Griffiths) (a) (b) By Gauss’ law: By direct integration:

a The electric field due to a Thin Spherical Shell(uniform charge density) A thin spherical shell has a total charge Q distributed uniformly over its surface. Find the electric field at points (i) outside and (ii) inside the shell.

a The electric field due to a Thin Spherical Shell(uniform charge density) (i) For r > a Gaussian surface : A concentric sphere of radius r r

a The electric field due to a Thin Spherical Shell(uniform charge density) (ii) For r < a Gaussian surface : A concentric sphere of radius r r

a b The electric field due to a Hollow Spherical Shell(non-uniform charge density) Pr. 2.15: Griffiths Charge density in the region a ≤ r ≤b: Find E in three regions: (i) r < a (ii) a < r < b (iii) r > b

a b Solution: Pr. 2.15: (i) r < a: E = 0 (ii) a < r < b: (iii) r > b:

- _ _ d + + + Pr. 2.18: Griffiths Two spheres , each of radius R, overlap partially. To show that the field in the region of overlap is constant. Find its value.

+ + + + + + + + The electric field due to a line of charge To find the electric field a distance r from the line of positive charge of infinite length and constant charge per unit length .

The electric field due to a line of charge Gaussian surface : A coaxial cylindrical surface of radius r and length l

+ + + + + + + + + + + + + + + + + + + + + + + + + + + + + An infinite sheet of charge To find the electric field due to the sheet with uniform surface charge density .

An infinite sheet Charge Gaussian surface: A small cylinder with axis  the sheet and whose ends each have an area A and are equidistant from the plane.

Two infinite, non-conducting sheets, parallel to each other + - • To calculate E at points • to the left of, • (b) in between, • (c) to the right of the two sheets.

+ + + + + + + + - - - - - - - - - - • To the left:: • Enet = 0. E+ E+ E+ E- E- E- • In the region between the sheets: • Enet = /o • towards the right. (a) (b) (c) (c)To the right: Enet = 0.

Pr. 2.17: Griffiiths z Infinite plane slabwith a volume charge density  2d y Find E as a function of y;with y = 0 at center. Plot E vs. y;with E +ve for y +ve. x

Solution: Pr. 2.17: ( for |y| < d) E - d y d ( for y > d)

Gauss’ Law and Conductors Conductor: charges free to move within the material. Electrostatic Equilibrium: when there is no net motion of charge within the conductor.

Electric Field is zero everywhere inside the conductor. WHY? If the E is NOT zero => free charges in the conductor accelerate. Motion of electrons, => the conductor NOT in electrostatic equilibrium. =>the existence of electrostatic equilibrium is consistent only with a zero field in the conductor.

Conductors in Electrostatic Equilibrium Under electrostatic conditions, any excess charge resides entirely on the surface of a solid conductor.

A cavity in a conductor If an electrically neutral material is scooped out from the conductor: no change in charge distribution on surface.

A cavity in a conductor - - - - - - - +q - - - - - - - If +q is placed in the cavity, -q is induced on the surface of the cavity.

Potential of a Conductor Inside a conductor E = 0: the potential everywhere in the conductor must be constant. The entire conductor is at the same potential

Electric Field outside the conductor A Gaussian surface: a small cylinder.

A thin conducting plate of area A • Charge q added to the plate • Each surface has a charge density  = (q/2)/A • EL = ER= /20 • Net E at A & C = /0 • E at B (interior point) = 0

A second plate carrying –q , brought near vicinity of the plate • Two surfaces of charges -q & +q are facing each other. • E due to each surface = /20 • Net E in between the plates = /0

-q +q a b c Pr. 27.4: Conducting sphere at the center of a spherical conducting shell To find E in the regions: (i) r < a (ii) a < r < b (iii) b < r < c (iv) r > c

-q +q a Soln: Pr. 27.4: (i) For r < a Gaussian surface : A concentric sphere of radius r

Soln: Pr. 27.4: (ii) For a < r < b Gaussian surface : A concentric sphere of radius r -q +q a b

-q +q a b c Soln: Pr. 27.4: (iii) For b < r < c Gaussian surface : A concentric sphere of radius r

-q +q a b c Soln: Pr. 27.4: (iv) For r > c Gaussian surface : A concentric sphere of radius r

Pr. 2.36: Griffiths a qa R b qb Two spherical cavities , are hollowed out from the interior of a (neutral) conducting sphere. i) Find the surface charges a ,b & R. ii) What is the field outside the conductor? iii) What is the field within each cavity?

Pr. 2.36: (cont’d) iv) What is the force on qaand qb? v) Which of the above answers change if a third charge qc ,were brought near the conductor? Ans. i) a qa R b qb

Solution Pr. 2.36: (cont’d) Ans. ii) Ans. iii) a qa R b qb Ans. iv) forces on qaand qb= 0.

Solution Pr. 2.36: (cont’d) Ans. v) R changes buta & b do not change. Eout changes butEa & Eb do not change. a qa forces on qa& qbstill = 0. R b qb

Application of Gauss’s Law (wire in pipe) Line Charge Densityls a Line Charge Densityl b Infinitely long cylindrical metallic shell with a line of charge coinciding with the axis of the cylindrical shell. To find E in three regions

(i) For r < a L a Gaussian surface : A coaxial cylinder of radius r and lengthL

(ii) For a < r < b L a b Gaussian surface : A coaxial cylinder of radius r and lengthL

(iii) For r > b a b L Gaussian surface : A coaxial cylinder of radius r and lengthL

Ex. 28.47: Two conducting spheres: 1 & 2 R1 = 5.88 cm, R2 = 12.2 cm q1 = q2 = 28.6 nC Far apart; subsequently connected by a wire To find (a) the final charge on each sphere (b) potential of each sphere Ans (a): 38.6 nC & 18.6 nC, (b): 2850 V

Application of Gauss’ Law to calculate Electric field: