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 E is parallel to dA at each point. => E . dA = E dA
(Surface Area of sphere)
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: +Q a P +Q a P (i) outside the sphere (ii) inside the sphere.
Gaussian surface : A concentric sphere of radius r (i) For r > a P +Q Gaussian surface : A concentric sphere of radius r a r
Gaussian surface : A concentric sphere of radius 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:
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.
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
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
Find E in three regions: 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: a b Find E in three regions: (i) r < a (ii) a < r < b (iii) r > b
Solution: Pr. 2.15: (i) r < a: E = 0 (ii) a < r < b: (iii) r > b: a b
Pr. 2.18: Griffiths Two spheres , each of radius R, overlap partially. + - _ + d 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 slab with 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 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= /20 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 = /20 Net E in between the plates = /0
To find E in the regions: (i) r < a (ii) a < r < b 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 a c +q -q b
Gaussian surface : A concentric sphere of radius r Soln: Pr. 27.4: (i) For r < a Gaussian surface : A concentric sphere of radius r a +q -q
Gaussian surface : A concentric sphere of radius r -q Soln: Pr. 27.4: (ii) For a < r < b Gaussian surface : A concentric sphere of radius r -q +q a b
Gaussian surface : A concentric sphere of radius r Soln: Pr. 27.4: (iii) For b < r < c Gaussian surface : A concentric sphere of radius r a c +q -q b
Gaussian surface : A concentric sphere of radius r Soln: Pr. 27.4: (iv) For r > c Gaussian surface : A concentric sphere of radius r a c +q -q b
Pr. 2.36: Griffiths Two spherical cavities , are hollowed out from the interior of a (neutral) conducting sphere. i) Find the surface charges a ,b & R. qa qb a b R ii) What is the field outside the conductor? iii) What is the field within each cavity?
iv) What is the force on qa and qb? Pr. 2.36: (cont’d) iv) What is the force on qa and 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) Ans. iv) qa R b qb Ans. iv) forces on qa and qb= 0.
Solution Pr. 2.36: (cont’d) Ans. v) R changes but a & b do not change. Eout changes but Ea & Eb do not change. a qa forces on qa & qb still = 0. R b qb
Application of Gauss’s Law (wire in pipe) Line Charge Density ls a Line Charge Density l 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
Gaussian surface : A coaxial cylinder of radius r and length L (i) For r < a L a Gaussian surface : A coaxial cylinder of radius r and length L
Gaussian surface : A coaxial cylinder of radius r and length L (ii) For a < r < b L a b Gaussian surface : A coaxial cylinder of radius r and length L
Gaussian surface : A coaxial cylinder of radius r and length L (iii) For r > b a b L Gaussian surface : A coaxial cylinder of radius r and length L
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