1. 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.

2. 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.

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

4. The Electric Field Due to a Point Charge
E is parallel to dA at each point.
=> E . dA = E dA

5. (Surface Area of sphere)

6. 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.

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

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

9. For r < a
For r > a

10. 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).

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

12. 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.

13. 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

14. 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

15. 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

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

17. 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. _ +

18. 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 .

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

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

21. 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.

22. 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. + -

23. + - 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.

24. 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

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

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

27. 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.

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

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

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

31. 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

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

33. 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

34. 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

35. 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

36. 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

37. 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

38. 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

39. 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

40. 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?

41. 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

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

43. 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

44. 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

45. 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

46. 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

47. 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

48. 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