1. Physics

2. ELECTROSTATICS - 3

3. Session Objectives Application of Gauss’ law
Electric potential (intro.)

4. Gaussian surface Any imaginary closed surface.
Gaussian surfaces are useful in computing
Electric field (usually uniform) and flux
Gaussian surfaces are very useful for finding electric field with symmetrical charge distribution

5. Electric field due to an infinitely long straight charged wire
Due to cylindrical symmetry of charge, electric field is away from the centre of the wire
The net electric flux,

6. r + Charge enclosed by the cylinder, q = l h By Gauss’ law,
1. Gaussian surfaces should not pass through discrete charges, as electric field is not well defined in any location
2. Gaussian surfaces can easily pass through continuous charge distribution.

7. Solved Example – 1
An infinite line charge produces a field of 9×104 N C–1 at a distance of 2 cm. Calculate the linear charge density.
Solution:
E = 9×104 N C–1, r = 2×10–2 m

8. Solved Example –2 – + An electric dipole consists of charges +1.6 nC
and –1.6 nC separated by a distance of 2×10–3 m.
It is placed near a long line charge of linear
charge density 5×10–4 Cm–1 as shown in the
figure. If the negative charge is at a distance of
2 cm from the line charge, then, find the net force
on the dipole – +
2 cm

9. Solution = 5×10–4 Cm–1, r1 = 2×10–2 m, r2 =2.2×10–2 m,
q1 =-1.6×10–9 C, q2 =1.6×10–9 C
Electric field due to a line charge at a distance r from it,
Field at the point of negative charge,
Force on the negative charge

10. Solution Contd. F1 is towards the line charge
Electric field at the point of positive charge is – + F1 F2
Force on the positive charge
F2 is away from the line charge
Net force on the dipole,
F =F1 – F2 = 0.72 – 0.65 =0.07 N, towards the line charge

11. Solved Example - 3 r A long cylindrical wire carries a positive charge
of linear charge density 4×10–7 Cm–1. An
electron revolves round the wire in a circular path
under the influence of the electrostatic force.
Find the KE of the electron. r
Solution
Let the electron revolves in a circular path
of radius r. Electrostatic force on the electron
provides the necessary centripetal force.
Field at a distance r from the wire of charge,

12. Solution Contd. Force on the electron, Required centripetal force
Kinetic energy of the electron,

13. Electric field due to infinite plane sheet of charge
Thin, infinite, nonconducting sheet having uniform surface charge density, as a result of which surface of the sheet

14. Proof. At R and S, Electric flux at plane faces,
Electric flux over the curved surface is zero, as no field lines crosses it
Total electric flux,

15. Charge enclosed by the Gaussian surface,
By Gauss’ law,
E is independent of the distance from the plane sheet

16. Two infinite parallel sheets of charge
Due to uniform surface charge density, surface of the sheet
Region II
Region I

17. Two infinite parallel sheets of charge
Region III

18. Special Case The sheets have equal and opposite charge density If and
, then I II
III
In regions I and III, E = 0
In region II,

19. Solved Example – 4
Two large thin metal plates with surface charge densities
of opposite signs but equal magnitude of ×10–20
Cm–2 are placed parallel and close to each other.
What is the field
To the left of the plates?
(ii) To the right of the plates?
(iii) Between the plates?
Solution:
Electric field exists only in the region between the plates. Therefore,
(i), (ii) E = 0

20. Applying Gauss’ Law – Spherical Symmetry
Electric field due a charged sphere of charge q and radius R
(i) Field at an outside point,
i.e., r > R
Gaussian surface
Note that E is same at all points
on the Gaussian surface
The field is the same as if the whole charge is placed at the centre of the shell

21. (ii) At a point on the surface,
r = R
(iii) At a point inside the spherical shell
r < R
Gaussian surface
No charge is enclosed by the
Gaussian surface.

22. r q + + r + + + + + + + +
Distance of gaussian surface is from the centre. Radius incorrectly shown.
Variation of electric field intensity from the centre of the shell
with distance

23. Solved Example –5
The uniform surface charge density on a spherical copper shell is s. What is the electric field strength on the surface of the shell?
Solution:
The electric field on the surface of a uniformly charged
spherical conductor is given by,

24. Solved Example –6
A spherical charged conductor has a uniform surface charge density s . The electric field on its surface is E. If the radius of the sphere is doubled keeping the surface density of charge unchanged, what will be the electric field on the surface of the new sphere ?
Solution:
The electric field on the surface of a uniformly charged
spherical conductor is given by,
Thus, E is independent of the radius of the sphere. As s is constant, E remains the same.

25. Solved Example –7
The uniform surface density of a spherical conductor is and the electric field on its surface is E1. The uniform surface density of an infinite cylindrical conductor is and the electric field on its surface is E2. Is the expression correct?
Solution:
The electric field on the surface of a charged
spherical conductor is given by,
(i)

26. Solution Contd. The electric field on the surface of a charged
cylinder is given by,
( r = radius and l = length )
(ii)
From equations (i) and (ii) we get,

27. Solved Example – 8
A spherical shell of radius 10 cm has a charge 2×10–6 C distributed uniformly over its surface. Find the electric field
(a) Inside the shell
(b) Just outside the shell
(c) At a point 15 cm away from the centre
Solution:
q = 2 ×10–6 C, R = 0.1 m, r = 0.15 m
(a) Inside the shell, electric field is zero

28. Solution Contd.

29. Electric potential
Electric potential at a point - the work done in bringing unit positive charge from infinity to that point against the electric forces.
SI unit – Volt (V)
One volt - the electric potential at a point if one joule of work is done in bringing unit positive charge from infinity to that point.

30. Solved Example - 9
If a positive charge be moved against the electric field, then what will happen to the energy of the system?
Solution:
If a positive charge be moved against the electric field, then energy will be used from an outside source.

31. Solved Example -10
If 80 J of work is required to transfer 4 C charge from infinity to a point, find the potential at that point
Solution:
W =80 J, q = 4 C, V =?

32. Thank You