Friday, February 8, 2008

IIT JEE Physics Formula Revision 30. Gauss's Law

1. ∆Φ = E ∆s cos θ

∆Φ is the flux of the electric field (E) through the surface ∆s. θ is the angle between positive normal to the surface and electric field.

Using the techniques of integration flux over a surface is:

Φ = ∫ E.∆s
Flux over a closed surface

Φ = ∮ E.∆s (* Intregration over a closed surface)


2. Solid angle

Ω = S/r²

Ω = solid angle (dimensionless figure)

S = the area of the part of sphere intercepted by the cone
r = radius of the sphere assumed on which we are assuming the cone

A complete circle subtends an angle 2 π

Any closed surface subtends a solid angle 4 π at the centre.

How much is the angle subtended by a closed plane curve at an external point? Zero.


3. Gauss's law

The flux of the net electric field through a closed surface equals the net charge enclosed by the surface divided by ε0.

In symbols

E.∆s (* Intregration over a closed surface) =
qin0.

Where
qin = charge enclosed by the closed surface
ε0 = emittivity of the free space

4. Electric field due to a uniformly charged sphere
A total charge Q is uniformly distributed in a spherical volume of radius R. what is the electric field at a distance r from the centre of the charge distribution outside the sphere?
E = Q/4π ε0


5. Field at an internal point

At centre E = O

At any other point inside (rless than R) radius of the sphere

E = Qr/4π ε0

6. c. Electric field due to a linear charge distribution

The linear charge density (charge per unit length) is λ.

Electric field at a distance r from the linear charge distribution

E = λ/2π ε0r

7. d. Electric field due to a plane sheet of charge

Plane sheet with charge density (charge per unit area) σ.

Field at distance d from the sheet = E = σ/2ε0

We see that the field is uniform and does not depend on the distance from the charge sheet. This is true as long as the sheet is large as compared to its distance from P.

8. Electric field due to a charged conducting surface

To find the field at a point near this surface but outside the surface having charge density σ.

E = σ/ ε0



Updated 12 October 2008


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1 comment:

amiya said...

Good idea, but please be elaborative on the formulae and also try to give problem examples and their answers.