GAUSS' THEOREMMMMMMMMMMMMM[2021-22].pptx

Gauss theorem problem sums with theory

1. GENERAL PROPERTIES OF CONDUCTORS: 1) Any charge added onto a conductor always resides entirely along its outer surface . (This is because like charges would repel one another and thus try and get arranged as far away from one another as possible) 2) Electric field within the enclosed volume of a charged conductor( leaving aside the outer surface) is always zero. (In case there were to be a non-zero field inside, the free electrons within the conductor would move around under the effect of this field and thus get arranged in a manner so as to reduce the field to 0 everywhere inside to ensure that static condition finally prevails.) Good conductors contain free electrons per cubic metre .

2. [Refer to the next slide for an explanation.]

3. +Q

4. -Q

5. Electric flux is a measure of the quantity of lines of force passing through a surface. (directed perp to the surface towards the observer)

6. +2 μ C -4 μ C +3 μ C -1 μ C -5 μ C +2 μ C S (closed surface) Net elec. Flux through S= (1/ ε₀ ). [+2-4-1+3] units. = (1/ ε₀ ).[algebraic sum of the enclosed charges] Note: In case the space enclosed by the surface S is filled with a dielectric (of dielectric constant k), = (1/ ε₀ k). [algebraic sum of the enclosed charges]   Note: The word ‘ dielectric ’ signifies a medium composed of polar molecules.

7. + = elec. Flux passing through area da Closed surface integral => Hence the force on any other charge )  

8. (Solid or hollow) Net flux through S = We construct a spherical Gaussian surface S of radius r(>R), concentric to the spherical conductor. (The charge distribn appears identical from all points on the surface S) for field points lying outside (and on) the surface of the conductor.

9. Gaussian surface S is slightly larger in radius as compared to the outer surface of the charged conductor; difference in the 2 radii is infinitesimally small. r=R +Q Here also, net flux through S = E   (Since r = R)

10. +Q

11. E v/s r graph has a discontinuity at r =   V v/s r graph has no discontinuity

12. As per Gauss’ theorem, net flux through S =  

13. (Since each point on the curved surface of S is identically placed relative to the charge distribution.) = radially outward pointing unit vector   NOTE: If the charge distribn is – ve , =  

15. Note: The independence of E w.r.t. r is on condition that r is much smaller than the dimensions of the charged sheet and field points restricted to locations close to the centre of the sheet. If the charge distribn is – ve , =  

16. The field exists only in between the two parallel sheets when the sheets possess charges of opposite kind having equal surface densities. No field exists in between the parallel sheets if they are charged identically.

17. Q. What would be the net flux passing through a hollow right circular cone having a point charge + q placed at the centre of the circular base?

18. Q. What would be the net flux passing through a hollow right circular cone having a point charge + q placed at the centre of the circular base? Flux through the complete structure = (q/ ϵ ₀) Thus flux through upper cone= q/(2 ϵ ₀) + q

19. . cos0 (since θ =0)

20. 10.

21. (Since potential along the outer surface of a conductor=potential anywhere within its volume)