Sunday, January 29, 2017

PROOF OF GAUSS'S THEOREM

Let a charge of q coulombs be placed inside a hollow enclosure as shown in figure:


Let a small surface area dA of this enclosure subtend a solid angle d๐œ” at q and be at a distance d from the charge q.

The flux density D' (considered in the direction of the radius vector d) will be:

=q4ฯ€d2 Coulombs/ sq.meter  or  C/m2

and its component D, normal to the surface will be

=q4ฯ€d2cos ๐œƒ  C/m2

where ๐œƒ is the angle between D and D'.

Now, the flux d๐œ“ crossing normally the surface of area dA = D โคฌ area



or        dฯˆ=q4ฯ€d2cos ฮธโ‹…dA    coulombs


By definition,

solid angle=surfacarearadius

โˆด    dฯ‰=dA  cos ฮธd2


โˆด    dฯˆ=q4ฯ€โ‹…dฯ‰


or the total flux ๐œ“ , crossing the total surface of the imagined enclosure is

โˆซdฯˆ=q4ฯ€ โˆซdฯ‰

ฯˆ=q4ฯ€ร—4ฯ€


or      ฯˆ=coulombs


Thus the total electric flux due to a charge at any point in an enclosure is equal to the charge enclosed in that enclosure.

If there are a number of charges +๐‘žโ‚ , +๐‘žโ‚‚ , -๐‘žโ‚ƒ  and  +๐‘žโ‚„ placed in an enclosure, the total electric flux due to all the charges will be

๐œ“ = ๐‘žโ‚ + ๐‘žโ‚‚ - ๐‘žโ‚ƒ + ๐‘žโ‚„

or     ๐œ“ = ๐›ด๐‘ž