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 d𝜔 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:

=q4πd2 Coulombs/ sq.meter  or  C/m2

and its component D, normal to the surface will be

=q4πd2cos 𝜃  C/m2

where 𝜃 is the angle between D and D'.

Now, the flux d𝜓 crossing normally the surface of area dA = D ⤬ area



or        dψ=q4πd2cos θ⋅dA    coulombs

By definition,

solid angle=surface arearadius

∴    dω=dA  cos θd2


∴    dψ=q4π⋅dω


or the total flux 𝜓 , crossing the total surface of the imagined enclosure is

∫dψ=q4π ∫dω

ψ=q4π×4π


or      ψ=q 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 +𝑞₁ , +𝑞₂ , -𝑞₃  and  +𝑞₄ placed in an enclosure, the total electric flux due to all the charges will be

𝜓 = 𝑞₁ + 𝑞₂ - 𝑞₃ + 𝑞₄

or     𝜓 = 𝛴𝑞

PROPERTIES OF LINES OF FORCE

Electrostatic lines of force possess the following properties:

(1) They originate from a positive charge and terminate on a negative charge.

(2) They are always normal to the surface of the body at the point from where they originate or terminate.

(3) A unit positive charge when placed in close neighbourhood of a positively charged body, will follow the path of a line of force and will move to the negatively charged body.

(4) A line of force is such that a tangent to it at any point indicates a direction of the electric intensity at that point.

RELATION BETWEEN POTENTIAL GRADIENT AND ELECTRIC INTENSITY

Potential gradient g is defined as:

g=dVdx

                                ..........................................(1)

Electric field E is defined as:

E=−dVdx

                                .........................................(2)    

From equation (1) and (2)

E=−g

This shows that the potential gradient g at a point in the electric field is negative in magnitude of E at that point.

From the point of view of magnitude, both E and g are equal and this can be proved as under:

Since                                    

V=Wg

∴  1 volt=1 joule1 coulomb


But      1 Joule=1 newton×1 meter


∴  1 volt=1 newton×1 meter1 coulomb


or          1 volt1 meter=1 newton1 coulomb


∴  1 volt / meter=1 newton / coulomb


thus     unit of potential gradient=unit of electric intensity