Electrostatics Gauss Law
According to this theorem, the surface integral ofel3ectrostic filed E produced by any sources over any closed surface S enclosing a volume V in vacuum total electric flux over the closed surface S in vacuum is 1/ ∈0 times the total charge (Q) contained inside S.
φ E = ØE. Ds = Q/ ∈0
The charges inside S may be point charges as well as continuous charge distributions. There is no contribution to troll electric flux from the charges outside S. further the location of Q inside S does not affect the value of surface integral.
The surface chosen to calculate the surface integral is called faussiansur face while selecting such a surface we shall avoid charges on S itself.
Proof of gauss’s theorem (for spherically symmetric surfaces only)
Suppose an isolated positive point charge q is situated at the centre O of a sphere of radius.
According to coulomb’s law, electric field intensity at any point p on the surface of the sphere is
E = q/4π ∈0 r/r2
Where r is unit vector directed form O to P. consider a small area element ds of the sphere around P. let it be represented by the vector ds = n ds where n is unit vector along outdrawn normal to the area element.
∴ E ds = (q/4π ∈0 r/r2) (n ds)
E. ds = q/4π ∈0 ds/r2 r . n
As normal to a sphere at every point is along the radius vector at that point therefore, r . n = 1
E ds = q/4π ∈0 ds/r
Integrating over the closed surface are of the sphere we get
?E.ds = q/4π ∈0 r2 ?ds = q/4π ∈0 r2 x total area of surface of sphere = q/4π ∈0 r2 (4π r3) = q/ ∈0
∅ε = ?E. Ds = q/ ∈0
Which proves gauss’s theorem?
If there are point charges q1. q2, q3 …qn lying inside the furnace, each will contribute to the electric flux independent of the other (superposition principle).
∴ ∅e = ∅E1 + ∅E2 + ∅E3 + … ∅En = q1/∈0 + q2/∈0 + q3/∈0 + …. qn/∈0 = 1/∈0 (q1 + q2 + … + qn)
∴ ∅E = Q/∈0
Where Q = ∑qi is the algebraic sum of all the charges inside the closed surface.
Hence total electric flux over a closed surface in vacuum is 1/∈0 time the total charge within the surface regardless of how the charges may be distributed.
If the medium surrounding the charge has a dielectric constant k then
∅E = Q/k ∈0 = Q/∈r ∈0 = Q/∈
Where k = ∈r = ∈/ ∈0
If there is no net charge within the closed surface when Ω = 0
∴ ∅E = 0
It means that the total electric flux thorough a closed surface is zero if no charge is enclosed by the surface.
It can be shown that charges situated outside the closed surface make no contribution to total electric flux over the surface.
1. Remembers that gauss’s theorem holds goods for any closed surface, regardless of its shape or size.
2. The surface that we choose for the application of gauss’s law is called the Gaussian surface. However, we usually choose a spherical Gaussian surface, because this choice has three simplifying features:
(i) The dot product E ds = E ds cos O = E (ds), because at all points on spherical Gaussian surface, θ = 0
(ii) Magnitude of E is constant at all points on the spherical Gaussian surface. Therefore, E can be brought out of the integral sing.
(iii) The remaining integral is merely the surface area of the sphere (=4π r2) which is obtained without actually doing the integration.
3. Take care to see that the Gaussian surface chosen does not pass through any discrete charge. This is because electric field is not well defined at the location of the charge.
4. In the situation when the surface is so chosen that there are some charges inside and some outside, the electric field E (whose flux is calculated) is due to all the charges, both inside and outside the closed surface .however the term (q) represents only the total charge inside the closed surface.
5. Gauss’s theorem is used most commonly for symmetric charge configurations.
6. Gauss’s theorem is based on inverse square dependence on distance.
7. From gauss’s theorem we can calculate the number of electric lines of force that radiate outwards from one coulomb of positive charge in vacuum. AS
∅ E = q / ∈0 therefore when q – 1 coulomb,
∅ E = 1 8.85 x 10-12 = 1.13 x 1011
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