what is gauss law. Derive gauss law from coulombs law
Answers
Thus Gauss' law states that electric flux through any closed surface is equal to the net charge enclosed inside the surface divided by permittivity of vacuum.
The flux passing through the area element dS ,that is,
d φ =E.dS= EdS cos 00=EdS
Hence, the total flux through the entire Gaussian sphere is obtained as,
Φ=∫EdS
Or φ=E∫dS
But ∫dS is the total surface area of the sphere and is equal to 4πr2,that is,
Φ=E(4πr2) (1)
But according to Gauss’s law for electrostatics
Φ=q/ε0 (2)
Where q is the charge enclosed within the closed surface
By comparing equation (1) and (2) ,we get
E(4πr2)=q/ε0
Or E=q/4πε0r2 (3)
The equation (3) is the expression for the magnitude of the intensity of electric field E at a point,distant r from the point charge +q.
In vector form, E=1/4πε0 q/r2 =1/4πε0qr/r3
In a second point charge q0be placed at the point at which the magnitude of E is computed ,then the magnitude of the force acting on the second charge q0would be
F=q0E
By substituting value of E from equation (3),we get
F=qoq/4πε0r2 (4)
The equation (4) represents the Coulomb’s Law and it is derived from gauss law.
HERE IS UR ANSWERS :)-
➫Gauss's Theorem:-
According to Gauss's theorem total number of electric lines of force passing normally through a closed surface of ray shape in an electric field(i.e., total electric flux) is equal to 1/ۥ times the total charge present within that surface.
➫Derivations of Gaussic Law:-
Let +q charge is placed at a point O and a point P lies at distance r from the point O. Imagine a sphere of radius r and centre O. Thus, point P lies on the surface of the sphere. Now, the surface of the sphere will be have as Gaussian surface. Therefore, the intensity of electric field on the surface at all the points will be equal in magnitude and will be directed radially outward.
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