Gauss Law Class 12 Notes Pdf Full Chapter Download

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Gauss Theorem

According to Gauss’ theorem, the value of the electric flux related to a charge is always equal to the ratio of the charge and the electronegativity of the vacuum.

=> In Gauss theorem, the always closed imaginary is related to the Gaussian surface
=> Gaussian surface sphere can be cylindrical or symmetrical shape

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Origin of Gauss Theorem

Let the charge q be located in vacuum, at a distance r from the charge, at a point p, the intensity of the electric field produced by the unit positive charge will be E and its corresponding flux fi, so according to Gauss’ theorem

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Q-1 What will be the number of electric force lines emerging from a 1 coulomb of charge in vacuum?

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Q-2  Find the number of lines of force emanating from the Gaussian surface in vacuum?

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Q-3 What will be the number of electric force lines associated with an electric dipole?

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Applications of Gaussian Theorem

According to Gauss’ theorem, electric flux is generated on a conducting surface and from a charged object.
With the help of Gauss theorem for different bodies, the intensity of the electric field is determined.

(A) The intensity of the electric field produced by an infinitely linear charged wire

Suppose a charged wire of infinite length is located in vacuum, the charge is distributed uniformly over the length L of this wire, due to which linear charge density (Lemda) is produced on the wire, so the electric field produced by the wire with the help of cylindrical Gaussian surface is The intensity will be E.

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(B) Intensity of electric field produced by infinitely charged dielectric (dielectric) layer 

Let the area A and surface charge density (sigma) of a charged dielectric be located in vacuum, so the electric field intensity produced by the charged layer will be E.

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Note – The charged layer is assumed to be bounded by an imaginary closed cylindrical Gaussian surface.

(C) Electric field intensity produced by a charged conducting sphere/spherical shell/shell

Let a conducting sphere with radius R and charge q be the electric field intensity E at a different position o from the midpoint of this conductor.

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Note – Therefore, the electric field intensity is found to be more at the true point than the outer point through the conducting sphere, but is found to be zero at the inner point.

(D) The electric field intensity produced by the charged conducting layer (plaque)

Let a charged conducting strip with area A and charge q be located in a vacuum, so the electric field intensity produced by the conducting strip is E .

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(E) The intensity of the electric field suddenly produced by the sphere

Let q of a dielectric sphere of radius R distributes charge throughout its volume, which produces volume charge density, so the electric field intensity at different points through the dielectric sphere will be E.

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Force and electric pressure on a charged conducting surface

Suppose a charged conductor on which q charge is uniformly distributed over the surface area, the surface charge density (sigma) is equally present on the conductor, so the force produced on the conductor al n dA and the second pressure will be

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Energy Density (Ud) :-

The potential energy present in a fixed volume is called energy density or the potential energy present in a unit volume of a conductor is called energy density.

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Hence, the energy density of the conductor represents its electric pressure.

The equilibrium of a charged bubble of soap

The surface tension of a charged soap bubble is T, radius R and charge q. On the soap bubble, due to surface pressure and charge due to surface tension, electricity arises due to charge, so the pressure on the bubble will be more.

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Q – What is the energy density Ud in vacuum at a place with electric field intensity ?

 

 

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