Gauss’s Law

IMPORTANT

Gauss’s Law: Overview

This Topic covers sub-topics such as Electric Flux, Gauss Theorem in Electrostatics, Proof of Gauss Law, Area Vector and, Electric Field due to a Long Charged Cylinder Using Gauss's Law

Important Questions on Gauss’s Law

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The SI unit of electrical flux is:

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Which law is used to derive the expression for the electric field between two uniformly charged large parallel sheets with surface charge densities σ and σ respectively:

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Applying Gauss theorem, the expression for the electric field intensity at a point due to an infinitely long, thin, uniformly charged straight wire is

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A hollow charged metal sphere has radius r. If the potential difference between its surface and a point at a distance 3r from the centre is V, then electric field intensity at a distance 3r is:

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An electric field E is present r to a square plate of side length L. What is the flux passing through the plate?

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A charge Q is distributed uniformly in a sphere (solid). Then the electric field at any point r, where r<R ( r is radius of sphere) varies as_____.

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Two infinitely long parallel conducting plates having surface charge densities +σ and -σ respectively are separated by a small distance. The medium between the plates is vacuum, if εo is dielectric permittivity of vacuum, then the electric field in the region between the plates is-

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A cylinder of radius R and length L is placed in a uniform electric field E parallel to the cylinder axis. The total flux for the surface of the cylinder is given by

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E.ds=Qnetεo

E signifies the _____ passing through a certain area dS. Qnet is the total amount of charge and εosignifies the permittivity of the vacuum.

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Derive Gauss's law from Coulomb's law.

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Assertion: Gauss's law can't be used to calculate electric field near an electric dipole.
Reason: Electric dipole don't have symmetrical charge distribution.

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Area vector is a vector quantity associated with each plane figure whose magnitude is

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The black shapes in the figure below are closed surfaces. The electric field lines are in red. For which case, the net flux through the surfaces is non-zero?

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Let the electrostatic field, E at distance, r from a point charge, q not be an inverse square but instead an inverse cubic, e.g. E=kqr3r^, here k is a constant. Consider the following two statements: (I) Flux through a spherical surface enclosing the charge is, ϕ=qenclosedε0. (II) A charge placed inside a uniformly charged shell will experience a force. Which of the above statements are valid?

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A point electric charge Q is placed at a corner of a cube as shown in the figure. What is the electric flux passing through ABCD of the cube? (0 is permittivity of free space)

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A point charge Q, is placed at the center of a cube of side a, in vacuum (permitivity ϵ0. The flux of the electric field through the shaded face is given by

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A sphere of radius R carries a positive charge density (ρ) that increases linearly with radial distance r from the centre (ρr). The radial dependence of the magnitude of electric field inside the sphere is given by

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A positive point charge Q is placed (on the axis of the disc) at a distance of 4R above the center of a disc of radius R as shown in situation I. The magnitude of electric flux through the disc is ϕ. Now, a hemispherical shell of a radius R is placed over the disc such that it forms a closed surface as shown in situation II. The flux through the curved surface in the situation II taking direction of area vector along outward normal as positive is

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In the figure shown, there is a large sheet of charge of uniform surface charge density σ. A charge particle of charge -q and mass m is projected from a point A on the sheet with a speed u with angle of projection such that it lands at maximum distance from A on the sheet. Neglecting gravity, find the time of flight.

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The total flux through the faces of the cube with side of length α if a charge q is placed at corner A of the cube is
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