Applying Gauss’s law

A charge q is placed at the centre of the open end of cylindrical vessel. The flux of the electric field through the surface of the vessel

According to Gauss Theorem, electric field of an infinitely long straight wire is proportional to

Electric field at a distance r from the axis of a infinitely long charged cylindrical conductor is

At a point lying just outside of a charged conductor, the intensity of electric field is 1/ε₀ times the surface charge density where ε₀ is the permittivity of free space. This statement is called

The intensity of electric field due to a point charge Q at a distance r from it is given by (k = 1/4πϵₒ):

For a given surface the Gauss's law is stated as . From this we can conclude that 

Three infinite long plane sheets carrying uniform charge densities  ,   and  are placed parallel to the x-z plane at y = a, y = 3a and y = 4a as shown in Fig. The electric field at point P is

The inward and outward electric flux for a closed surface in units of    are respectively  and . Then the total charge inside the surface is [where   permittivity constant]

If the electric flux entering and leaving an enclosed surface respectively is  and the electric charge inside the surface will be

The electric flux for Gaussian surface A that enclose the charged particles in free space is (given  )

It is not convenient to use a spherical Gaussian surface to find the electric field due to an electric dipole using Gauss's theorem because  

and are point charges located at points as shown in the figure and S is a spherical Gaussian surface of radius R. Which of the following is true according to the Gauss' law   

Two infinite plane parallel sheets separated by a distance d have equal and opposite uniform charge densities . Electric field at a point between the sheets is

The electric field intensity at a point lying inside a charged spherical shell (conducting) is

Electric flux through a closed surface is 'x' times the net charge enclosed by the surface. Value of 'x' is equal to: