Electric Flux & Gauss's Law - Free MCQ Practice Test with solutions, JEE

Electric flux = electric field intensity x area. The dimension of field intensity is [M L T^-3 I^-1] and the dimension of the area is [L²]. Therefore, the dimension of flux = [M L³ T^-3 I^-1]. This can also be justified as flux=potential x length. By putting the dimensions of potential and length, we can get the same result.

We can assume 8 imaginary cubes, all of side a placed together, sharing a common corner point and forming a bigger cube of side 2a. The point charge is placed at the common point of the cubes. According to Gauss’s Law, net electric flux coming out of the bigger cube =  Therefore, flux coming out from each of the smaller cubes i.e., cubes of side a is .

In the case of a non-conducting sphere, an electric charge is uniformly distributed throughout the sphere. Inside the sphere, the electric field at a point will be only due to the electric charge inside that point. There will be no effect of the remaining charge outside that point. In the case of the conducting sphere, the electric field at every point inside the sphere is zero.

According to Gauss’s Law, the total number of electric field lines coming out of a charge q is =where εo is the absolute permittivity of air. Its value is 8.85 * 10^-12. Therefore the number of lines coming out from a 1C charge = 1/8.85 * 10^-12.

Gauss law relates the electric flux density and the charge density. Thus it can be used to compute the radius of the Gaussian surface. Permittivity and permeability are constants for a particular material.

According to Gauss’s Law, the electric field outside a closed surface will be only because of the total charge enclosed inside the surface. In this case, there are two positive and negative charges of the same magnitude inside the surface. Therefore the net charge inside the surface is zero. Therefore, the electric field outside the surface will also be zero. There will be no effect of the second charge q₂

We know that the electric field and area both are vector quantity and the electric flux is expressed as E.s (taking dot product). But if the angle between the two vectors is θ then the formula becomes E^-.s^-.cosθ. Cosθ will be maximum if theta is zero, in all other cases, the value of Cosθ is less than 1. Therefore flux will be the maximum if the angle is 0 degrees.

According to Gauss’s principle, if there is no charge bound inside a surface, net electric flux coming out of the surface will always be 0. In this case, the charge is kept outside the surface, so it will generate no field lines and hence no flux inside the surface. The situation will be the same if an electric dipole is placed inside a surface as the dipole has equal positive and negative charges; the total charge inside the surface becomes 0.

The Gauss law exists for all materials. Depending on the Gaussian surface of the material, we take the coordinate system accordingly. Suppose if the material is a coaxial cable, the Gaussian surface is in the form of a cylinder. Thus we take the cylinder/circular coordinate system.

Since 1m does not enclose any cylinder (three Gaussian surfaces of radius 2m, 4m, 5m exists), the charge density and charge becomes zero according to Gauss law. Thus flux density is also zero.