Question

What is the electric flux through a cube of side 1 cm which encloses an electric dipole?

Answer 1

Hint: These are conceptual questions and easily answered if you remember the basic concepts of electric flux and electric dipoles such as the concept of Gauss law and the formula of Gauss law.i.e. \[{\phi _E} = \oint {E.ds} = \dfrac{Q}{{{\varepsilon _0}}}\].

Complete Step-by-Step solution:
Since in this question, we are going through the terms electric dipole so we should know what the electric dipole is; The electric dipole is the pair of q and q scales, separated by 2a distance, which has equal and opposite charges. Direction from -q to q is dipole direction.
According to the Gauss law of electrostatics, electric flux through any closed surface is given by \[{\phi _E} = \oint {E.ds} = \dfrac{Q}{{{\varepsilon _0}}}\] (this is equation 1) here E is electrostatic field, Q = total charge enclosed by the surface and ${\varepsilon _0}$is the absolute electric permittivity of free space
In the given case a cube encloses an electric dipole
Therefore the net charge enclosed within the cube is 0 .i.e. Q = 0.
Substituting the value of Q in equation 1
$ \Rightarrow $\[{\phi _E} = \dfrac{0}{{{\varepsilon _0}}}\] = 0
Hence, the electric flux through the cube is 0.

Note: In the above question Gauss law made the above question much clear so let’s know what Gauss law is about; Gauss law states that the total electric flux passing through a closed surface is equal to the charge enclosed divided by the permittivity, In this law the electric flux in an area is defined as the electric field multiplied by the area of the surface projected in a plane and perpendicular to the field.