An electric dipole of dipole moment 20*10^-6c.m. Is enclosed by a closed surface .what is the net flux coming out of the surface?

Where is the answer ? Can anyone tell me?..

To find the net flux coming out of a closed surface enclosing an electric dipole, we can use Gauss's Law. Gauss's Law states that the net electric flux through any closed surface is equal to the total charge enclosed divided by the permittivity of free space (ε₀).

In this case, the electric dipole moment is given as 20*10^-6 C.m. The electric dipole moment (p) is the product of the charge (q) and the separation between the charges (d). Therefore, we can express the electric dipole moment as p = q * d.

Let's assume that the charges in the dipole are +q and -q, separated by a distance d. So, p = q * d.

Now, we need to find the total charge enclosed by the surface. Since the dipole consists of equal and opposite charges, the net charge enclosed by the surface is zero. Thus, the numerator becomes zero.

Using Gauss's Law, the formula for the net flux (Φ) becomes:

Φ = q_enclosed / ε₀

Since there is no net charge enclosed, the net flux coming out of the surface is zero.

Therefore, the net flux coming out of the surface in this case is zero.