An electric dipole is placed at the centre of an imaginary sphere. Which of the following option is/are correct?

Solution:

Let an electric dipole consist of two charges, +q and -q, separated by a small distance, placed at the center of an imaginary sphere.

1. Electric field on the sphere: The electric field due to an electric dipole at a point (r, θ) is given by $E = \frac{1}{4\pi\epsilon_0} \frac{p}{r^3} \sqrt{1 + 3\cos^2\theta}$. For any point on the surface of the sphere, r is constant (radius of the sphere). The electric field lines originate from the positive charge and terminate at the negative charge. These field lines will pass through the surface of the sphere. Except for specific points (e.g., infinitely far away or exactly at the center of the dipole, which is not on the surface), the electric field due to a dipole is generally non-zero. Thus, the electric field is not zero anywhere on the sphere. So, statement (1) is correct.

2. Electric potential on the sphere: The electric potential due to an electric dipole at a point (r, θ) is given by $V = \frac{1}{4\pi\epsilon_0} \frac{p \cos\theta}{r^2}$. The potential V depends on the angle $\theta$ (angle between the dipole axis and the position vector to the point). If the dipole is placed at the center of the sphere, there will be points on the sphere where $\cos\theta = 0$ (i.e., points on the equatorial plane of the dipole). At these points, the electric potential will be zero. Therefore, the electric potential is not zero anywhere on the sphere is an incorrect statement. So, statement (2) is incorrect.

3. Net flux through the sphere: According to Gauss's Law, the net electric flux ($\Phi$) through any closed surface is given by $\Phi = \frac{Q_{enclosed}}{\epsilon_0}$, where $Q_{enclosed}$ is the total charge enclosed within the surface. For an electric dipole, the total charge enclosed within the sphere is $Q_{enclosed} = (+q) + (-q) = 0$. Therefore, the net flux through the sphere is zero. So, statement (3) is correct.

Since both statements (1) and (3) are correct, option (4) is the correct answer.