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

Options

A. Gauss’s law fails in this case
B. This problem does not have spherical symmetry
C. Coulomb’s law is more fundamental than Gauss’s law
D. Spherical Gaussian surface will alter the dipole moment

Answer

To answer the question of why it is not convenient to use a spherical Gaussian surface to find the electric field due to an electric dipole using Gauss’s theorem, we can break down the reasoning step by step.

### Step-by-Step Solution:

1. **Understanding Gauss's Law**: Gauss's law states that the electric flux through a closed surface is proportional to the charge enclosed by that surface. Mathematically, it is expressed as:
\[ \Phi_E = \oint \mathbf{E} \cdot d\mathbf{A} = \frac{Q_{\text{enc}}}{\epsilon_0} \]
where \( \Phi_E \) is the electric flux, \( \mathbf{E} \) is the electric field, \( d\mathbf{A} \) is the differential area vector, \( Q_{\text{enc}} \) is the enclosed charge, and \( \epsilon_0 \) is the permittivity of free space.

2. **Electric Dipole Characteristics**: An electric dipole consists of two equal and opposite charges separated by a small distance. The electric field due to a dipole varies with distance and direction, and it does not exhibit spherical symmetry.

3. **Spherical Symmetry Requirement**: For Gauss's law to be conveniently applied, the Gaussian surface must exhibit symmetry that allows for a uniform electric field over the surface. This is typically the case for point charges or uniformly charged spheres, where the electric field is the same at all points on the surface.

4. **Electric Field of a Dipole**: The electric field lines of a dipole are not uniform and do not radiate outward symmetrically as they do for a point charge. Instead, they have a more complex distribution that varies with the angle and distance from the dipole.

5. **Inapplicability of Spherical Gaussian Surface**: When a spherical Gaussian surface is placed around a dipole, the electric field is not constant across the surface. The angle between the electric field vector and the area vector \( d\mathbf{A} \) varies at different points on the surface, leading to complications in calculating the electric flux. Therefore, the integral \( \oint \mathbf{E} \cdot d\mathbf{A} \) becomes difficult to evaluate.

6. **Conclusion**: Since the electric field of a dipole does not have spherical symmetry, it is not convenient to use a spherical Gaussian surface to find the electric field due to an electric dipole using Gauss’s theorem. Instead, other methods, such as direct application of Coulomb's law or the dipole field equations, are more suitable.