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
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
- 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
Similar Questions
An electric charge $q$ is placed at the centre of a cube of side $\alpha $. The electric flux on one of its faces will be
An electric charge $q$ is placed at the centre of a cube of side $\alpha $. The electric flux on one of its faces will be
- [AIIMS 2001]
A cube of side $l$ is placed in a uniform field $E$, where $E = E\hat i$. The net electric flux through the cube is
A cube of side $l$ is placed in a uniform field $E$, where $E = E\hat i$. The net electric flux through the cube is
The figure shows some of the electric field lines corresponding to an electric field. The figure suggests
The figure shows some of the electric field lines corresponding to an electric field. The figure suggests
A charge $q$ is placed at the centre of the open end of cylindrical vessel. The flux of the electric field through the surface of the vessel is
A charge $q$ is placed at the centre of the open end of cylindrical vessel. The flux of the electric field through the surface of the vessel is
$\mathrm{C}_1$ and $\mathrm{C}_2$ are two hollow concentric cubes enclosing charges $2 Q$ and $3 Q$ respectively as shown in figure. The ratio of electric flux passing through $\mathrm{C}_1$ and $\mathrm{C}_2$ is :
$\mathrm{C}_1$ and $\mathrm{C}_2$ are two hollow concentric cubes enclosing charges $2 Q$ and $3 Q$ respectively as shown in figure. The ratio of electric flux passing through $\mathrm{C}_1$ and $\mathrm{C}_2$ is :
- [JEE MAIN 2024]