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
$\mathrm{C}_1$ 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]
When electric flux is said to be positive, negative or zero ?
When electric flux is said to be positive, negative or zero ?
If a charge $q$ is placed at the centre of a closed hemispherical non-conducting surface, the total flux passing through the flat surface would be
If a charge $q$ is placed at the centre of a closed hemispherical non-conducting surface, the total flux passing through the flat surface would be
- [JEE MAIN 2022]
Each of two large conducting parallel plates has one sided surface area $A$. If one of the plates is given a charge $Q$ whereas the other is neutral, then the electric field at a point in between the plates is given by
Each of two large conducting parallel plates has one sided surface area $A$. If one of the plates is given a charge $Q$ whereas the other is neutral, then the electric field at a point in between the plates is given by
Draw electric field by negative charge.
Draw electric field by negative charge.