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
Sketch the electric field lines for a uniformly charged hollow cylinder shown in figure.
Sketch the electric field lines for a uniformly charged hollow cylinder shown in figure.
Two charged thin infinite plane sheets of uniform surface charge density $\sigma_{+}$ and $\sigma_{-}$ where $\left|\sigma_{+}\right|>\left|\sigma_{-}\right|$ intersect at right angle. Which of the following best represents the electric field lines for this system
Two charged thin infinite plane sheets of uniform surface charge density $\sigma_{+}$ and $\sigma_{-}$ where $\left|\sigma_{+}\right|>\left|\sigma_{-}\right|$ intersect at right angle. Which of the following best represents the electric field lines for this system
- [JEE MAIN 2020]
If the electric field intensity in a fair weather atmosphere is $100 \,V / m$, then the total charge on the earth's surface is ............ $C$ (radius of the earth is $6400\,km$ )
If the electric field intensity in a fair weather atmosphere is $100 \,V / m$, then the total charge on the earth's surface is ............ $C$ (radius of the earth is $6400\,km$ )
How does the electric field lines depend on area ?
How does the electric field lines depend on area ?
The electric flux for Gaussian surface A that enclose the charged particles in free space is (given $q_1$ = $-14\, nC$, $q_2$ = $78.85\, nC$, $q_3$ = $-56 \,nC$)
The electric flux for Gaussian surface A that enclose the charged particles in free space is (given $q_1$ = $-14\, nC$, $q_2$ = $78.85\, nC$, $q_3$ = $-56 \,nC$)