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
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 $6400\,km$ )
Gauss’s law states that
Gauss’s law states that
- [AIIMS 2017]
The electric field components in Figure are $E_{x}=\alpha x^{1 / 2}, E_{y}=E_{z}=0,$ in which $\alpha=800 \;N / C\, m ^{1 / 2} .$ Calculate
$(a)$ the flux through the cube, and
$(b)$ the charge within the cube. Assume that $a=0.1 \;m$
The electric field components in Figure are $E_{x}=\alpha x^{1 / 2}, E_{y}=E_{z}=0,$ in which $\alpha=800 \;N / C\, m ^{1 / 2} .$ Calculate
$(a)$ the flux through the cube, and
$(b)$ the charge within the cube. Assume that $a=0.1 \;m$
A point charge of $+\,12 \,\mu C$ is at a distance $6 \,cm$ vertically above the centre of a square of side $12\, cm$ as shown in figure. The magnitude of the electric flux through the square will be ....... $\times 10^{3} \,Nm ^{2} / C$
A point charge of $+\,12 \,\mu C$ is at a distance $6 \,cm$ vertically above the centre of a square of side $12\, cm$ as shown in figure. The magnitude of the electric flux through the square will be ....... $\times 10^{3} \,Nm ^{2} / C$
- [JEE MAIN 2021]
Expression for an electric field is given by $\vec{E}=4000 x^2 \hat{i} \frac{V}{m}$. The electric flux through the cube of side $20\,cm$ when placed in electric field (as shown in the figure) is $.........V cm$.
Expression for an electric field is given by $\vec{E}=4000 x^2 \hat{i} \frac{V}{m}$. The electric flux through the cube of side $20\,cm$ when placed in electric field (as shown in the figure) is $.........V cm$.
- [JEE MAIN 2023]