Gauss’s law states that
Gauss’s law states that
- [AIIMS 2017]
- A
the total electric flux through a closed surface is $\frac{1}{\varepsilon_0}$ times the total charge placed near the closed surface.
- B
the total electric flux through a closed surface is $\frac{1}{\varepsilon_0}$ times the total charge enclosed by the closed surface.
- C
the total electric flux through an open surface is $\frac{1}{\varepsilon_0}$ times the total charge placed near the open surface.
- D
the line integral of electric field around the boundary of an open surface is $\frac{1}{\varepsilon_0}$ times the total charge placed near the open surface.
Similar Questions
Explain the electric field lines and the magnitude of electric field.
Explain the electric field lines and the magnitude of electric field.
An infinitely long uniform line charge distribution of charge per unit length $\lambda$ lies parallel to the $y$-axis in the $y-z$ plane at $z=\frac{\sqrt{3}}{2} a$ (see figure). If the magnitude of the flux of the electric field through the rectangular surface $A B C D$ lying in the $x-y$ plane with its center at the origin is $\frac{\lambda L }{ n \varepsilon_0}\left(\varepsilon_0=\right.$ permittivity of free space $)$, then the value of $n$ is
An infinitely long uniform line charge distribution of charge per unit length $\lambda$ lies parallel to the $y$-axis in the $y-z$ plane at $z=\frac{\sqrt{3}}{2} a$ (see figure). If the magnitude of the flux of the electric field through the rectangular surface $A B C D$ lying in the $x-y$ plane with its center at the origin is $\frac{\lambda L }{ n \varepsilon_0}\left(\varepsilon_0=\right.$ permittivity of free space $)$, then the value of $n$ is
- [IIT 2015]
A metallic solid sphere is placed in a uniform electric field. The lines of force follow the path(s) shown in figure as
A metallic solid sphere is placed in a uniform electric field. The lines of force follow the path(s) shown in figure as
- [IIT 1996]
A cubical volume is bounded by the surfaces $x =0, x = a , y =0, y = a , z =0, z = a$. The electric field in the region is given by $\overrightarrow{ E }= E _0 \times \hat{ i }$. Where $E _0=4 \times 10^4 NC ^{-1} m ^{-1}$. If $a =2 cm$, the charge contained in the cubical volume is $Q \times 10^{-14} C$. The value of $Q$ is $...........$
Take $\left.\varepsilon_0=9 \times 10^{-12} C ^2 / Nm ^2\right)$
A cubical volume is bounded by the surfaces $x =0, x = a , y =0, y = a , z =0, z = a$. The electric field in the region is given by $\overrightarrow{ E }= E _0 \times \hat{ i }$. Where $E _0=4 \times 10^4 NC ^{-1} m ^{-1}$. If $a =2 cm$, the charge contained in the cubical volume is $Q \times 10^{-14} C$. The value of $Q$ is $...........$
Take $\left.\varepsilon_0=9 \times 10^{-12} C ^2 / Nm ^2\right)$
- [JEE MAIN 2023]
If $\oint_s \vec{E} \cdot \overrightarrow{d S}=0$ over a surface, then:
If $\oint_s \vec{E} \cdot \overrightarrow{d S}=0$ over a surface, then:
- [NEET 2023]