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
The flat base of a hemisphere of radius $a$ with no charge inside it lies in a horizontal plane. A uniform electric field $\vec E$ is applied at an angle $\frac {\pi }{4}$ with the vertical direction. The electric flux through the curved surface of the hemisphere is
The flat base of a hemisphere of radius $a$ with no charge inside it lies in a horizontal plane. A uniform electric field $\vec E$ is applied at an angle $\frac {\pi }{4}$ with the vertical direction. The electric flux through the curved surface of the hemisphere is
- [AIEEE 2012]
A charged particle $q$ is placed at the centre $O$ of cube of length $L$ $(A\,B\,C\,D\,E\,F\,G\,H)$. Another same charge $q$ is placed at a distance $L$ from $O$.Then the electric flux through $BGFC$ is
A charged particle $q$ is placed at the centre $O$ of cube of length $L$ $(A\,B\,C\,D\,E\,F\,G\,H)$. Another same charge $q$ is placed at a distance $L$ from $O$.Then the electric flux through $BGFC$ is
- [AIEEE 2002]
What is called Gaussian surface ?
What is called Gaussian surface ?
$Assertion\,(A):$ A charge $q$ is placed on a height $h / 4$ above the centre of a square of side b. The flux associated with the square is independent of side length.
$Reason\,(R):$ Gauss's law is independent of size of the Gaussian surface.
$Assertion\,(A):$ A charge $q$ is placed on a height $h / 4$ above the centre of a square of side b. The flux associated with the square is independent of side length.
$Reason\,(R):$ Gauss's law is independent of size of the Gaussian surface.
- [AIIMS 2015]
For a given surface the Gauss's law is stated as $\oint {E \cdot ds} = 0$. From this we can conclude that
For a given surface the Gauss's law is stated as $\oint {E \cdot ds} = 0$. From this we can conclude that