Gauss law is applicable to:
A) Any surface
B) Asymmetrical surfaces only
C) Symmetrical surfaces only
D) Imaginary surfaces only
Answer
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Hint: To answer this question, we should know what Gauss’s Law is. The definition itself will help us to get the answer. Gauss’ law is applicable to any closed surface with some charge distribution. The Gaussian surface is the surface in three dimensional space through which the flux line of a vector field passes.
Complete answer:
Let’s start with the definition of gauss’s law. Gauss’s law states that, The total electric flux coming out of a closed surface is equal to the charge enclosed divided by the permittivity of free space. Also, the electric flux in an area is the electric field multiplied by the area of the surface that is projected in a plane and is perpendicular to the field.
Mathematically,
$\phi = \dfrac{q}{{{\varepsilon _0}}}$ and
Where $\phi $ is the flux from the surface, ${\varepsilon _0}$ is the permittivity of the free space, $q$ is the charge enclosed in the given area, $E$ is the electric field and $ds$ is the perpendicular area.
So we can say that,
Here ${\varepsilon _0}$ is the permittivity of the free space, $q$ is the charge enclosed in the given area, $E$ is the electric field and $ds$ is the perpendicular area. So from the above derivation we can say that the gauss’s law is for any area which is enclosed. It is applicable to a symmetric surface as well as to an asymmetric surface. Also, we can apply Gauss's law for an imaginary surface which has charge so as to know the amount of flux coming out of it.
Hence option (A) is the correct option.
Note: The gauss’s law is a modification of coulomb’s law. From the gauss law, we can conclude that only the inside the surface area contributes to the flux. Charge outside the surface has no role in providing flux. Here, the integral of the electric field is over a closed loop and not an open loop. This means that all surfaces on which the Gauss’s law is applicable should be closed. Best example is a sphere can be a surface where the Gauss’s law is applicable, however a hemisphere with an open base cannot. Although we can always assume an imaginary closed Gaussian surface enveloping the hemisphere.
Complete answer:
Let’s start with the definition of gauss’s law. Gauss’s law states that, The total electric flux coming out of a closed surface is equal to the charge enclosed divided by the permittivity of free space. Also, the electric flux in an area is the electric field multiplied by the area of the surface that is projected in a plane and is perpendicular to the field.
Mathematically,
$\phi = \dfrac{q}{{{\varepsilon _0}}}$ and
Where $\phi $ is the flux from the surface, ${\varepsilon _0}$ is the permittivity of the free space, $q$ is the charge enclosed in the given area, $E$ is the electric field and $ds$ is the perpendicular area.
So we can say that,
Here ${\varepsilon _0}$ is the permittivity of the free space, $q$ is the charge enclosed in the given area, $E$ is the electric field and $ds$ is the perpendicular area. So from the above derivation we can say that the gauss’s law is for any area which is enclosed. It is applicable to a symmetric surface as well as to an asymmetric surface. Also, we can apply Gauss's law for an imaginary surface which has charge so as to know the amount of flux coming out of it.
Hence option (A) is the correct option.
Note: The gauss’s law is a modification of coulomb’s law. From the gauss law, we can conclude that only the inside the surface area contributes to the flux. Charge outside the surface has no role in providing flux. Here, the integral of the electric field is over a closed loop and not an open loop. This means that all surfaces on which the Gauss’s law is applicable should be closed. Best example is a sphere can be a surface where the Gauss’s law is applicable, however a hemisphere with an open base cannot. Although we can always assume an imaginary closed Gaussian surface enveloping the hemisphere.
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