Question

What is the nature of the Gaussian surface involved in Gauss law of electrostatic?

(A) Scalar
(B) Electrical
(C) Magnetic
(D) Vector

Answer 1

Hint:
- We should know about the difference of scalar and vector.
- We should know what the Gaussian surface is .
- Gauss’s law of electrostatic tells that the total amount of electric flux out of a closed surface is equal to the enclosed charge divided by the permittivity.
- Scalar quantity is one dimensional.
- Every vector quantity obeys only vector algebraic equations. But scalar quantity does not flow vector algebraic equations. It obeys a simple mathematical algebraic equation.

Complete step by step solution:
We know according to the Gauss’s law, the surface integral of electric field($\vec E$) over closed surface ($\vec S$) enclosing a volume ($V$) is the total amount of electric flux that is $\dfrac{1}{{\mathop \varepsilon \nolimits_0 }}$ times of the total charge($\mathop q\nolimits_{in} $) enclosed by the surface.
Mathematically, we can express the Gauss’s law as, $\oint {\vec E \cdot d\vec s = \dfrac{{\mathop q\nolimits_{in} }}{{\mathop \varepsilon \nolimits_0 }}} $
A Gaussian surface is an imaginary closed surface around a charge in a $3$ dimensional space. The total amount of flux is calculated by using this surface.
Now, the Gaussian surface is an imaginary surface. So it can never be electrical and magnetic in nature. The Gaussian surface has a direction outwards the enclosed path. It obeys vector algebra rules. It is in 3D space. So clearly it is a vector in nature.

Hence the correct answer is option (D) Vector.

Note:
- The Gaussian surface is not electrical or magnetic in nature. But the flux of the electric field and magnetic field is calculated through it.
- This is an imaginary enclosed surface and its direction is always outward the surface.
- As it is an imaginary surface no charge can lie on this surface.
If the electric field is normal to the Gaussian surface, then the total amount of flux will be zero.