Question

To understand the meaning of the variables in Gauss's law, andthe conditions under which the law is applicable.

Gauss's law is usually written

$\Phi_E=\oint \vec{E} \cdot d\vec{A} = \frac{q_{\rm encl}}{\epsilon_0},$

where $\epsilon_0 = 8.85 \times 10^{-12}\, {\rm C^2/(N\cdot m^2)}$ isthe permittivity of vacuum.

How should theintegral in Gauss's law be evaluated?

	around theperimeter of a closed loop
	over the surfacebounded by a closed loop
	over a closedsurface

Answer 1

Concepts and reason

The concept related to solve this problem is the gauss’s law. First explain the gauss’s law. Next, explain how integral in gauss’s law should be evacuated.

Fundamentals

Gauss’s Law: The flux of the net electric field through a closed surface equals the net charge enclosed by the surface divided by${\varepsilon _0}$. Mathematically, the statement can be written as,

$\oint {\vec E \cdot d\vec A} = \frac{{{q_{encl}}}}{{{\varepsilon _0}}}$

Here, $\vec E$ is the electric field, $d\vec A$ is the infinitesimal surface element, ${q_{encl}}$ is the charge, and ${\varepsilon _0}$ is the permittivity of vacuum.

Evaluating the integral only over the perimeter of a closed loop will not be enclosing any field lines through the surface hence the integral in Gauss’s law cannot be evaluated around the perimeter of a closed loop.

Evaluating the integral only over the surface bound by a closed loop will include the field lines going out or inside the surface but will not be enclosing any charges inside the surface hence the integral in Gauss’s law cannot be evaluated around the surface bounded by a closed loop.

The small circle in the integral means closed, $d\vec A$ means the integrating over a closed surface. The statement of gauss’s law was referred to determine if the integral in gauss’s law can be evaluated around a closed surface.

Ans:

The integral in gauss’s law can be evaluated over a closed surface.