Question

In finding the electric field using Gauss law the formula $\left|\overrightarrow{E}\right|= \frac{q_{enc}}{\epsilon_{0}\left|A\right|}$ is applicable. In the formula $\epsilon_{0}$ is permittivity of free space, $A$ is the area of Gaussian surface and $ q_{enc}$ is charge enclosed by the Gaussian surface. This equation can be used in which of the following situation ?

• A. For any choice of Gaussian surface.
• B. Only when the Gaussian surface is an equipotential surface.
• C. Only when the Gaussian surface is an equipotential surface and $\left|\overrightarrow{E}\right|$ is constant on the surface.
• D. Only when $\left|\overrightarrow{E}\right| = $ constant on the surface.

Answer 1

The Correct Option is C

To solve this problem, we need to understand the conditions under which Gauss's law can be applied using the formula:

$\left|\overrightarrow{E}\right|= \frac{q_{enc}}{\epsilon_{0}\left|A\right|}$

Gauss's Law Description:

Gauss's law relates the electric flux passing through a closed surface (Gaussian surface) to the charge enclosed by that surface. Mathematically, it can be stated as:

$\Phi_{E} = \oint \overrightarrow{E} \cdot d\overrightarrow{A} = \frac{q_{enc}}{\epsilon_{0}}$

where $\Phi_{E}$ is the electric flux, $d\overrightarrow{A}$ is an infinitesimal area vector on the surface, and $q_{enc}$ is the charge enclosed.

Conditions for Using the Given Formula:
The formula $\left|\overrightarrow{E}\right|= \frac{q_{enc}}{\epsilon_{0}\left|A\right|}$ assumes that:

• The electric field $|\overrightarrow{E}|$ is constant in magnitude over the Gaussian surface.
• The Gaussian surface is an equipotential surface such that the angle between $d\overrightarrow{A}$ and $|\overrightarrow{E}|$ is zero everywhere, making the dot product constant.

Explanation of Options:

• Option 1: For any choice of Gaussian surface.
  This option is incorrect because the formula cannot be generally applied to any Gaussian surface unless $|\overrightarrow{E}|$ is constant on it.
• Option 2: Only when the Gaussian surface is an equipotential surface.
  This option is insufficient because, apart from being an equipotential surface, $|\overrightarrow{E}|$ must also be constant.
• Option 3: Only when the Gaussian surface is an equipotential surface and $|\overrightarrow{E}|$ is constant on the surface.
  This option is correct as it satisfies both necessary conditions.
• Option 4: Only when $|\overrightarrow{E}| = $ constant on the surface.
  This is incorrect as being constant alone does not ensure the surface is equipotential.

Conclusion:
The correct answer is: Only when the Gaussian surface is an equipotential surface and $|\overrightarrow{E}|$ is constant on the surface.