Question

If finding the electric field around a surface is given by $\left|\overrightarrow{E}\right|=\frac{q_{enc}}{{\epsilon }_0\left|A\right|}$ is applicable. In the formula ${\epsilon }_0$ is permittivity of free space, $A$ is area of Gaussian and $q_{enc}$ is charge enclosed by the Gaussian surface. This equation can be used in which of the following situation?

• A. For any Gaussian surface.
• B. Only when Gaussian surface is an equipotential surface.
• C. Only when electric field is constant on the surface.
• D. Equipotential surface and electric field is constant on the surface.

Answer 1

The correct option is D

Equipotential surface and electric field is constant on the surface.

Step 1: Given data

Given that the formula $\left|\overrightarrow{E}\right|=\frac{q_{enc}}{{\epsilon }_0\left|A\right|}$ is applicable

$\therefore \left|\overrightarrow{E}\right|\left|A\right|=\frac{q_{enc}}{{\epsilon }_0}\to \left(1\right)$

Step 2: Formula used

According to Gauss's law,

$$\begin{gathered} \oint \overrightarrow{E_1}·\overrightarrow{dA}=\frac{q_{enc}}{{\epsilon }_0} \\ \Rightarrow \oint \left|\overrightarrow{E_1}\right|\left|\overrightarrow{dA}\right|\cos \theta =\frac{q_{enc}}{{\epsilon }_0}\to \left(2\right) \end{gathered}$$

where $\overrightarrow{E}$ is electric field, $\overrightarrow{dA}$ is elemental area, $q_{enc}$ is charge enclosed by the Gaussian surface and ${\epsilon }_0$ is permittivity of free space.

Step 3: Comparing the equations

From $\left(1\right)$ and $\left(2\right)$, we get

$\left|\overrightarrow{E}\right|\left|A\right|=\oint \left|{\overrightarrow{E}}_1\right|\left|\overrightarrow{dA}\right|\cos \theta$

This equation is valid when $\overrightarrow{E_1}$ is constant and $\left|\overrightarrow{E_1}\right|=\left|\overrightarrow{E}\right|$ and also when $\theta =0°$,i.e., when $\cos \theta =1$

Then the equation becomes,

$\left|\overrightarrow{E}\right|\left|A\right|=\left|\overrightarrow{E}\right|\oint \left|\overrightarrow{dA}\right|$

$\theta =0°\Rightarrow$components of the electric field parallel to the elemental surface , $E_1\sin \theta =0$

The potential gradient over the surface is zero, therefore, the surface is equipotential.

Therefore, the surface is equipotential and the electric field is constant on the surface.

Hence, option D is correct