Question

In finding the electric field using Gauss Law the formula $|E|=\frac{q_{enc}}{{\epsilon }_0|A|}$ 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. The equation can be used in which of the following situation?

Answer 1

According to Gauss' Law,

$\oint \overrightarrow{E}.\overrightarrow{dA}=\frac{q_{enc}}{{\in }_0}$

Where $\overrightarrow{E}$ is the electric field and

$\overrightarrow{dA}$ is the elemental area vector for any small elemental area on the gaussian surfae.

$\oint |\overrightarrow{E}||\overrightarrow{dA}|\cos \theta =\frac{q_{enc}}{{\in }_0}...\left(1\right)$

Given that the formula

$|\overrightarrow{E}|=\frac{q_{enc}}{{\in }_0|A|}$

is applicable.

$|\overrightarrow{E}||A|=\frac{q_{enc}}{{\in }_0}$

(Using equation 1)
$|\overrightarrow{E}||A|=\oint |\overrightarrow{E}||\overrightarrow{dA}|\cos \theta ...\left(2\right)$

For constant E $\oint |\overrightarrow{E}||\overrightarrow{dA}|\cos \theta$

becomes $|\overrightarrow{E}|\oint |\overrightarrow{dA}|\cos \theta =|\overrightarrow{E}||A|\cos \theta$ Hence, the given relation will be valid when $\overrightarrow{E}$ is a constant throughout the gaussian surface and angle $\left(\theta \right)$ between the area vector and the Electric field vector is same at all points and equal to 0°

For constant E$\oint |\overrightarrow{E}||\overrightarrow{dA}|\cos \theta$

becomes $|\overrightarrow{E}|\oint |\overrightarrow{dA}|=|\overrightarrow{E}||A|.$
The condition $\theta =0°$ means that the components of the electric field parallel to the surface is zero.

Since, $E=\frac{dv}{dx},$ Where V is the Electric potential.
The potential gradient over the surface is zero. Hence the surface is equipotential.

Hence, option (d) is correct.