Question

Total normal electric field =

• A. ${ϵ}_0\overrightarrow{E}.\overrightarrow{dS}$
• B. $\overrightarrow{E}.\overrightarrow{dS}$
• C. ${ϵ}_0\overrightarrow{E}$
• D. ${ϵ}_0\overrightarrow{dS}$

Answer 1

The correct option is A ${ϵ}_0\overrightarrow{E}.\overrightarrow{dS}$

Total normal electric field $=$ Normal electric field $\left(\overrightarrow{E}\right)\times$ area vector $\overrightarrow{dS}$

Normal electric field $={\epsilon }_{0}E\cos \theta$

So, total normal electric field $={\epsilon }_{0}E\cos \theta \times \overrightarrow{dS}$

$={\epsilon }_0\overrightarrow{E}.\overrightarrow{dS}$

where $E\cos 0$ is the component of electric field intensity parallel to area vector $\overrightarrow{dS}$.