Question

Total normal electric induction through any imaginary closed surface is equal to

• A. the total quantity of field enclosed in that surface.
• B. the total quantity of flux enclosed in that surface.
• C. the total quantity of charge enclosed in that surface.
• D. All of the above

Answer 1

The correct option is C the total quantity of charge enclosed in that surface.

Total normal electric induction through any closed surface $=\oint$ Normal electric induction $\times$ area vector $\overrightarrow{dS}$

$=\oint {\epsilon }_0\left(E\cos \theta \right)\times dS$

$=\oint {\epsilon }_0(\overrightarrow{E}.\overrightarrow{dS})$ ...(i)

From Gauss Law,

$=\oint \overrightarrow{E}.\overrightarrow{dS}=\frac{q_{enclosed}}{{\epsilon }_0}$ ...(ii)

$q_{enclosed}=$ Total charge enclosed in the surface

Comparing (i) and (ii), we get, total normal electric induction is equal to total charge enclosed in that surface.