Question

Consider Gauss's law $\oint \overrightarrow{E}.\overrightarrow{ds}=\frac{q}{{\in }_0}$

Then, for the situation shown in figure at the Gaussian surface

• A. $\overrightarrow{E}$ due to $q_2$ would be zero
• B. $\overrightarrow{E}$ due to both $q_1$ and $q_2$ would be non-zero
• C. $\varphi$ due to both $q_1$ and $q_2$ would be non-zero
• D. $\varphi$ due to $q_2$ would be zero

Answer 1

Answer: (B), (D)

$\varphi$ due to $q_2$ would be zero

Let us first draw the electric field lines coming out of charges $q_1$ and $q_2$

As the electric field lines coming out of charge $q_2$ are entering the gaussian surface so the electric field due to $q_2$ would be non-zero.

In case of charge $q_1$ also the electric field due charge $q_1$ are crossing the gaussian surface so the electric field due to $q_2$ would also be non-zero.

From the above figure we can say that due to charge $q_2$
The number of field lines entering the gaussian surafce = The number of field lines leaving the gaussian surface
$\therefore \varphi$ due to $q_2=0$
The charge $q_1$ is placed inside the gaussian surface which means that the eletric field lines are only leaving the gaussian surface and not entering inside it.
$⟹\varphi$ due charge $q_1$ will be non-zero.