Question

Change $Q_1$ and $Q_2$ lie inside and outside respectively of a closed surface $S$. Let $E$ be the field at any point on $S$ $\varphi$ be the flux of $E$ (electric field) over $S$.

	List-I		List-II
(P)	If $Q_1=0$ and $Q_2\ne 0$ then	(1)	Both $E$ and $\varphi$ will change
(Q)	If $Q_1\ne 0$ and $Q_2=0$ then	(2)	$E\ne 0$ and $\varphi \ne 0$
(R)	If $Q_2$ changes	(3)	net electric field will change but $\varphi$ will not change
(S)	If $Q_1$ changes	(4)	$E\ne 0$ but $\varphi =0$

• A. P - 4, Q - 2, R - 3, S - 1
• B. P - 1, Q - 2, R - 4, S - 3
• C. P - 2, Q - 4, R - 1, S - 3
• D. P - 3, Q - 4, R - 1, S - 2

Answer 1

The correct option is A P - 4, Q - 2, R - 3, S - 1

Given,

$Q_1$ is inside and $Q_2$ is outside.

We know,

$\varphi =\frac{Q_{in}}{{\epsilon }_0}=\frac{Q_1}{{\epsilon }_0}$

And Electric field on the surface will depend on both the Charges.

$\text{Case 1}$

$\varphi =\frac{0}{{\epsilon }_0}=0$

$E\ne 0$ as there'll be field due to $Q_2$

$P\to 4$

$\text{Case 2}$

$\varphi \ne 0$

And also $E\ne 0$ as there'll be field due to $Q_1$

$Q\to 2$

$\text{Case 3}$

As $\varphi$ is independent of $Q_2$

Net Electric field will change but $\varphi$ won't change.

$R\to 3$

$\text{Case 4}$

Here $\varphi$ changes as $Q_1$ Changes.

And also Electric field will change of course.

$R\to 1$