Question

Charges $Q_1$ and $Q_2$ lie inside and outside respectively of a closed surface S. Let E be the field at any point on S and $\psi$ be the flux of E over S.

• A. If $Q_1$ changes, both E and $\psi$ will change.
• B. If $Q_2$ changes, E will change but $\psi$ will not change.
• C. If $Q_1=0$ and $Q_2\ne 0$ then $E\ne 0$ but $\psi =0$.
• D. If $Q_1\ne 0$ and $Q_2=0$ then $E=0$ but $\psi \ne 0$.

Answer 1

Answer: (A), (B), (C)

The correct options are
A If $Q_1$ changes, both E and $\psi$ will change.
B If $Q_2$ changes, E will change but $\psi$ will not change.
C If $Q_1=0$ and $Q_2\ne 0$ then $E\ne 0$ but $\psi =0$.

The electric flux $\left(\psi \right)$ only depends on the charge enclosed by the closed surface and the electric field $\left(E\right)$ is the resultant effect due to all charges.
So here $\psi$ only depends on $Q_1$.i.e $\psi =\frac{Q_1}{{ϵ}_0}$
if $Q_1$ change , both $\psi$ and $E$ will change.
if $Q_2$ change , $\psi$ will not change but $E$ will change.
if $Q_1=0$ and $Q_2\ne 0$ , then $\psi =0$ and $E\ne 0$
if,$Q_1\ne 0$ and $Q_2=0$ , then $\psi \ne 0$ and $E\ne 0$
Ans: (A),(B),(C)