In the circuit shown below,the switch $S$ is connected to position $P$ for a long time so that the charge on the capacitor becomes $q_1 \mu C$. Then $S$ is switched to position $Q$. After a long time,the charge on the capacitor is $q_2 \mu C$.
$(1)$ The magnitude of $q_1$ is
$(2)$ The magnitude of $q_2$ is
Give the answer of question $(1)$ and $(2)$

Four equal capacitors are connected to a $10 \, V$ battery as shown in the figure. The potentials of $A$ and $B$ are:

Statement $(A)$: Inside a charged hollow metal sphere,$E = 0$ and $V \neq 0$. ($E$ = electric field,$V$ = electric potential).
Statement $(B)$: The work done in moving a positive charge on an equipotential surface is zero.
Statement $(C)$: When two like charges are brought closer,their mutual electrostatic potential energy will increase.

$A$ parallel plate capacitor of capacitance $C$ is connected to a battery and is charged to a potential difference $V$. Another capacitor of capacitance $2C$ is connected to another battery and is charged to a potential difference $2V$. The charging batteries are now disconnected and the capacitors are connected in parallel to each other in such a way that the positive terminal of one is connected to the negative terminal of the other. The final energy of the configuration is

Calculate the amount of charge on the capacitor of $4\, \mu \text{F}$ in the given circuit. The internal resistance of the battery is $1\, \Omega$. (in $\mu \text{C}$)

Three capacitors $A, B$ and $C$ are connected with a battery of $emf$ $\varepsilon$. All capacitors are identical initially. If a dielectric slab is inserted between the plates of capacitor $A$ slowly with the help of an external force,then: