Four capacitors with capacitances $C_1 = 1 \ \mu F, C_2 = 1.5 \ \mu F, C_3 = 2.5 \ \mu F$ and $C_4 = 0.5 \ \mu F$ are connected as shown and are connected to a $30 \ V$ source. The potential difference between points $a$ and $b$ is: (in $V$)

$A$ part of a circuit is shown in the figure. If $V_A - V_B = 4 \, V$,then find the charge on the $4 \, \mu F$ capacitor connected to point $A$. (in $, \mu C$)

$A$ capacitor of capacity $C$ is connected to a battery of potential $V$ in parallel. The distance between its plates is reduced to half at once,while the battery remains connected. Then,the energy supplied by the battery to charge the capacitor to potential $V$ again is:

Six point charges are kept at the vertices of a regular hexagon of side $L$ and centre $O$,as shown in the figure. Given that $K = \frac{1}{4 \pi \varepsilon_0} \frac{q}{L^2}$,which of the following statement$(s)$ is (are) correct?
$(A)$ The electric field at $O$ is $6K$ along $OD$
$(B)$ The potential at $O$ is zero
$(C)$ The potential at all points on the line $PR$ is same
$(D)$ The potential at all points on the line $ST$ is same.

In the circuit shown in the figure,$K_1$ is open. The charge on capacitor $C$ in steady state is $q_1$. Now,the key $K_1$ is closed and at steady state,the charge on $C$ is $q_2$. The ratio of charges $q_1/q_2$ is

$A$ capacitor $C_1 = 1\,\mu F$ can withstand a maximum voltage of $V_1 = 6\,kV$,and a capacitor $C_2 = 3\,\mu F$ can withstand a maximum voltage of $V_2 = 4\,kV$. If these two capacitors are connected in series,what is the maximum voltage (in $kV$) that can be applied to the combination?