As shown in the figure,four capacitors each of capacitance $25 \, \mu F$ are connected. The voltmeter reads $200 \, V$. Find the charge on each plate of the capacitors.

The charge on the $4 \mu F$ capacitor in the given circuit is (in $\mu C$)

Two thin dielectric slabs of dielectric constants $K_1$ and $K_2$ $(K_1 < K_2)$ are inserted between the plates of a parallel plate capacitor,as shown in the figure. The variation of the electric field $E$ between the plates with distance $d$ as measured from plate $P$ is correctly shown by:

In the circuit shown,the charge on the $5\, \mu F$ capacitor is $........\mu C$.

$A$ parallel plate capacitor $A$ is filled with a dielectric whose dielectric constant varies with applied voltage as $K = V$. An identical capacitor $B$ of capacitance $C_0$ with air as dielectric is connected to a voltage source $V_0 = 30\,V$ and then connected to the first capacitor $A$ after disconnecting the voltage source. Find the charge and voltage on capacitor $A$.

$A$ parallel plate capacitor of capacity $C_0$ is charged to a potential $V_0$. $(i)$ The energy stored in the capacitor when the battery is disconnected and the plate separation is doubled is $E_1$. (ii) The energy stored in the capacitor when the charging battery is kept connected and the separation between the capacitor plates is doubled is $E_2$. Then,the value of $E_1 / E_2$ is: