In the circuit shown in the figure,four capacitors are connected to a battery. $A$ parallel plate capacitor has an electric field of $10^5 \ V/m$ between its plates. If the charge on the capacitor plate is $1 \ \mu C$,then the force on each capacitor plate is.....$N$.

In the given circuit, what are the potential differences between points $A$ and $B$ $(V_{AB})$ and between points $B$ and $C$ $(V_{BC})$?

$A$ circuit is connected as shown in the figure with the switch $S$ open. When the switch is closed,the total amount of charge that flows from $Y$ to $X$ is

Two parallel plate capacitors $C_1$ and $C_2$ each having capacitance of $10 \mu F$ are individually charged by a $100 \, V$ $D.C.$ source. Capacitor $C_1$ is kept connected to the source and a dielectric slab is inserted between its plates. Capacitor $C_2$ is disconnected from the source and then a dielectric slab is inserted in it. Afterwards,the capacitor $C_1$ is also disconnected from the source and the two capacitors are finally connected in parallel combination. The common potential of the combination will be $......... \, V$. (Assuming dielectric constant $K = 10$)

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:

$A$ particle of charge $Q$ and mass $M$ moves in a circular path of radius $R$ in a uniform magnetic field of magnitude $B$. The same particle now moves with the same speed in a circular path of same radius $R$ in the space between the cylindrical electrodes of a cylindrical capacitor. The radius of the inner electrode is $R/2$ while that of the outer electrode is $3R/2$. Then the potential difference between the capacitor electrodes must be