Take the potential of the point $\mathrm{B}$ as shown in the figure to be $100 \mathrm{V}.$ 

$\left(\mathrm{a}\right)$ Find the potentials at the point $\mathrm{C}$ and $\mathrm{D}$.

$\left(\mathrm{b}\right)$  If an uncharged capacitor is connected between $\mathrm{C} \mathrm{and} \mathrm{D}$, then find the amount of charge that will appear on this capacitor

 

Find the potential difference between the points $\mathrm{A}$ and $\mathrm{B}$ and between the points $\mathrm{B}$ and $\mathrm{C}$ of figure in steady state.

Consider the situation shown in the figure. The switch $\mathrm{S}$ is open for a long time and then closed and again steady state reached then

$\left(\mathrm{a}\right)$ Find the charge flown through the battery after the switch $\mathrm{S}$ is closed.

$\left(b\right)$ Find the work done by the battery after the switch $\mathrm{S}$ is closed.

$\left(c\right)$ Find the change in energy stored in the system of capacitors.

$\left(d\right)$ Find the heat developed in the system after the switch $\mathrm{S}$ is closed.

The parallel plates of a capacitor have an area $0.2 {\mathrm{m}}^2$ and are ${10}^{-2} \mathrm{m}$ apart. The original potential difference between them is $3000 \mathrm{V}$, and it decreases to $1000 \mathrm{V}$ when a sheet of dielectric is inserted between the plates filling the full space. Compute: (${\in }_0$ $= 9 \mathrm{x} {10}^{-12}$ S. I. units) 

$\left(\mathrm{i}\right)$ Original capacitance ${\mathrm{C}}_0$.

$\left(\mathrm{ii}\right)$ The charge $\mathrm{Q}$ on each plate.

$\left(\mathrm{iii}\right)$ Capacitance $\mathrm{C}$ after insertion of the dielectric.

$\left(\mathrm{iv}\right)$ Dielectric constant $\mathrm{K}.$

$\left(\mathrm{v}\right)$ Permittivity $\in$ of the dielectric.