B M Sharma Solutions for Chapter: Capacitor and Capacitance, Exercise 7: Archives

Two capacitors $C_1$ and $C_2$ are charged to $120 \mathrm{V}$ and $200 \mathrm{V}$, respectively. It is found that by connecting them together the potential on each one can be made zero. Then

A parallel plate capacitor is made of two circular plates separated by a distance of $5 \mathrm{mm}$ and with a dielectric of dielectric constant $2.2$ between them. When the electric field in the dielectric is $3\times {10}^4 \mathrm{V} {\mathrm{m}}^{-1}$, the charge density of the positive plate will be close to

In the given circuit, charge $Q_2$ on the $2 \mathrm{\mu F}$ capacitor changes as $C$ is varied from $1 \mathrm{\mu F}$ to $3 \mathrm{\mu F}. Q_2$ as a function of $C$ is given properly by: (figures are drawn schematically and are not to scale)

In the given circuit, a charge of $+80 \mathrm{\mu C}$ is given to the upper plate of the $4 \mathrm{\mu F}$ capacitor. Then in the steady state, the charge on the upper plate of the $3 \mathrm{\mu F}$ capacitor is

A parallel plate capacitor having plates of area $S$ and plate separation $d$, has capacitance $C$, in air. When two dielectrics of different relative permittivity (${\epsilon }_1=2$ and ${\epsilon }_2=4$) are introduced between the two plates as shown in the figure, the capacitance becomes $C_2$. The ratio $\frac{C_2}{C_1}$ is

In the circuit shown in the figure, there are two parallel plate capacitors each of capacitance $C$. The switch $S_1$ is pressed first to fully charge the capacitor $C_1$ and then released. The switch $S_2$ is then pressed to charge the capacitor $C_2.$ After some time, $S_2$ is released and then $S_3$ is pressed. After some time,

A parallel plate capacitor has a dielectric slab of dielectric constant $K$ between its plates that covers $\frac{1}{3}$ of the area of its plates, as shown in the figure. The total capacitance of the capacitor is $C$ while that of the portion with dielectric in between is $C_1.$ When the capacitor is charged, the plate area covered by the dielectric gets charge $Q_1$ and the rest of the area gets charge $Q_2.$ The electric field in the dielectric is $E_1$ and that in the other portion is $E_{2^{·}}$ Choose the correct option/ options, ignoring edge effects.

Three identical capacitors $C_1\text{,} C_2$ and $C_3$ have a capacitance of $1.0 \mu \mathrm{F}$ each and they are initially uncharged. They are connected in a circuit as shown in the figure and $C_1$ is then completely filled with a dielectric material of relative permittivity ${\epsilon }_{\mathrm{r}}$. The cell electromotive force (emf) is, $V_0=8 \mathrm{V}$. First, the switch $S_1$ is closed while the switch $S_2$ is kept open. When the capacitor $C_3$ is fully charged, $S_1$ is opened and $S_2$ is closed simultaneously. When all the capacitors reach equilibrium, the charge on $C_3$ is found to be $5 \mu \mathrm{C}$. The value of ${\epsilon }_{\mathrm{r}}$ is,