Embibe Experts Solutions for Chapter: Capacitance, Exercise 4: Exercise-4

The$V$ vs $x$ plot for six identical metal plates of cross-sectional area $A$ is as shown. The equivalent capacitance between $2$ and $5$ is (adjacent plates are placed at a separation $d$),

The plates of a parallel plate condenser are being moved away with a constant speed $v$. If the plate separation at any instant of time is $d$, then the rate of change of capacitance with time is proportional to,

A parallel plate capacitor of capacitance $C$ is as shown. A thin metal plate $A$ is placed between the plates of the given capacitor in such a way that its edges touch the two plates as shown. The capacity now becomes,

A capacitor of capacitance $C_0$ is charged to a potential $V_0$ and then isolated. A small capacitor $C$ is then charged from $C_0$, discharged and charged again. The process is repeated $n$ times. Due to this, the potential of the larger capacitor is decreased to $V$. Then, the value of $C$ is,

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,

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 $Q_2$. Choose the correct option ignoring edge effects.

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