The capacitance of an air filled parallel plate capacitor is $10\,p F$. The separation between the plates is doubled and the space between the plates is then filled with wax giving the capacitance a new value of $40 \times {10^{ - 12}}farads$. The dielectric constant of wax is
$12$
$10$
$8$
$4.2$
. Three identical capacitors $C _1, C _2$ and $C _3$ have a capacitance of $1.0 \mu F$ each and they are uncharged initially. They are connected in a circuit as shown in the figure and $C _1$ is then filled completely with a dielectric material of relative permittivity $\varepsilon_{ r }$. The cell electromotive force (emf) $V_0=8 V$. 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 $5 \mu C$. The value of $\varepsilon_{ r }=$. . . . .
A container has a base of $50 \mathrm{~cm} \times 5 \mathrm{~cm}$ and height $50 \mathrm{~cm}$, as shown in the figure. It has two parallel electrically conducting walls each of area $50 \mathrm{~cm} \times 50 \mathrm{~cm}$. The remaining walls of the container are thin and non-conducting. The container is being filled with a liquid of dielectric constant $3$ at a uniform rate of $250 \mathrm{~cm}^3 \mathrm{~s}^{-1}$. What is the value of the capacitance of the container after $10$ seconds? [Given: Permittivity of free space $\epsilon_0=9 \times 10^{-12} \mathrm{C}^2 \mathrm{~N}^{-1} \mathrm{~m}^{-2}$, the effects of the non-conducting walls on the capacitance are negligible]
In one design of capacitor thin sheets ot metal of area $80\ mm \times 80\ mm$ sandwich between them a piece of paper whose thickness is $40\ μm$. The relative permittivity of the paper is $4.0$ and its dielectric strength is $20\ MVm^{-1}$. Calculate the maximum charge that can be put on the capacitor
[permittivity of free space $ = 9 \times 10^{-12}\ Fm^{-1}$]
Give examples of polar and non-polar molecules.
A capacitor with plate separation $d$ is charged to $V$ volts. The battery is disconnected and a dielectric slab of thickness $\frac{d}{2}$ and dielectric constant ' $2$ ' is inserted between the plates. The potential difference across its terminals becomes