A parallel plate capacitor of capacitance $C$ has spacing $d$ between two plates having area $A$. The region between the plates is filled with $N$ dielectric layers, parallel to its plates, each with thickness $\delta=\frac{ d }{ N }$. The dielectric constant of the $m ^{\text {th }}$ layer is $K _{ m }= K \left(1+\frac{ m }{ N }\right)$. For a very large $N \left(>10^3\right)$, the capacitance $C$ is $\alpha\left(\frac{ K \varepsilon_0 A }{ d \;ln 2}\right)$. The value of $\alpha$ will be. . . . . . . .
[ $\epsilon_0$ is the permittivity of free space]
$1$
$3$
$5$
$6$
Consider the arrangement shown in figure. The total energy stored is $U_1$ when key is closed. Now the key $K$ is made off (opened) and two dielectric slabs of relative permittivity ${ \in _r}$ are introduced between the plates of the two capacitors. The slab tightly fit in between the plates. The total energy stored is now $U_2$. Then the ratio of $U_1/U_2$ is
Two identical parallel plate capacitors of capacitance $C$ each are connected in series with a battery of emf, $E$ as shown below. If one of the capacitors is now filled with a dielectric of dielectric constant $k$, then the amount of charge which will flow through the battery is (neglect internal resistance of the battery)
Initially the circuit is in steady state. Now one of the capacitor is filled with dielectric of dielectric constant $2$ . Find the heat loss in the circuit due to insertion of dielectric
The capacitance of a parallel plate capacitor is $5\, \mu F$ . When a glass slab of thickness equal to the separation between the plates is introduced between the plates, the potential difference reduces to $1/8$ of the original value. The dielectric constant of glass is
A parallel plate capacitor filled with a medium of dielectric constant $10$ , is connected across a battery and is charged. The dielectric slab is replaced by another slab of dielectric constant $15$ . Then the energy of capacitor will ......................