While a capacitor remains connected to a battery and dielectric slab is applied between the plates, then
Potential difference between the plates is changed
Charge flows from the battery to the capacitor
Electric field between the plates increases
Energy store in the capacitor decreases
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]
A parallel plate capacitor with plate area $A$ and plate separation $d$ is filled with a dielectric material of dielectric constant $K =4$. The thickness of the dielectric material is $x$, where $x < d$.
Let $C_1$ and $C_2$ be the capacitance of the system for $x =\frac{1}{3} d$ and $x =\frac{2 d }{3}$, respectively. If $C _1=2 \mu F$ the value of $C _2$ is $........... \mu F$
A parallel plate capacitor has capacitance $C$. If it is equally filled with parallel layers of materials of dielectric constants $K_1$ and $K_2$ its capacity becomes $C_1$. The ratio of $C_1$ to $C$ is
An uncharged parallel plate capacitor having a dielectric of constant $K$ is connected to a similar air-cored parallel capacitor charged to a potential $V$. The two share the charge and the common potential is $V'$. The dielectric constant $K$ is
Following operations can be performed on a capacitor : $X$ - connect the capacitor to a battery of $emf$ $E.$ $Y$ - disconnect the battery $Z$ - reconnect the battery with polarity reversed. $W$ - insert a dielectric slab in the capacitor