Write the formula of capacitance of capacitor having dielectric constant $\mathrm{K} = 2$.
A parallel plate capacitor with air between the plate has a capacitance of $15 pF$. The separation between the plate becomes twice and the space between them is filled with a medium of dielectric constant $3.5.$ Then the capacitance becomes $\frac{ x }{4}\,pF$.The value of $x$ is $............$
The two metallic plates of radius $r$ are placed at a distance $d$ apart and its capacity is $C$. If a plate of radius $r/2$ and thickness $d$ of dielectric constant $6$ is placed between the plates of the condenser, then its capacity will be
The electric field between the plates of a parallel plate capacitor when connected to a certain battery is ${E_0}$. If the space between the plates of the capacitor is filled by introducing a material of dielectric constant $K$ without disturbing the battery connections, the field between the plates shall be
Capacitance of a parallel plate capacitor becomes $4/3$ times its original value if a dielectric slab of thickness $t = d/2$ is inserted between the plates ($d$ is the separation between the plates). The dielectric constant of the slab is
Three identical capacitors $\mathrm{C}_1, \mathrm{C}_2$ and $\mathrm{C}_3$ have a capacitance of $1.0 \mu \mathrm{F}$ each and they are uncharged initially. They are connected in a circuit as shown in the figure and $\mathrm{C}_1$ is then filled completely with a dielectric material of relative permittivity $\varepsilon_{\mathrm{r}}$. The cell electromotive force (emf) $V_0=8 \mathrm{~V}$. 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 $\mathrm{C}_3$ is found to be $5 \mu \mathrm{C}$. The value of $\varepsilon_{\mathrm{r}}=$. . . .
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