A capacitor is kept connected to the battery and a dielectric slab is inserted between the plates. During this process
No work is done
Work is done at the cost of the energy already stored in the capacitor before the slab is inserted
Work is done at the cost of the battery
Work is done at the cost of both the capacitor and the battery
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 $\left(\varepsilon_1=2\right.$ and $\left.\varepsilon_2=4\right)$ are introduced between the two plates as shown in the figure, the capacitance becomes $C _2$. The ratio $\frac{ C _2}{ C _1}$ is
In a parallel plate capacitor the separation between the plates is $3\,mm$ with air between them. Now a $1\,mm$ thick layer of a material of dielectric constant $2$ is introduced between the plates due to which the capacity increases. In order to bring its capacity to the original value the separation between the plates must be made......$mm$
A parallel plate air capacitor has a capacitance $C$. When it is half filled with a dielectric of dielectric constant $5$, the percentage increase in the capacitance will be.....$\%$
Voltage rating of a parallel plate capacitor is $500\,V$. Its dielectric can withstand a maximum electric field of ${10^6}\,\frac{V}{m}$. The plate area is $10^{-4}\, m^2$ . What is the dielectric constant if the capacitance is $15\, pF$ ? (given ${ \in _0} = 8.86 \times {10^{ - 12}}\,{C^2}\,/N{m^2}$)
The potential gradient at which the dielectric of a condenser just gets punctured is called