The capacity of a condenser in which a dielectric of dielectric constant $5$ has been used, is $C$. If the dielectric is replaced by another with dielectric constant $20$, the capacity will become

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 $............$

A parallel plate capacitor is to be designed with a voltage rating $1\; k\,V ,$ using a material of dielectric constant $3$ and dielectric strength about $10^{7}\; V\,m ^{-1} .$ (Dielectric strength is the maximum electric field a material can tolerate without breakdown, i.e., without starting to conduct electricity through partial ionisation.) For safety, we should like the field never to exceed, say $10 \%$ of the dielectric strength. What minimum area (in $cm^2$) of the plates is required to have a capacitance of $50\; pF ?$

A parallel plate capacitor with width $4\,cm$, length $8\,cm$ and separation between the plates of $4\,mm$ is connected to a battery of $20\,V$. A dielectric slab of dielectric constant $5$ having length $1\,cm$, width $4\,cm$ and thickness $4\,mm$ is inserted between the plates of parallel plate capacitor. The electrostatic energy of this system will be......... $\in_{0}\,J$. (Where $\epsilon_{0}$ is the permittivity of free space)

A parallel plate air capacitor has a capacitance of $100\,\mu  F$. The plates are at a distance $d$ apart. If a slab of thickness $t(t \le d)$and dielectric constant $5$ is introduced between the parallel plates, then the capacitance will be.......$\mu F$

A combination of parallel plate capacitors is maintained at a certain potential difference When a $3\, mm$ thick slab is introduced between all the plates, in order to maintain the same potential difference, the distance between the plates is increased by $2.4\, mm$. Find the dielectric constant of the slab.