A parallel plate capacitor is to be designed, using a dielectric of dielectric constant $5$, so as to have a dielectric strength of $10^9\;Vm^{-1}$ . If the voltage rating of the capacitor is $12\;kV$, the minimum area of each plate required to have a capacitance of $80\;pF$ is
$10.5 \times 10^{-6} \;m ^2$
$25.0 \times 10^{-5}\; m ^2$
$12.5 \times 10^{-5}\; m ^2$
$21.7 \times 10^{-6} \;m ^2$
There are two identical capacitors, the first one is uncharged and filled with a dielectric of constant $K$ while the other one is charged to potential $V$ having air between its plates. If two capacitors are joined end to end, the common potential will 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
On which the extant of polarisation depend ?
If the distance between parallel plates of a capacitor is halved and dielectric constant is doubled then the capacitance will become
The capacitance of an air capacitor is $15\,\mu F$ the separation between the parallel plates is $6\,mm$. A copper plate of $3\,mm$ thickness is introduced symmetrically between the plates. The capacitance now becomes.........$\mu F$