A capacitor of capacitance $9 n F$ having dielectric slab of $\varepsilon_{ r }=2.4$ dielectric strength $20\, MV / m$ and $P.D. =20 \,V$ then area of plates is ....... $\times 10^{-4}\, m ^{2}$

When a dielectric material is introduced between the plates of a charges condenser, then electric field between the plates

A parallel plate air-core capacitor is connected across a source of constant potential difference. When a dielectric plate is introduced between the two plates then :

A parallel plate capacitor of capacitance $12.5 \mathrm{pF}$ is charged by a battery connected between its plates to potential difference of $12.0 \mathrm{~V}$. The battery is now disconnected and a dielectric slab $\left(\epsilon_{\mathrm{r}}=6\right)$ is inserted between the plates. The change in its potential energy after inserting the dielectric slab is_______.$\times 10^{-12} \mathrm{~J}$.

The capacity of a parallel plate condenser is $10\,\mu F$ without dielectric. Dielectric of constant $2$ is used to fill half the distance between the plates, the new capacitance in $\mu F$ is

A parallel plate capacitor with air between the plates has capacitance of $9\ pF$. The separation between its plates is '$d$'. The space between the plates is now filled with two dielectrics. One of the dielectrics has dielectric constant $k_1 = 3$ and thickness $\frac{d}{3}$ while the other one has dielectric constant $k_2 = 6$ and thickness $\frac{2d}{3}$ . Capacitance of the capacitor is now.......$pF$