A parallel plate capacitor with plate area $A$ and plate separation $d$ is filled with a dielectric material of dielectric constant $K =4$. The thickness of the dielectric material is $x$, where $x < d$.

Let $C_1$ and $C_2$ be the capacitance of the system for $x =\frac{1}{3} d$ and $x =\frac{2 d }{3}$, respectively. If $C _1=2 \mu F$ the value of $C _2$ is $........... \mu F$

Initially the circuit is in steady state. Now one of the capacitor is filled with dielectric of dielectric constant $2$ . Find the heat loss in the circuit due to insertion of dielectric

A capacitor of capacitance $15 \,nF$ having dielectric slab of $\varepsilon_{r}=2.5$ dielectric strength $30 \,MV / m$ and potential difference $=30\; volt$ then the area of plate is ....... $ \times 10^{-4}\; m ^{2}$

With the rise in temperature, the dielectric constant $K$ of a liquid

Assertion : If the distance between parallel plates of a capacitor is halved and dielectric constant is three times, then the capacitance becomes $6\,times$.

Reason : Capacity of the capacitor does not depend upon the nature of the material.

The area of the plates of a parallel plate capacitor is $A$ and the gap between them is $d$. The gap is filled with a non-homogeneous dielectric whose dielectric constant varies with the distance $‘y’$ from one plate as : $K = \lambda \ sec(\pi y/2d)$, where $\lambda $ is a dimensionless constant. The capacitance of this capacitor is