The area of the plates of a parallel plate condenser is $A$ and the distance between the plates is $10\,mm$. There are two dielectric sheets in it, one of dielectric constant $10$ and thickness $6\,mm$ and the other of dielectric constant $5$ and thickness $4\,mm$. The capacity of the condenser is
$\frac{{12}}{{35}}{\varepsilon _0}A$
$\frac{2}{3}{\varepsilon _0}A$
$\frac{{5000}}{7}{\varepsilon _0}A$
$1500\;{\varepsilon _0}A$
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
The expression for the capacity of the capacitor formed by compound dielectric placed between the plates of a parallel plate capacitor as shown in figure, will be (area of plate $ = A$)
A slab of material of dielectric constant $K$ has the same area as the plates of a parallel-plate capacitor but has a thickness $(3/4)d$, where $d$ is the separation of the plates. How is the capacitance changed when the slab is inserted between the plates?
The outer sphere of a spherical air capacitor is earthed. For increasing its capacitance
Separation between the plates of a parallel plate capacitor is $d$ and the area of each plate is $A$. When a slab of material of dielectric constant $k$ and thickness $t(t < d)$ is introduced between the plates, its capacitance becomes