In the reported figure, a capacitor is formed by placing a compound dielectric between the plates of parallel plate capacitor. The expression for the capacity of the said capacitor will be (Given area of plate $=A$ )
$\frac{25}{6} \frac{{K} \varepsilon_{0} {A}}{{d}}$
$\frac{15}{34} \frac{{K\varepsilon}_{0} {A}}{{d}}$
$\frac{15}{6} \frac{{K} \varepsilon_{0} {A}}{{d}}$
$\frac{9}{6} \frac{{K} \varepsilon_{0} {A}}{{d}}$
A capacitor with air as the dielectric is charged to a potential of $100\;volts$. If the space between the plates is now filled with a dielectric of dielectric constant $10$, the potential difference between the plates will be......$volts$
A capacitor of capacity $'C'$ is connected to a cell of $'V'\, volt$. Now a dielectric slab of dielectric constant ${ \in _r}$ is inserted in it keeping cell connected then
A capacitor is half filled with a dielectric $(K=2)$ as shown in figure A. If the same capacitor is to be filled with same dielectric as shown, what would be the thickness of dielectric so that capacitor still has same capacity?
A parallel plate capacitor is made of two square plates of side $a$, separated by a distance $d\,(d < < a)$. The lower triangular portion is filled with a dielectric of dielectric constant $K$, as shown in the figure. Capacitance of this capacitor is
A parallel plate capacitor $\mathrm{C}$ with plates of unit area and separation $\mathrm{d}$ is filled with a liquid of dielectric constant $\mathrm{K}=2$. The level of liquid is $\frac{\mathrm{d}}{3}$ initially. Suppose the liquid level decreases at a constant speed $V,$ the time constant as a function of time $t$ is Figure: $Image$