Two capacitors, each having capacitance $40\,\mu F$ are connected in series. The space between one of the capacitors is filled with dielectric material of dielectric constant $K$ such that the equivalence capacitance of the system became $24\,\mu F$. The value of $K$ will be.

A parallel plate capacitor is filled with $3$ dielectric materials of same thickness, as shown in the sketch. The dielectric constants are such that $k_3 > k_2 > k_1$. Let the magnitudes of the electric field in and potential drops across each dielectric be $E_3$, $E_2$,$ E_1$, $\Delta V_3$, $\Delta V_2$ and $\Delta V_1$, respectively. Which one of the following statement is true ?

Between the plates of a parallel plate condenser, a plate of thickness ${t_1}$ and dielectric constant ${k_1}$ is placed. In the rest of the space, there is another plate of thickness ${t_2}$ and dielectric constant ${k_2}$. The potential difference across the condenser will be

A parallel plate capacitor of area ' $A$ ' plate separation ' $d$ ' is filled with two dielectrics as shown. What is the capacitance of the arrangement?

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}$

A parallel plate capacitor is to be designed with a voltage rating $1\; k\,V ,$ using a material of dielectric constant $3$ and dielectric strength about $10^{7}\; V\,m ^{-1} .$ (Dielectric strength is the maximum electric field a material can tolerate without breakdown, i.e., without starting to conduct electricity through partial ionisation.) For safety, we should like the field never to exceed, say $10 \%$ of the dielectric strength. What minimum area (in $cm^2$) of the plates is required to have a capacitance of $50\; pF ?$