Seven capacitors each of capacitance $2 \mu \mathrm{~F}$ are connected in a configuration to obtain an effective capacitance $\frac{6}{13} \mu \mathrm{~F}$. The combination which will achieve this will have
Two identical capacitors have the same capacitance ' $C$ '. One of them is charged to potential $\mathrm{V}_1$ and other to $\mathrm{V}_2$. The negative ends of capacitors are connected together. When positive ends are also connected, the decrease in energy of the combined system is
The potential difference that must be applied across the series and parallel combination of 4 identical capacitors is such that the energy stored in them becomes the same. The ratio of potential difference in series to parallel combination is
Air capacitor has capacitance of $1 \mu \mathrm{~F}$. Now the space between two plates of capacitor is filled with two dielectrics as shown in figure. The capacitance of the capacitor is [ $\mathrm{d}=$ distance between two plates of capacitor, $\mathrm{K}_1$ and $\mathrm{K}_2$ are dielectric constants of first dielectric and second dielectric respectively]
