Seven capacitors each of capacitance $2 \mu \mathrm{~F}$ are to be connected in a configuration to obtain an effective capacitance $\left(\frac{10}{11}\right) \mu \mathrm{F}$. The combination is

Two identical metal plates are given charges $q_1$ and $q_2\left(q_2 < q_1\right)$ respectively. If they are now brought close together to form a parallel plate capacitor with capacitance ' C ', the potential difference ' $V$ ' between the plates is

Five capacitors, each of capacity ' $C$ ' are connected as shown in the figure. The resultant capacity between A and B is $14 \mu \mathrm{~F}$. The capacity of each capacitor is

Two capacitors of $100 \mu \mathrm{~F}$ and $50 \mu \mathrm{~F}$ are connected in parallel. If the potential difference across $100 \mu \mathrm{~F}$ is 20 V and across $50 \mu \mathrm{~F}$ is 40 V , then the common potential of the parallel combination will be (same polarities of the capacitor connected together)