A capacitance of 2 $$\mu $$F is required in an electrical circuit across a potential difference of 1.0 kV. A large number of 1 $$\mu $$F capacitors are available which can withstand a potential difference of not more than 300 V. The minimum number of capacitors required to achieve this is:

Figure shows a network of capacitors where the numbers indicates capacitances in micro Farad. The value of capacitance C if the equivalent capacitance between point A and B is to be 1 $$\mu $$F is :

Three capacitors each of 4 $$\mu $$F are to be connected in such a way that the effective capacitance is 6 $$\mu $$F. This can be done by connecting them :

A combination of capacitors is set up as shown in the figure. The magnitude of the electric field, due to a point charge $$Q$$ (having a charge equal to the sum of the charges on the $$4$$ $$\mu \,F$$ and $$9$$ $$\mu \,F$$ capacitors), at a point distance $$30$$ $$m$$ from it, would equal :