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 :

In the given circuit, charges $${Q_2}$$ on the $$2\mu F$$ capacitor changes as $$C$$ is varied from $$1\,\mu F$$ to $$3\mu F.$$ $${Q_2}$$ as a function of $$'C'$$ is given properly by:
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A parallel plate capacitor is made of two circular plates separated by a distance $$5$$ $$mm$$ and with a dielectric of dielectric constant $$2.2$$ between them. When the electric field in the dielectric is $$3 \times {10^4}\,V/m$$ the charge density of the positive plate will be close to: