Electrostatic P...

On removing the dielectric from a charged condenser, its energy

Three condensers each of capacitance 2 F, are connected in series. The resultant capacitance will be :

When two condensers of capacitance $$1\mu F$$ and $$2\mu F$$ are connected is series then the effective capacitance will be :

The minimum number of condensers each of capacitance of $$2 \mu F$$, in order to obtain resultant capacitance of $$5\mu F$$ will be :

Three condenser of capacity $$2\mu F, 4\mu F$$ and $$8 \mu F$$ respectively, are first connected in series and then connected in parallel. The ratio of equivalent capacitances in two cases will be :

What will be area of pieces of paper in order to make a paper condenser of capacitance $$0.04 \mu F$$, if the dielectric constant of paper is $$2.5$$ and its thickness is $$0.025 mm$$ ?

When dielectric medium of constant k is filled between the plates of a charged parallel-plate condenser, then the energy stored becomes, as compared to its previous value, 

A capacitor of capacitance C is connected to battery of emf $$V_0$$. Without removing the battery, a dielectric of strength $$\varepsilon _r$$ is inserted between the parallel plates of the capacitor C, then the charge on the capacitor is :

The capacitance of a parallel plate capacitor in air is $$2\ \mu F$$. If a dielectric medium is placed between the plates then the potential difference reduces to $$\dfrac{1}{6}$$ of the original value. The dielectric constant of the medium is :

The capacitance of parallel-plate capacitor is $$4\mu F$$. If a dielectric material of dielectric constant 16 is placed between the plates then the new capacitance will be :