A parallel-plate capacitor of capacitance $40 \mu \mathrm{~F}$ is connected to a 100 V power supply. Now the intermediate space between the plates is filled with a dielectric material of dielectric constant $\mathrm{K}=2$. Due to the introduction of dielectric material, the extra charge and the change in the electrostatic energy in the capacitor, respectively, are

A capacitor is made of a flat plate of area A and a second plate having a stair-like structure as shown in figure. If the area of each stair is $$\frac{A}{3}$$ and the height is $$d$$, the capacitance of the arrangement is :

A capacitor has air as dielectric medium and two conducting plates of area $$12 \mathrm{~cm}^2$$ and they are $$0.6 \mathrm{~cm}$$ apart. When a slab of dielectric having area $$12 \mathrm{~cm}^2$$ and $$0.6 \mathrm{~cm}$$ thickness is inserted between the plates, one of the conducting plates has to be moved by $$0.2 \mathrm{~cm}$$ to keep the capacitance same as in previous case. The dielectric constant of the slab is : (Given $$\epsilon_0=8.834 \times 10^{-12} \mathrm{~F} / \mathrm{m}$$)