An electron is made to enter symmetrically between two parallel and equally but oppositely charged metal plates, each of 10 cm length. The electron emerges out of the electric field region with a horizontal component of velocity $10^6 \mathrm{~m} / \mathrm{s}$. If the magnitude of the electric field between the plates is $9.1 \mathrm{~V} / \mathrm{cm}$, then the vertical component of velocity of electron is (mass of electron $=9.1 \times 10^{-31} \mathrm{~kg}$ and charge of electron $=1.6 \times 10^{-19} \mathrm{C}$ )

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 $$\epsilon_0=8.834 \times 10^{-12} \mathrm{~F} / \mathrm{m}$$)