Two identical charged spheres are suspended by strings of equal lengths. The strings make an angle of \$30^{0}\$ with each other. When suspended in a liquid of density 0.8 \$\mathrm{g}\mathrm{c}\mathrm{m}^{-3}\$, the angle remains the same. lf density of the material of the sphere is 1.6 \$\mathrm{g}\mathrm{c}\mathrm{m}^{-3}\$, the dielectric constant of the liquid is:

From fig,\$tan\theta= qE/mg\$\$tan15^o=\dfrac{k q^2}{d^2 mg}\$\$tan15^o=\dfrac{k q^2}{1.6 Vgd^2 }\$ .....[\$V\$ is the volume]...........(1)When system is suspended in liquid,\$tan15^o=\dfrac{k q^2}{K(mg- \rho Vg)d^2 }\$ \$tan15^o=\dfrac{k q^2}{K(1.6-0.8)Vgd^2 }\$ ......(2) from (1) and (2) we get,\$\dfrac{k q^2}{K(1.6-0.8)Vgd^2 }\$\$=\dfrac{k q^2}{1.6 Vgd^2 }\$ \$\therefore K=2=\$Dielectric constant of liquid.

A plane electromagnetic wave in a non magnetic dielectric medium is given by \$\bar{E} = \bar{E_0} ( 4 \times 10^{-7} \times - 50 t)\$ with distance being in meter and time in seconds. The dielectric constant of the medium is:

Given equation of wave \$\vec{E} = \vec{E}_0 (4 \times \vec{10} x - 50 t)\$Comparing with general equation of wave,\$\vec{E} = \vec{E}_0 (kx - wt),\$we get,\$w = 50\ rad/s.\$ \$k = 4 \times 10^{-7} m^{-1}\$Thus velocity of wave, \$v = \dfrac{w}{k} = \dfrac{50}{4 \times 10^{-7}} = 1.25 \times 10^8 m/s\$so, refractive index of medium,\$\mu = \dfrac{e}{v} = \dfrac{3 \times 10^8}{1.25 \times 10^8} = 2.4\$using, \$u^2 = km.ke\$ [km and ke are magnetic and dielectric di - constants]as medium is non diamagnetic, \$km = 1\$\$\Rightarrow ke = \mu^2 = (2.4)^2 \Rightarrow ke = 5.76\$

Class 12 Electrostatic Potential and Capacitance

Q: What are polar and non polar dielectric ,give one example each other ?Define di-electric constants.

NON POLAR DIELECTRICS:Are made of non-polar molecules.The center of mass of positive charges coincides with the center of mass of the negative charges, in the molecule.In its normal state (i.e., in the absence of any external force, esp electric field), each molecule has zero dipoles andhas a symmetric shape.POLAR DIELECTRICS:Are made of polar molecules.The center of mass of positive charges does not coincide with the center of mass of the negative charges, in the molecule.In its normal state (i.e., in the absence of any external force, esp electric field), each molecule has some intrinsic permanent dipole andhas an asymmetric shape.The dielectric constant is the ratio of the permittivity of a substance to the permittivity of free space. It is an expression of the extent to which a material concentrates electric flux, and is the electrical equivalent of relative magnetic permeability.

Q: Define dielectric constant of a medium. What is its \$S.I\$. unit?

The Dielectric constant (K) of the medium is the ratio of the permittivity of the substance (\varepsilon) to permittivity of the free space \$\left( \varepsilon _ { 0 } \right) .\$i.e \$\mathrm { K } = \varepsilon / \varepsilon _ { 0 }\$As Permittivity of medium and permittivity of free space both have same units(F/m ie Farad/meter) dielectric constant becomes dimensionless quantity.

Q: Define the relative permittivity of a medium.

The amount by which the force between the charge (in presence of medium) is relatively less than that in a vacuum, is called the dielectric constant of that medium or relative permittivity (\$\epsilon_r\$)\$\epsilon_r=\dfrac{\text{Force between charges in air or vaccum}}{\text{Force between charges in medium}}\$\$\therefore\,\,\varepsilon_r=\dfrac{\varepsilon}{\varepsilon_0}\$

Q: Dielectric constant for a metal is:

Permittivity of metals is very high comparable to the permittivity of free space. So dielectric constant for metal is infinite.

Q: \$64\$ small drops of mercury each of radius \$r\$ and change \$q\$ coalesce to from a big drop. The ratio of the surface charge density of each small drop with that of big drop is:

Formulae:\$\dfrac{\sigma_{small} }{\sigma_{big} }=\dfrac{q}{Q} \times \dfrac{R^2}{r^2}\$\$=\dfrac{q}{nq}\times \dfrac{(n^{\frac{1}{3}}r)^2}{r^2}\$\$=n^{-\frac{1}{3}}\$\$=64^{-\frac{1}{3}}\$\$=\dfrac{1}{4}\$hence the ratio is \$1:4\$

Q: Two parallel plate capacitor of capacitances C and 2C are connected in parallel and charged to a potential difference V . If the batteryis disconnected and the space between the plate of the capacitor of capacince C is completely filled with a material of dielectric constant K, then the potential difference across the capacitor will be come

The original capacitance of the parallel combination of C and 2 C is = C +2C = 3C Total charge Q = 3CV When the capacitor C is filled with a dielectric, its capacitance becomes KC. Therefore, after the capacitor is filled with dielectric the capacitance of the combination, C' = KC + 2C = C (K + 2). However, the charge remains the same, i.e. Q = 3CV Therefore, the potential difference across the capacitors will be