Dielectric substances and polarization - class-XII

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

Displacement current, ${i _d} = {\varepsilon _ \circ }A\frac{{dE}}{{dt}}$
where, $A$ is area and $E$ is electric field
$\begin{array}{l}E = \frac{e}{d} = \frac{{4\sin (100\pi t)}}{d}\{i _d} = {\varepsilon _ \circ }A \times \frac{1}{d}\left[ {\frac{d}{{dt}}\left( {4\sin 100\pi t} \right)} \right]\ = \frac{{{\varepsilon _ \circ }A}}{d}.4(cos100\pi t) \times 100\pi \ = C.400\pi .cos100\pi t\{\left( {{i _d}} \right) _{\max }} = \left( {10 \times {{10}^{ - 6}}} \right) \times 400\pi A\ = 4\pi  \times {10^{ - 3}}A = 4\pi mA\end{array}$

The dielectric is a material through which no electric current passes. Here the given materials-plastic, mica and porcelain are all the  dielectric because current can not pass through them.

Ammonia and $HCl$ are polar molecules since they have a net dipole moment towards a particular direction. Both water and benzene are non-polar molecules. But water is a conductor of electricity, whereas benzene is a dielectric (insulator).

Breakdown voltage for a dielectric is the maximum voltage at which, applied electric field becomes so high that electrons break up from the molecules of dielectric and becomes free inside the dielectric due to this dielectric becomes conductive. 

The capacitance of a parallel plate capacitor filled with the given material of dielectric constant $\kappa$ is given by:

The value of dielectric constant for air at room temperature $ (25^{o}C , or 77^{o}F )$ is $1.00059$ .

Polar dielectric are those in which center of positive and negative charge is not same due to their shapes. 

The capacitance of a parallel plate capacitor completely filled with a given material of dielectric constant $K$ is given by:

Dielectric constant or relative permittivity is the materialistic property of any material .Relative permittivity of a material is the ratio of permittivity of the material to the permittivity of free space,

Electric field intensity is inversely proportional to the square of the distance between the point under consideration and the charge. Hence, electric field reduces when the charge moves away or gets separated. When a dielectric is polarized,  negative charge get attracted towards the source of the electric field and positive charge gets repelled from the source. The dipole of the dielectrics get aligned in a direction opposite to the electric field. This produces a  reverse electric field and the intensity of the applied electric field reduces .

All the given materials are very good conductors having high value of dielectric constant. Silicon dioxide is used in MOSFET, silicon nitride in VLSI (very large scale integration) fabrication of IC and titanium oxide in ceramics.

We know that Dielectric material refers to nonconductors or insulators in which an electric field can be sustained with a minimum dissipation of power.

The relative permittivity of a material is ratio of its (absolute) permittivity to the permittivity of vacuum. For air it is almost 1. 

Given equation of wave 

The dielectric constant K is defined as the ratio of the force in vacuum (or air) to the force in the medium. K = F_air / F_medium = 10^-4 / (2.5 * 10^-5) = 10 / 2.5 = 4.

The dielectric constant k of a medium is defined as the ratio of the capacitance of a capacitor filled with that medium (M) to the capacitance of the same capacitor with a vacuum or air (N). Therefore, M represents the capacity with the dielectric between the plates.

The relative permittivity of water id defined as:

For equilibrium in air: tan(theta) = F_e / (mg). In liquid: tan(theta) = F_e' / (mg - F_buoyant). Since theta is same, F_e/mg = F_e'/(mg - F_b). Substituting F_e' = F_e/K and F_b = Vsigmag, we get K = rho / (rho - sigma) = 1.6 / (1.6 - 0.8) = 1.6 / 0.8 = 2.

C_air = epsilon_0 * A / d = 8 pF. C_wax = K * epsilon_0 * A / (d/2) = 2 * K * C_air = 80 pF. Thus 2 * K * 8 = 80, so K = 5.

The dielectric strength of dry air is the maximum electric field it can withstand without experiencing electrical breakdown, which is approximately 3 x 10^6 V/m, equivalent to 3000 V/mm.

The dielectric constant of a metal is effectively 1 because metals are perfect conductors. In an ideal conductor, the electric field inside is zero in electrostatic conditions, and the polarization response is such that εr = 1. This is because free electrons can move instantly to cancel any internal field. The apparent 'infinite' capacitance in parallel plate capacitors comes from geometry, not material properties.

Integrating the potential contribution from concentric ring elements of charge dq over the disc yields the electric potential at the center. Solving the integration with the given surface charge density results in alpha equal to 4.

The electrostatic force between charges in a medium is inversely proportional to the dielectric constant K of the medium. The ratio of the force in air to the force in the medium is therefore K:1.

The electric field near the surface of a charged conducting sphere is given by E = sigma / epsilon_0, where sigma is the surface charge density. Since the spheres are charged to the same surface charge density, the electric fields near their surfaces are equal, yielding a ratio of 1:1.

Using the equilibrium condition for suspended charges in a medium: K = rho / (rho - sigma).

The force in a medium is F_medium = F_air / K. With K=6, the force decreases by 6 times.

The electrostatic force between two charges in a medium of dielectric constant K is given by F = (1 / (4 * pi * epsilon_0 * K)) * (q1 * q2 / r^2). For the force to remain the same when placed in an oil of dielectric constant 4, the effective distance r' must be r / sqrt(K). Since r = 1 m and K = 4, r' = 1 / sqrt(4) = 0.5 m.

Induced charge Q_i = Q_free * (1 - 1/K). Q_free = C_air * V = 4uF * 50V = 200uC. Q_i = 200 * (1 - 1/4) = 200 * 0.75 = 150uC.

The capacitor is divided into two capacitors joined in series. So (b) is correct. Also as K for a metal is $\infty$, therefore (d) is also correct.

we know that only non conductor can be used as a dilectric.So only A and B are correct option.

The capacitance of a parallel plate capacitor filled with the given material of dielectric constant $K$ is given by:

We know that dielectric is a medium or substance with a dielectric property; an insulator.So, from the option Mineral oil,Electrets and also Sulfur hexafluoride can be used as dilectric.

The capacitance of a parallel plate capacitor filled with dielectric of dielectric constant $K$ is,

The capacitance of an air filled parallel plate capacitor is given by:

Oppositely charged plates of a capacitor attract each other due to the electrostatic force between the plates. Each plate experiences an attractive force toward the other plate.

Formula 

The total charge of a thin disc with radially varying surface charge density sigma = sigma_0 * (R / r) is found by integrating dq = sigma * 2 * pi * r * dr from r = 0 to r = R. Substituting sigma gives the integral of 2 * pi * sigma_0 * R * dr, which evaluates to 2 * pi * sigma_0 * R^2.