Class 12th

2.5 Capacitance between the parallel plates of the capacitor, C = 8 pF

Let us assume, initially, distance between the parallel plates was d and it was filled with air.

Dielectric constant of air, k = 1

Capacitance C is given by the formula,

C = $\frac{k{\epsilon }_{0}A}{d}$ = $\frac{{\epsilon }_{0}A}{d}$ , since k = 1…………………….(1)

Where A = area of each plate

${\epsilon }_0$ = permittivity of free space = 8.854 $*{10}^{-12}$ $C^2{N}^{-1}$ $m^{-2}$

d = distance between two plates

If the distance between two plates is reduced to half, then distance between two plates $d_1$ = $\frac{d}{2}$

Dielectric constant of a new substance, $k_1$ = 6

Then the resistance between two plates, $C_1$ = $\frac{6{\epsilon }_{0}A}{d/2}$ = $\frac{12{\epsilon }_{0}A}{d}$ ……………(2)

From equation (1) and (2) we get

$C_1=12C$ = 12 $*8$ pF = 96 pF

2.1

Let the charges be

$q_1$ = 5 $*{10}^{-8}$ C and $q_2$ = -3 $*{10}^{-8}$ C

Distance between the two charges, d = 16 cm = 0.16 m

Consider a point P on the line joining the two charges

R = distance of the point P from the charge $q_1$

Let the electrical potential (V) at point P is zero

Potential at point P caused by charges $q_1$ and $q_2$ respectively.

V = $\frac{1}{4\pi {\epsilon }_0}*\frac{q_1}{r}$ $+$ $\frac{1}{4\pi {\epsilon }_0}*\frac{q_2}{(d-r)}$ ………………………….(1)

Where ${\epsilon }_0$ = permittivity of free space

For V = 0, the equation (1) becomes

$\frac{1}{4\pi {\epsilon }_0}*\frac{q_1}{r}=-\frac{1}{4\pi {\epsilon }_0}*\frac{q_2}{(d-r)}$ or $\frac{q_1}{r}$ = $-\frac{q_2}{(d-r)}$ or $\frac{5\mathrm{}*{10}^{-8}}{r}$ = $-\frac{-3\mathrm{}*{10}^{-8}}{(d-r)}$

$\frac{5\mathrm{}*{10}^{-8}}{r}$ = $\frac{3\mathrm{}*{10}^{-8}}{(d-r)}$ or $\frac{r}{(d-r)}$ = $\frac{5\mathrm{}*{10}^{-8}}{3\mathrm{}*{10}^{-8}}$ or $\frac{r}{(d-r)}$ = $\frac{5}{3}$

3r = 5d -5r or r = (5d/8) = 0.1 m = 10 cm

Therefore the potential is zero at a distance of 10 cm from the charge $q_1$

Suppose the point P is outside the system of two charges, as shown in the following figure.

V = $\frac{1}{4\pi {\epsilon }_0}*\frac{q_1}{s}$ $+$ $\frac{1}{4\pi {\epsilon }_0}*\frac{q_2}{(s-d)}$ ……………………………(2)

For V = 0, we get

$\frac{1}{4\pi {\epsilon }_0}*\frac{q_1}{s}=-\frac{1}{4\pi {\epsilon }_0}*\frac{q_2}{(s-d)}$ or $\frac{q_1}{s}=-\frac{q_2}{(s-d)}$ or $\frac{5\mathrm{}*{10}^{-8}}{s}=-\frac{-3\mathrm{}*{10}^{-8}}{(s-d)}$

$\frac{5}{s}$ = $\frac{3}{s-0.16}$ or 5s – 0.8 = 3s or s = $\frac{0.8}{2}$ = 0.4 m = 40 cm

Therefore, the potential is zero at a distance 40 cm from $q_1$ , outside the system of charge.

8.15 Long distance radio broadcasts use shortwave bands because only these bands can be refracted by ionosphere.

It is necessary to use satellites for long distance TV transmissions because television signals are of high frequencies and high energies. Thus these signals are not reflected by ionosphere. Hence, satellites are helpful in reflecting TV signals. Also they help in long distance TV transmissions.

With reference to X-ray astronomy, X-rays are absorbed by the atmosphere. However, visible and radio waves can penetrate. Hence, optical and radio telescopes are built on the ground, while X-ray astronomy is possible only with the help of satellites orbiting the Earth.

The small ozone layer on the top of the atmosphere is crucial for human survival because it absorbs harmful ultraviolet radiation present in sunlight and prevents from reaching Earth's surface.

In the absence of an atmosphere, there would be no greenhouse effect on the surface of the Earth. As a result, the temperature of the Earth decreases rapidly, making it chilly and difficult for human survival.

A global nuclear war on the surface of the Earth would have disastrous consequences. Post nuclear war, the Earth will experience severe winter as the war will produce clouds of smoke that would cover maximum parts of the sky, thereby preventing solar light to reach the atmosphere. Also, it will lead to the depletion of the ozone layer.

8.13 A body at a particular temperature produces a continuous spectrum of wavelengths. In case of a black body, the wavelength corresponding to maximum intensity of radiation is given according to Plank's law. It can be given by the relation,

T = 0.29 cmK

${\mathrm{}\lambda }_m=\frac{0.29}{T}$ cm K, where ${\mathrm{}\lambda }_{\mathit{m}}$

= maximum wavelength and T = temperature

Thus, the temperature for different wavelengths can be obtained as:

For ${\mathrm{}\lambda }_{\mathit{m}}$  = ${10}^{-4}$ cm, T = 2900 $°K$

For ${\mathrm{}\lambda }_{\mathit{m}}$  = ${5*10}^{-5}$ cm, T = 5800 $°K$

For ${\mathrm{}\lambda }_{\mathit{m}}$  = ${10}^{-6}$  cm, T = 290 $*{10}^3$ $°K$ and so on

The numbers obtained tell us that temperature ranges are required for obtaining radiations in different parts of an electromagnetic spectrum. As the wavelength decreases, the corresponding temperature increases.