Class 12th

1.23 Electric field produced by the infinite line charge at a distance d having linear charge density$\lambda$ 
 is given by

E = $\frac{\lambda }{2\pi {\epsilon }_{0}d}$, where

E = Electric field = 9 $*{10}^4$ N/C

${?}_0$ = Permittivity of free space = 8.854  $*{10}^{-12}$ $C^2{N}^{-1}$$m^{-2}$

d = 2 cm = 0.02 m

Hence, $\lambda$ = 9  $*{10}^4$ $*2*\pi *$8.854 $*{10}^{-12}*$
 0.02 = 10$\mu C/m$ 

1.22

(a) Diameter of the sphere, d = 2.4 m, Radius, r = 1.2 m

Surface charge density,  = 80 μC/$m^2$  = 80$*{10}^{-6}$  C/ $m^2$

Total charge on the surface of the sphere is given by Q =  A, where A = surface area of the sphere = 4 $\pi r^2$

Hence Q = 80 $*{10}^{-6}*4*\pi *{1.2}^2$ C = 1.447 $*{10}^{-3}$ C

(b) The total electric flux ( $<math><msub><mrow><mrow><mi>\phi </mi></mrow></mrow><mrow><mrow><mi>T</mi><mi>o</mi><mi>t</mi><mi>a</mi><mi>l</mi></mrow></mrow></msub></math>$ ) is given by ${\phi }_{Total}$ = $\frac{Q}{{?}_0},$

where ${?}_0$ = Permittivity of free space = 8.854$*{10}^{-12}$$C^2{N}^{-1}$ $m^{-2}$,

Q = 1.447 $*{10}^{-3}$ C

$<math><msub><mrow><mrow><mi>\phi </mi></mrow></mrow><mrow><mrow><mi>T</mi><mi>o</mi><mi>t</mi><mi>a</mi><mi>l</mi></mrow></mrow></msub></math>$ = 
 $\frac{1.447\mathrm{}*{10}^{-3}}{8.854\mathrm{}*{10}^{-12}}$ $C^{-1}N$ $m^2$= 1.63  $*{10}^8$ $C^{-1}N$$m^2$

1.21 Electric field intensity, E, at a distance d, from the centre of a sphere containing net charge q is given by the relation,

E =  $\frac{1}{4\pi {?}_0}$$*\frac{q}{d^2}$

Where ${?}_0$= Permittivity of free space = 8.854 *${10}^{-12}$ $C^2{N}^{-1}$$m^{-2},$

q= Net charge

E = 1.5 $*{10}^3$ N/C

d = 2r = 20 cm = 0.2 m

q = E $4\pi {?}_0*d^2$${}^{}$ = 1.5 $*{10}^3*4*\pi *$ 8.854 $*{10}^{-12}*{0.2}^2$ C = 6.67 *${10}^{-9}$C

= 6.67 nC

The net charge on the sphere is 6.67 nC

10.21 Consider that a single slit of width $d$ is divided in to $n$   smaller slits.

Therefore width of each slit,

$d^{\text{'}}=\frac{d}{n}$

Angle of diffraction is given by the relation,

$?=\frac{\frac{d}{d}}{d}?$ = $\frac{?}{d}$

Now, each of these infinitesimally small slit sends zero intensity in the direction of Ø.

Hence, the combination of these slits will give zero intensity.

1.20 Electric flux, $\phi$= –1.0 $*{10}^3$ N$m^2$/C

Radius of Gaussian surface, r = 10 cm = 0.1 m

(a) Electric flux piercing through a surface depends on the net charge enclosed inside a body, not on the size of the body. Hence, if the radius is doubled, the net flux passing does not change. The net flux passing will remain as -1  N$*{10}^3$ N$m^2$/C

(b) The relation between point charge and the electric flux is given by $\phi$=$\frac{q}{{?}_0},$

Where ${?}_0$= Permittivity of free space = 8.854 *${10}^{-12}$ $C^2{N}^{-1}$$m^{-2}$

Hence point charge q = φ$*$ ${?}_0$= –1.0 $*{10}^3*$ 8.854 $*{10}^{-12}$ C = - 8.854 $*{10}^{-9}$ C

= - 8.854$*{10}^{-3}\mu$C

10.20 Weak radar signals sent by a low flying aircraft can interfere with the TV signal received by the antenna. Hence, TV signal may get distorted, resulting in shaking of picture on the TV.

This is because superposition follows from the linear character of a differential equation that governs wave motion. If $y_1$ and $y_2$ are the solutions of the second order wave equation, then any linear combination of $y_1$ and $y_2$ will also be the solution of the wave equation.

10.19 Wavelength of the light beam, $?$ = 500 nm = 500 $*{10}^{-9}$ m

Distance of the screen from the slit, D = 1 m $n$

Distance of the first minimum from the centre of the screen, x = 2.5 mm = 2.5 $*{10}^{-3}$ m

Let the width of the slit be = d

From the equation

$n?=x\frac{d}{D}$ we get d = $\frac{n?D}{x}$

$d=\frac{1*500\mathrm{}*{10}^{-9}*1}{2.5\mathrm{}*{10}^{-3}}$  = 2 $*{10}^{-4}$ m = 0.2 mm

Hence, the width of the slot is 0.2 mm

10.18 Distance between the towers, d = 40 km

Height of the line joining the hills, h = 50 m

Thus, the radial spread of the radio wave should not exceed 40 km

Since the hill is located halfway between the towers,

Fresnel's distance $Z_P$  = $\frac{40}{2}$ = 20 km = 2 $*{10}^{4}m$

Aperture can be taken as a = h = 50 m

Fresnel's distance is given by the relation,

$Z_P$ = $\frac{a^2}{?}$ or

$?=\frac{a^2}{Z_P}$ =  $\frac{{50}^2}{2\mathrm{}*{10}^4}$ = 0.125 m = 12.5 cm

Therefore, the wavelength of the radio wave is 12.5 cm

10.16 Wavelength of the light used, $?=$ 6000 nm = 600 $*{10}^{-9}$ m

Angular width of the fringe, $?=0.1°$ = 0.1 $*\frac{?}{180}$  rad

Angular width of a fringe is related to slit spacing (d) as $?=\frac{?}{d}$

d = $\frac{?}{?}$ = $\frac{600*{10}^{-9}}{0.1\mathrm{}*\frac{?}{180}}$ = 3.44 $*{10}^{-4}$ m