Ncert Solutions Physics Class 12th

1.27 Dipole moment of the system, p = q$*dl$ = ${10}^{-7}$ Cm

Rate of increase of electric field per unit length,$\frac{dE}{dl}$ =${10}^5$ N$C^{-1}$ 

Force experienced by the system is given by the relation, F = qE = q$*\frac{dE}{dl}*dl$ 

= (q  $*dl)*\frac{dE}{dl}$ =p $*\frac{dE}{dl}$ =$-{10}^{-7}*$ ${10}^5$= ${-10}^{-2}$ N

The force is  N in the negative z-direction i.e. opposite to the direction of electric field. Hence, the angle between electric field and dipole moment is 180 $°$
.

Torque ( $\tau$ ) is given by the relation, $\tau$ = pE $\sin ?180°$ = 0

Therefore, the torque experienced by the system is zero.

1.25 Excess electrons on an oil drop, n = 12

Electric field intensity, E = 2.55 $*{10}^{4}N{C}^{-1}$$\mathit{}$

Density of oil, $\rho =$ 1.26 g / ${cm}^3$$$ = 1.26$*{10}^3$  g/ $m^3$

Acceleration due to gravity, g = 9.81 m/$s^2$ 

Charge of an electron, e = 1.60$*{10}^{-19}$C

Let the radius of the oil drop be r

Force (F) due to electric field (E) is equal to the weight of the oil drop (W)

F = W

Eq = mg

Ene = $\frac{4}{3}\pi r^3*\rho *g$

$$$r^3$ = $\frac{3*E*n*e}{4*\pi *\rho *g}$=$\frac{3*2.55\mathrm{}*{10}^4*12*1.60\mathrm{}*{10}^{-19}}{4*\pi *1.26*{10}^3*9.81}$

r = 9.815 $*{10}^{-7}$ m = 9.815 $*{10}^{-4}$ mm

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

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

1.19 Net electric flux ( ${\phi }_{Net}$ ) through the cubic surface is given by

${\phi }_{Net}$ = $\frac{q}{{?}_0}$

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

q= 2.0 μC

${\phi }_{Net}$ =$\frac{2*{10}^{-6}}{8.854\mathrm{}*{10}^{-12}}$ $C^{-1}N$ $m^2=2.25*{10}^5$