Ncert Solutions Physics Class 12th

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

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

1.18 The square can be considered as one face of a cube of edge 10 cm with a centre where charge q is placed.

According to Gauss's theorem for a cube, total electric flux is through all its six faces.

  ${\phi }_{Tota\mathrm{l\; =}}$$\frac{q}{{?}_0}$ 

$\mathrm{H}\mathrm{e}\mathrm{n}\mathrm{c}\mathrm{e},\mathrm{}\mathrm{e}\mathrm{l}\mathrm{e}\mathrm{c}\mathrm{t}\mathrm{r}\mathrm{i}\mathrm{c}\mathrm{}\mathrm{f}\mathrm{l}\mathrm{u}\mathrm{x}\mathrm{}\mathrm{t}\mathrm{h}\mathrm{r}\mathrm{o}\mathrm{u}\mathrm{g}\mathrm{h}\mathrm{}\mathrm{o}\mathrm{n}\mathrm{e}\mathrm{}\mathrm{f}\mathrm{a}\mathrm{c}\mathrm{e}\mathrm{}\mathrm{o}\mathrm{f}\mathrm{}\mathrm{t}\mathrm{h}\mathrm{e}\mathrm{}\mathrm{c}\mathrm{u}\mathrm{b}\mathrm{e}$ i.e. through the square is $\phi =\frac{{\phi }_{Total}}{6}$ = $\frac{q}{{6?}_0}$ ,Where${?}_0$= Permittivity of free space = 8.854
 $*{10}^{-12}$ $C^2{N}^{-1}$$m^{-2}$

We have q = +10 $\mu C$

Hence,  = $\phi$ = $\frac{q}{{6?}_0}$ =$\frac{10*{10}^{-6}}{6*8.854\mathrm{}*{10}^{-12}}$= 188238.83 N$m^2{C}^{-1}$  = 1.88$*{10}^5$ N$m^2{C}^{-1}$  

1.17  (a) Net outward flux through the surface of the box , $\phi$ = 8.0 * ${10}^3$ ${\mathrm{Nm}}^2/$C

For a body containing net charge q, flux is given by the relation, 
 $\phi$ =$\frac{q}{{?}_0}$ 

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

Hence q =  $\phi$ $*{?}_0$= 8.0 *${10}^3*$ 8.854 $*{10}^{-12}$ C = 7.0832 $*{10}^{-8}$ C

= 7.0832 $*{10}^{-2}\mu C$

(b) Net flux piercing out through a body depends on the net charge contained in the body. If net flux is zero, then it can be inferred that the net charge inside the body is zero. The body may have equal amount of positive and negative charges.

1.16 When the cube side is oriented so that its faces are parallel to the coordinate planes, number of field lines entering the cube is equal to the number of field lines piercing out of the cube. A as a result, net flux through the cube is zero.