Ncert Solutions Physics Class 12th

2.16 Electric field on one side of a charged body is $E_1$ and electric field on the other side of the same body is $E_2.$ If infinite plane charged body has a uniform thickness, then electric field due to one surface of the charged body is given by,

$\stackrel{?}{E_1}$ = $\frac{\sigma }{2{\epsilon }_0}\stackrel{?}{n}$ ………….(i)

Where,

$\stackrel{?}{n}$ = unit vector normal to the surface at a point

$\sigma$ = surface charge density at that point

Electric field due to the other surface of the charged body is given by

$\stackrel{?}{E_2}$ = $\frac{\sigma }{2{\epsilon }_0}\stackrel{?}{n}$ ………….(ii)

$\mathrm{E}\mathrm{l}\mathrm{e}\mathrm{c}\mathrm{t}\mathrm{r}\mathrm{i}\mathrm{c}\mathrm{}\mathrm{f}\mathrm{i}\mathrm{e}\mathrm{l}\mathrm{d}\mathrm{}\mathrm{a}\mathrm{t}\mathrm{}\mathrm{a}\mathrm{n}\mathrm{y}\mathrm{}\mathrm{p}\mathrm{o}\mathrm{i}\mathrm{n}\mathrm{t}$ due to the two surfaces,

$\stackrel{?}{E_2}-\stackrel{?}{E_1}=\frac{\sigma }{2{\epsilon }_0}\stackrel{?}{n}+\frac{\sigma }{2{\epsilon }_0}\stackrel{?}{n}=\frac{\sigma }{{\epsilon }_0}\stackrel{?}{n}$ ……(iii)

$\mathrm{S}\mathrm{i}\mathrm{n}\mathrm{c}\mathrm{e}\mathrm{}\mathrm{i}\mathrm{n}\mathrm{s}\mathrm{i}\mathrm{d}\mathrm{e}\mathrm{}\mathrm{a}\mathrm{}\mathrm{c}\mathrm{l}\mathrm{o}\mathrm{s}\mathrm{e}\mathrm{d}\mathrm{}\mathrm{c}\mathrm{o}\mathrm{n}\mathrm{d}\mathrm{u}\mathrm{c}\mathrm{t}\mathrm{o}\mathrm{r},\mathrm{}\stackrel{?}{{\mathrm{E}}_1}=0,$

$\stackrel{?}{E}$ = $\stackrel{?}{E_2}$ = $\frac{\sigma }{{\epsilon }_0}\stackrel{?}{n}$

Therefore, the electric field just outside the conductor is $\frac{\sigma }{{\epsilon }_0}\stackrel{?}{n}$

When a charged particle is moved from one point to the other on a closed loop, the work done by the electrostatic field is zero. Hence, the tangential component of electrostatic field is continuous from one side of a charged surface to the other.

2.31 The force between two conducting spheres is not exactly given by the expression ${\mathit{Q}}_1{\mathit{Q}}_2$ /4 $\mathit{?}{\mathit{?}}_0{\mathit{r}}^2$ , because there is non-uniform charge distribution on the spheres.

Gauss's law will not be true, if Coulomb's law involved $\frac{1}{r^3}$ dependence, instead of $\frac{1}{r^2}$ , on r

Yes. If a small test charge is released at rest at a point in an electrostatic field configuration, then it will travel along the field lines passing through the point, only if the field lines are straight. This is because the field lines give the direction of acceleration and not the velocity.

Whenever the electron completes an orbit, either circular or elliptical, the work done by the field of a nucleus is zero.

No. Electric field is discontinuous across the surface of a charged conductor. However, electric potential is continuous.

The capacitance of a single conductor is considered as a parallel plate capacitor with one of its two plates at infinity.

Water has an unsymmetrical space as compared to mica. Since it has a permanent dipole moment, it has a greater dielectric constant than mica.

13.28 The equation for deuterium-tritium fusion is given as:

${}_1{}^{2}H+{}_1{}^{3}H\to {}_2{}^{4}He+n$

 It is given that

Mass of ( ${}_1{}^2\mathrm{H}),\mathrm{}{\mathrm{m}}_1$  = 2.014102 u

Mass of ( ${}_1{}^3\mathrm{H})$ , ${\mathrm{m}}_2=\mathrm{}$ 3.016049 u

Mass of ( ${}_2{}^{4}He),m_3$  = 4.002603 u

Mass of ( ${}_0{}^{1}n),m_4$ = 1.008665 u

Q-value of the given D-T reaction is:

Q = $\left[{\mathrm{m}}_1+{\mathrm{m}}_2-m_3-m_4\right]c^2$

= $\left[2.014102+3.016049-4.002603-1.008665\right]c^{2}u$

= 0.018883 $c^{2}u$

= 17.59 MeV

Radius of the deuterium and tritium, r $\approx 2.0fm$ = 2 $*{10}^{-15}$ m

Distance between the centers of the nucleus when they touch each other,

d = r +r = 4 $*{10}^{-15}$ m

Charge on the deuterium and tritium nucleus = e

Hence the repulsive potential energy between the two nuclei is given as:

V = $\frac{e^2}{4\pi {?}_{0}d}$

Where,

${?}_0$  = permittivity of free space

 It is given that $\frac{1}{4\pi {?}_0}$  = 9 $*{10}^9$ N $m^2{C}^{-2}$

Hence, V = $\frac{{(1.6*{10}^{-19})}^2}{4*{10}^{-15}}$ $*$  9 $*{10}^9$ = 5.76 $*{10}^{-14}$  J = $\frac{5.76*{10}^{-14}}{1.6*{10}^{-19}}$  eV = 360 keV

Hence, 360 keV of kinetic energy is needed to overcome the Coulomb repulsion between the two nuclei.

It is given that

KE = 2 $*\frac{3}{2}kT$

Where

k = Boltzmann constant = 1.38 $*{10}^{-23}$ $m^{2}kg{s}^{-2}{K}^{-1}$

T = Temperature required to trigger the reaction

Therefore T = $\frac{KE}{3k}$  = $\frac{5.76*{10}^{-14}}{3*1.38*{10}^{-23}}$  = 1.39 $*{10}^9$  K

Hence the gas must be heated to 1.39 $*{10}^9$ K to initiate the reaction.