Physics Ncert Solutions Class 12th

2.19

Let the charge in Proton 1 be $q_1$ = 1.6 $*{10}^{-19}$ C, in Proton 2, $q_2$ = 1.6 $*{10}^{-19}$ C and the charge in the electron, $q_3$ = $-$ 1.6 $*{10}^{-19}$ C

Distance between Proton 1 & Proton 2, $d_1$ = 1.5 Å = 1.5 $*{10}^{-10}$ m

Distance between Proton 1 and Electron, $d_2$ = 1.0 Å = 1.0 $*{10}^{-10}$ m

Distance between Proton 2 and Electron, $d_3$ = 1.0 Å = 1.0 $*{10}^{-10}$ m

The potential energy of the system,

V = $\frac{q_1{q}_2}{4\pi {\epsilon }_0{d}_1}$ + $\frac{q_2{q}_3}{4\pi {\epsilon }_0{d}_3}$ + $\frac{q_3{q}_1}{4\pi {\epsilon }_0{d}_2}$ ,

Where ${\mathrm{\epsilon }}_0=\mathrm{P}\mathrm{e}\mathrm{r}\mathrm{m}\mathrm{i}\mathrm{t}\mathrm{t}\mathrm{i}\mathrm{v}\mathrm{i}\mathrm{t}\mathrm{y}\mathrm{}\mathrm{o}\mathrm{f}\mathrm{}\mathrm{f}\mathrm{r}\mathrm{e}\mathrm{e}\mathrm{}\mathrm{s}\mathrm{p}\mathrm{a}\mathrm{c}\mathrm{e}$ = 8.854 $*{10}^{-12}$ $C^2{N}^{-1}$ $m^{-2}$

V = $\frac{1}{4\pi {\epsilon }_0}$ ( $\frac{q_1{q}_2}{d_1}$ + $\frac{q_2{q}_3}{d_2}$ + $\frac{q_3{q}_1}{d_2}$ )

= $\frac{1}{4\pi *8.854\mathrm{}*{10}^{-12}}*$ ( $\frac{1.6\mathrm{}*{10}^{-19}*1.6\mathrm{}*{10}^{-19}}{1.5\mathrm{}*{10}^{-10}}$ + $\frac{1.6\mathrm{}*{10}^{-19}*-1.6\mathrm{}*{10}^{-19}}{1.0\mathrm{}*{10}^{-10}}$ + $\frac{-1.6\mathrm{}*{10}^{-19}*1.6\mathrm{}*{10}^{-19}}{1.0\mathrm{}*{10}^{-10}}$ )

= 8.988 $*{10}^9*(1.707*{10}^{-28}$ $-2.56*{10}^{-28}-2.56*{10}^{-28})$

= $-3.067*{10}^{-18}$ J

= $-3.067*{10}^{-18}*\frac{1}{1.6\mathrm{}*{10}^{-19}}$ eV

= - 19.17 eV

2.18 The distance between electron – proton of a hydrogen atom, d = 0.53 Å = 0.53 $*{10}^{-10}$ m

Charge of an electron, $q_1$ = $-$ 1.6 $*{10}^{-19}$ C

Charge of a proton, $q_2$ = 1.6 $*{10}^{-19}$ C

Potential at infinity = 0

Potential energy of the system = Potential energy at infinity – Potential energy at a distance d

= 0 - $\frac{q_1{q}_2}{4\pi {\epsilon }_{0}d}$ , where ${\mathrm{\epsilon }}_0=\mathrm{P}\mathrm{e}\mathrm{r}\mathrm{m}\mathrm{i}\mathrm{t}\mathrm{t}\mathrm{i}\mathrm{v}\mathrm{i}\mathrm{t}\mathrm{y}\mathrm{}\mathrm{o}\mathrm{f}\mathrm{}\mathrm{f}\mathrm{r}\mathrm{e}\mathrm{e}\mathrm{}\mathrm{s}\mathrm{p}\mathrm{a}\mathrm{c}\mathrm{e}$ == 8.854 $*{10}^{-12}$ $C^2{N}^{-1}$ $m^{-2}$

= 0 - $\frac{1.6\mathrm{}*{10}^{-19}*1.6\mathrm{}*{10}^{-19}}{4\pi *8.854\mathrm{}*{10}^{-12}*0.53\mathrm{}*{10}^{-10}}$ = - 4.34 $*{10}^{-18}$ J = $\frac{4.34\mathrm{}*{10}^{-18}}{1.6\mathrm{}*{10}^{-19}}$ = - 27.13 eV

(Since 1 eV = 1.6 $*{10}^{-19}$ J)

Kinetic energy = $\frac{1}{2}$ of potential energy = $\frac{1}{2}$ $*-$ 27.13 eV = -13.57 eV

Total energy = - 13.57 – (-27.13) = 13.57 eV

Therefore, the minimum work required to free the electron is 13.57 eV

In case of zero potential energy, $d_1$ = 1.06 Å = 1.06 $*{10}^{-10}$ m

Potential energy of the system = Potential energy at $d_1$ - Potential energy at d

= $\frac{q_1{q}_2}{4\pi {\epsilon }_{0}d}$ $*\frac{1}{1.6\mathrm{}*{10}^{-19}}$ - 27.13 eV = $\frac{{(1.6\mathrm{}*{10}^{-19})}^2}{4\pi *8.854\mathrm{}*{10}^{-12}*1.06\mathrm{}*{10}^{-10}}*\frac{1}{1.6\mathrm{}*{10}^{-19}}-$ 27.13 = 13.56 -27.13 eV= -13.56 eV

2.15 Charge placed at the centre of the shell is +q. Hence a charge of magnitude of –q will be induced to the inner surface of the shell. Therefore, the total charge on the inner surface of the shell is –q.

Surface charge density at the inner surface of the shell is given by the relation,

${\sigma }_1=\frac{Totalcharge}{Innersurfacearea}=\frac{-q}{4\pi r_1^2}$ …………(1)

When a charge of magnitude of Q is placed on the outer shell and the radius of outer shell is $r_2$ . The total charge experienced by the outer shell will be Q + q. The surface charge density at the outer surface of the shell is given by

${\sigma }_2=\frac{Totalcharge}{Outersurfacearea}=\frac{Q+q}{4\pi r_2^2}$ …………(2)

The electric field intensity inside a cavity is zero, even if the shell is not spherical and has any irregular shape.

2.14

Two charges are at point A and B. O is the midpoint of line joining A & B

Magnitude of charge at point A, $q_1$ = 1.5 $\mu C$ = 1.5 $*{10}^{-6}C$ and

at point B, $q_2$ = 2.5 $\mu C=2.5*{10}^{-6}C$

Distance between A & B, d = 30 cm = 0.3 m

Let $V_1$ and $E_1$ be the electric potential and electric field respectively at O.

$V_1=$ Potential due to charge at A + Potential due to charge at B

$=\frac{q_1}{4\pi {\epsilon }_0\frac{d}{2}}$ + $\frac{q_2}{4\pi {\epsilon }_0\frac{d}{2}}$ = $\frac{1}{4\pi {\epsilon }_0\frac{d}{2}}$ ( $q_1+q_2)$

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

$V_1=$ $\frac{1}{4\pi *8.854\mathrm{}*{10}^{-12}*\frac{0.3}{2}}$ (1.5 $*{10}^{-6}+2.5*{10}^{-6})$ = 2.4 $*{10}^5$ V

$E_1=$ Electric field due to $q_2$ - Electric field due to $q_1$

$=$ $\frac{q_2}{4\pi {\epsilon }_0({\frac{d}{2})}^2}$ $-$ $\frac{q_1}{4\pi {\epsilon }_0({\frac{d}{2})}^2}$

$=$ $\frac{1}{4\pi {\epsilon }_0({\frac{d}{2})}^2}(q_2-q_1)$ = $\frac{1}{4\pi *8.854\mathrm{}*{10}^{-12}*({\frac{0.3}{2})}^2}(2.5*{10}^{-6}-1.5\mathrm{}*{10}^{-6})$

= 4.0 $*{10}^5$ V/m

Therefore, the potential at mid-point is 2.4 $*{10}^5$ V and the electric field is

4.0 $*{10}^5$ V/m and it is directed from larger charge to smaller charge.

 

Let us consider a point Z such that normal distance OZ = 10 cm = 0.1 m

Let $V_2$ and $E_2$ be the electric potential and electric field respectively at Z.

From the $?$ ABZ, we can write,

BZ = AZ = $\sqrt{{0.15}^2+{0.1}^2}$ = 0.18 m

$V_2=$ Potential due to charge at A + Potential due to charge at B

$=\frac{q_1}{4\pi {\epsilon }_{0}AZ}$ + $\frac{q_2}{4\pi {\epsilon }_{0}BZ}$ = $\frac{1}{4\pi {\epsilon }_{0}0.18}$ ( $q_1+q_2)$

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

$V_2=$ $\frac{1}{4\pi *8.854\mathrm{}*{10}^{-12}*0.18}$ (1.5 $*{10}^{-6}+2.5*{10}^{-6})$ = 2.0 $*{10}^5$ V

Electric field due to charge $q_{1}atZ,E_A$

$E_A=\frac{q_1}{4\pi {\epsilon }_0({AZ)}^2}$ = $\frac{1.5\mathrm{}*{10}^{-6}}{4*\pi *8.854\mathrm{}*{10}^{-12}*({0.18)}^2}$ = 0.416 $*{10}^6$ V/m

Electric field due to charge $q_{2}atZ,E_B$

$E_B=\frac{q_2}{4\pi {\epsilon }_0({BZ)}^2}$ = $\frac{2.5\mathrm{}*{10}^{-6}}{4*\pi *8.854\mathrm{}*{10}^{-12}*({0.18)}^2}$ = 0.693 $*{10}^6$ V/m

The resultant intensity at Z

$E_2$ = $\sqrt{{E_A}^2+{E_B}^2+2E_A{E}_{B}cos2\theta }$ where 2 $\theta$ is the $\angle AZB$

From the figure, we get $\cos ?\theta$ = $\frac{ZO}{ZB}$ = $\frac{0.1}{0.18}$

Hence $\theta$ = 56.25 $°$

Therefore, $E_2$ = $E_2=\sqrt{({0.416\mathrm{}*{10}^6)}^2+{(0.693\mathrm{}*{10}^6)}^2+2{\left(0.416\mathrm{}*{10}^6\right)*(0.693\mathrm{}*{10}^6)}_{}cos2*56.25\mathrm{}°}$ = 6.57 $*{10}^5$ V/m