Class 12th

2.36 We do not get an electric shock as we step out of our house because the original equipotential surfaces of open air changes, keeping our body and the ground at the same potential.

Yes, the person will get an electric shock if he touches the metal slab next morning. The steady discharging current in the atmosphere charges up the aluminium sheet. As a result, its voltage rises gradually. The raise in the voltage depends on the capacitance of the capacitor formed by the aluminium slab and the ground.

The occurrence of thunderstorm and lightning charges the atmosphere continuously. Hence, even with the presence of discharging current of 1800 A, the atmosphere is not discharged completely. The two opposing currents are in equilibrium and the atmosphere remains electrically neutral.

During lightning and thunderstorm, light energy, heat energy and sound enegy are dissipated in the atmosphere.

2.33 Potential rating of the capacitor, V = 1 kV = 1000 V

Dielectric constant of a material, ${?}_r$ = 3

Dielectric strength = ${10}^7$ V/m

For safety, the field intensity should not cross 10% of the dielectric strength, hence

Electric field intensity, E = 10 % of ${10}^7$ = ${10}^6$ V/m

Capacitance of the parallel plate capacitor, C = 50 pF = 50 $*{10}^{-12}$ F

Distance between the plates, d is given by d = $\frac{V}{E}$ = $\frac{1000}{{10}^6}$ m = ${10}^{-3}$ m

Capacitance is given by the relation

C = $\frac{{\epsilon }_0{?}_{r}A}{d}$ , where ${\epsilon }_0$ = permittivity of free space = 8.854 $*{10}^{-12}$ $C^2{N}^{-1}$ $m^{-2}$

50 $*{10}^{-12}$ = $\frac{8.854\mathrm{}*{10}^{-12}*3*A}{{10}^{-3}}$

A = 1.88 $*{10}^{-3}$ $m^2$ = 18.8 ${cm}^2$

Hence, the area of each plate is about 19 ${cm}^2$

2.30 Radius of the inner sphere, $r_2$ = 12 cm = 0.12 m

Radius of the outer sphere, $r_1$ = 13 cm = 0.13 m

Charges on the inner sphere, q= 2.5 $\mu C=2.5*{10}^{-6}$ C

Dielectric constant of the liquid, k = 32

Capacitance of the capacitor is given by the relation, C = $\frac{4\pi {\epsilon }_0{kr}_1{r}_2}{(r_1-r_2)}$

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

$C=$ $\frac{4\pi *8.854\mathrm{}*{10}^{-12}*32*0.13*0.12}{(0.13-0.12)}$ = 5.55 $*{10}^{-9}$ F

Potential of the inner surface is given by

V = $\frac{q}{C}$ = $\frac{2.5*{10}^{-6}}{5.55\mathrm{}*{10}^{-9}}$ = 450 V

Radius of the isolate sphere, r = 12 cm = 12 $*{10}^{-2}$ m

Capacitance on the isolated sphere is given by C' = 4 $\pi {\epsilon }_0$ r

= 4 $*\pi *$ 8.854 $*{10}^{-12}*12\mathrm{}*{10}^{-2}$ F

= 1.34 $*{10}^{-11}$ F

The capacitance of the isolated sphere is less in comparison to the concentric spheres. This is because the outer sphere of the concentric sphere is earthed, so the potential difference is less and the capacitance is more than the isolated sphere.

2.27 Capacitance of the charged capacitor, $C_1$ = 4 $\mu F=4*{10}^{-6}$ F

Supply voltage, $V_1$ = 200 V

Electrostatic energy stored in the $C_1$ capacitor, $E_1$ = $\frac{1}{2}{C}_1{V_1}^2$ = $\frac{1}{2}*4*{10}^{-6}*{200}^2$ J

= 0.08 J

Capacitance of the uncharged capacitor, $C_2$ = 2 $\mu F$ = 2 $*{10}^{-6}$ F

When $C_2$ is connected to $C_1$ , the potential acquired by $C_2$ be $V_2$

From the conservation of energy, the charge acquired by $C_1$ becomes the charge acquired by $C_1$ and $C_2.$

Hence $V_2*\left(C_1+C_2\right)=C_1{V}_1$

or $V_2$ = $\frac{C_1{V}_1}{\left(C_1+C_2\right)}$ = $\frac{4*{10}^{-6}*200}{(4*{10}^{-6}+2\mathrm{}*{10}^{-6})}$ = $\frac{400}{3}$ V

Electrostatic energy of the combination of two capacitors is given by

$E_2$ = $\frac{1}{2}$ ( $C_1+C_2)*V_2^2$ = $\frac{1}{2}*$ ( $4*{10}^{-6}+$ 2 $*{10}^{-6})*{\left(\frac{400}{3}\right)}^2$ = 5.33 $*{10}^{-2}$ J

Hence the amount of electrostatic energy lost by capacitor $C_1=E_1-E_2$

= 0.08 - 5.33 $*{10}^{-2}$ J = 0.0267 J = 2.67 $*{10}^{-2}$ J