Physics Ncert Solutions Class 12th

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

2.26 Area of the parallel plate capacitor, A = 90 ${cm}^2$ = 90 $*{10}^{-4}$ $m^2$

Distance between plate, d = 2.5 mm = 2.5 $*{10}^{-3}$ m

Potential difference across plates, V = 400 V

Capacitance of the capacitor is given by, C = $\frac{{\epsilon }_{0}A}{d},$

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 $*{10}^{-12}$ $C^2{N}^{-1}$ $m^{-2}$

Electrostatic energy stored in the capacitor is given by the relation

$E_1$ = $\frac{1}{2}$ C $V^2$ = $\frac{1}{2}$ $*\frac{{\epsilon }_{0}A}{d}*V^2$ = $\frac{1}{2}$ $*\frac{8.854\mathrm{}*{10}^{-12}*90\mathrm{}*{10}^{-4}}{2.5\mathrm{}*{10}^{-3}}*{400}^2$ = 2.55 $*{10}^{-6}$ J

Volume of the given capacitor, V' = A $*d$ = 90 $*{10}^{-4}*$ 2.5 $*{10}^{-3}$ $m^3$

= 2.25 $*{10}^{-5}$ $m^3$

Energy stored is given by u = $\frac{E_1}{\mathrm{V}\mathrm{\text{'}}}$ = $\frac{2.55\mathrm{}*{10}^{-6}}{2.25\mathrm{}*{10}^{-5}}$ = 0.113 J/ $m^3$

Also, u = $\frac{E_1}{\mathrm{V}\mathrm{\text{'}}}$ = $\frac{\frac{1}{2}C{V}^2}{Ad}$ = $\frac{\frac{1}{2}\mathrm{}*\frac{{\epsilon }_{0}A}{d}*V^2}{Ad}$ = $\frac{1}{2}{\epsilon }_0{\left(\frac{V}{d}\right)}^2$ = $\frac{1}{2}{\epsilon }_0{\mathrm{E}}^2$ , where E = $\frac{V}{d}$ electric intensity

2.25 Let C' be the equivalent capacitance for capacitors $C_2$ and $C_3$ connected in series.

Hence, $\frac{1}{C\text{'}}$ = $\frac{1}{200}$ + $\frac{1}{200}$ . So C' = 100 pF

Capacitors C' and $C_1$ are in parallel, if the equivalent capacitance be C”, then

C” = C' + $C_1$ = 100 + 100 = 200 pF

Now C” and $C_4$ are connected in series. If the total equivalent capacitance of the circuit be $C_{Total}$ , then $\frac{1}{C_{Total}}$ = $\frac{1}{C"}$ + $\frac{1}{C_4}$ = $\frac{1}{200}$ + $\frac{1}{100}$ , $C_{Total}$ = $\frac{200}{3}$ pF = $\frac{200}{3}*{10}^{-12}$ F

Let V” be the potential difference across C” and $V_4$ be the potential difference across $C_4$

Then, V” + $V_4$ = 300 V

Now, charge $Q_4$ on $C_4$ is given by $Q_4$ = $C_{Total}*V$ = $\frac{200}{3}*{10}^{-12}$ $*300$ = 2 $*{10}^{-8}$ C

$V_4=\frac{Q_4}{C_4}$ = $\frac{2\mathrm{}*{10}^{-8}}{100*{10}^{-12}}$ = 200. So V” = 100 V

So potential difference across C' and $C_1$ is V” = 100 V

Charge on $C_1$ is given by $Q_1$ = 100 $*{10}^{-12}*100$ C = 1 $*{10}^{-8}$ C

$C_2$ and $C_3$ are having same capacitance and have a potential difference of 100V together. Since $C_2$ and $C_3$ are in series, the potential difference is given by $V_2$ = $V_3$ = 50V

Hence the charge on $C_2$ is given by $Q_2$ = 200 $*{10}^{-12}*50$ = ${10}^{-8}$ C and the charge on $C_3$ is given by $Q_3$ = 200 $*{10}^{-12}*50$ = ${10}^{-8}$ C

Therefore, the equivalent capacitance of the circuit is $\frac{200}{3}$ pF and charge and voltage at all capacitance is given as

For $C_{1:}{Q}_1$ = ${10}^{-8}$ C, $V_1$ = 100 V

For $C_{2:}{Q}_2$ = ${10}^{-8}$ C, $V_2$ = 50 V

For $C_{3:}{Q}_3$ = ${10}^{-8}$ C, $V_3$ = 50 V

For $C_{4:}{Q}_4$ = 2 $*{10}^{-8}$ C, $V_4$ = 200 V