Class 12th

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

2.23 Requirements:

Capacitance, C = 2 $\mu F$

Potential difference, V = 1 kV = 1000 V

Available:

Each capacitor, capacitance, $C_1$ = 1 $\mu F$

Potential difference, $V_1$ = 400 V

Assumption:

A number of capacitors are connected in series and these series circuits are connected in parallel to each other. The potential difference across each row in parallel connection must be 1000 V and potential difference across each capacitor must be 400 V.

Hence, number of capacitors in each row is given as $\frac{1000}{250}$ = 2.5

So the number of capacitor in each row = 3

The equivalent capacitance in each row is given as, $\frac{1}{C_R}$ = $\frac{1}{1}+\frac{1}{1}$ + $\frac{1}{1}$ = 3, so $C_R=\frac{1}{3}$ $\mu$ F

Let there be 'n' rows, each having 3 capacitors, which are connected in parallel

The total equivalent capacitance of the circuit is given as

$C_{eff}$ = $\frac{1}{3}$ + $\frac{1}{3}$ + $\frac{1}{3}$ + $\frac{1}{3}$ + ……. upto n = $\frac{n}{3}$

It is given $C_{eff}$ = 2. So n = 6

So the final circuit will be 3 capacitors in each row and 6 rows connected in parallel, Total capacitors required = 3 $*6=18$

2.22 Four charges of same magnitudes are placed at points X, Y, Y and Z respectively, as shown in the figure.

It can be considered that the system of the electric quadrupole has three charges.

Charge +q placed at point X

Charge -2q placed at Y

Charge +q placed at point Z.

XY = YZ = a

YP = r

PX = r + a

PZ = r – a

Electrostatic potential caused by the system of three charges at point P is given by,

V = $\frac{1}{4\pi {\epsilon }_0}[\frac{q}{XP}$ - $\frac{2q}{YP}$ + $\frac{q}{ZP}$ ]

= $\frac{1}{4\pi {\epsilon }_0}[\frac{q}{r+a}$ - $\frac{2q}{r}$ + $\frac{q}{r-a}$ ]

= $\frac{q}{4\pi {\epsilon }_0}[\frac{r\left(r-a\right)-2\left(r-a\right)\left(r+a\right)+r(r+a)}{r\left(r-a\right)(r+a)}$ ]

= $\frac{q}{4\pi {\epsilon }_0}[\frac{r^2-ra-2r^2+2a^2+r^2+ra}{r\left(r-a\right)(r+a)}$ ]

= $\frac{q}{4\pi {\epsilon }_0}[\frac{2a^2}{r(r^2-a^2)}$ ]

= $\frac{2q{a}^2}{4\pi {\epsilon }_0{r}^3(1-\frac{a^2}{r^2})}$

Since $\frac{r}{a}?1,$ therefore $\frac{a}{r}?1$ . So $\frac{a^2}{r^2}$ is taken as negligible.

Therefore V = $\frac{2q{a}^2}{4\pi {\epsilon }_0{r}^3}$

It can be inferred that V $\propto \frac{1}{r^3}$ . It is known that for a dipole V $\propto \frac{1}{r^2}$ and for monopole V $\propto \frac{1}{r}$

2.21 Charge –q is located at (0,0,-a) and charge +q is located at (0,0,a). Hence, they form a dipole. Point (0,0,z) is on the axis of this dipole and point (x,y,0) is normal to the axis of the dipole. Hence electrostatic potential at point (x,y,0) is zero. Electrostatic potential at point (0,0,z) is given by,

V = $\frac{1}{4\pi {\epsilon }_0}$ ( $\frac{q}{z-a})+\frac{1}{4\pi {\epsilon }_0}$ ( $-\frac{q}{z+a})$ = $\frac{q(z+a-z+a)}{4\pi {\epsilon }_0(z^2-a^2)}$ = $\frac{2qa}{4\pi {\epsilon }_0(z^2-a^2)}$ = $\frac{p}{4\pi {\epsilon }_0(z^2-a^2)}$

Where ${\epsilon }_0$ = Permittivity of free space

p = Dipole moment of the system of two charges = 2qa

Distance r is much greater than half of the distance between the two charges. Hence, the potential (V) at a distance r is inversely proportional to the distance. i.e. V $\propto$ $\frac{1}{r^2}$

Electrostatic potential ( $V_1)$ at point (5,0,0) is given by

$V_1$ = $\frac{-q}{4\pi {\epsilon }_0}\frac{1}{\sqrt{{(5-0)}^2+{(-a)}^2}}+\frac{q}{4\pi {\epsilon }_0}\frac{1}{\sqrt{{(5-0)}^2+{\left(a\right)}^2}}$

= 0

Electrostatic potential ( $V_2)$ at point (-7,0,0) is given by

$V_2$ = $\frac{-q}{4\pi {\epsilon }_0}\frac{1}{\sqrt{{(-7)}^2+{(-a)}^2}}+\frac{q}{4\pi {\epsilon }_0}\frac{1}{\sqrt{{(-7)}^2+{\left(a\right)}^2}}$

= 0

Hence, no work is done in moving a small test charge from point (5,0,0) to point (-7,0,0) along the x-axis.

The answer does not change because work done by the electrostatic field in moving a test charge between the two points is independent of the path connecting the two points.