Physics 5 Marks Questions

By loop method application in the closed circuit ABCabDA

$-12+\frac{2\text{Q}}{2\mu\text{F}}=\frac{\text{Q}_1}{2\mu\text{F}}+\frac{\text{Q}_1}{4\mu\text{F}}\ \dots(1)$

In the close circuit ABCDA

$-12+\frac{\text{Q}}{2\mu\text{F}}+\frac{\text{Q}+\text{Q}_1}{4\mu\text{F}}=0\ \dots(2)$

From (1) and (2) 2Q + 3Q₁ = 48 ...(3)

And 3Q - q₁ = 48 and subtracting Q = 4Q₁, and substitution in equation

$2\text{Q}+3\text{Q}_1=48$

$\Rightarrow8\text{Q}_1+3\text{Q}_1=48$

$\Rightarrow11\text{Q}_1=48,\ \text{Q}_1=\frac{48}{11}$

$\text{V}_{\text{ab}}=\frac{\text{Q}_1}{4\mu\text{F}}=\frac{48}{11\times4}=\frac{12}{11}\text{V}$

The potential = 24 - 12 = 12

Potential difference $\text{V}=\frac{(2\times0+12\times4)}{2+4}=\frac{48}{6}=8\text{V}$

$\therefore$ The V_a - V_b = -8V

From the figure it is cleared that the left and right branch are symmetry and reversed, so the current go towards BE from BAFEB same as the current from EDCBE.

$\therefore$ The net charge Q = 0

$\therefore\text{V}=\frac{\text{Q}}{\text{C}}=\frac{0}{\text{C}}=0$

$\therefore\text{V}_{\text{ab}}=0$

$\therefore$ The potential at K is zero.

The net potential $=\frac{\text{Net charge}}{\text{Net capacitance}}=\frac{24+24+24}{7}$

$=\frac{72}{7}=10.3\text{V}$

$\therefore\text{V}_{\text{a}}-\text{V}_{\text{b}}=-10.3\text{V}$

By star Delta conversion

$\text{C}_{\text{eff}}=\frac{3}{8}+\Bigg[\frac{\big(3+\frac{1}{2}\big)\times\big(\frac{3}{2}+1\big)}{\big(3+\frac{1}{2}\big)+\big(\frac{3}2{+1\big)}}\Bigg]$

$=\frac{3}{8}+\frac{35}{24}=\frac{9+35}{24}=\frac{11}{6}\mu\text{F}$

$=\frac{3}{8}+\frac{16}{8}+\frac{3}{8}=\frac{11}{4}\mu\text{F}$

$\text{C}_{\text{ef}}=\frac{4}{3}+\frac{8}{3}+4=8\mu\text{F}$

$\text{Cef}=\frac{3}{8}+\frac{32}{12}+\frac{32}{12}+\frac{8}{6}$

$=\frac{16+32}{6}=8\mu\text{F}$

1. $\therefore$ C₁, C₁ are series & C₂, C₂ are series as the V is same at p & q. So no current pass through p & q.

$\frac{1}{\text{C}}=\frac{1}{\text{C}_1}=\frac{1}{\text{C}_2}$

$\Rightarrow\frac{1}{\text{C}}=\frac{1+1}{\text{C}_1\text{C}_2}$

$\text{C}_{\text{p}}=\frac{\text{C}_1}{2}=\frac{4}{2}=2\mu\text{F}$

And $\text{C}_{\text{q}}=\frac{\text{C}_2}{2}=\frac{6}{2}=3\mu\text{F}$

$\therefore\text{C}=\text{C}_{\text{p}}+\text{C}_{\text{q}}=2+3=5\mu\text{F}$

2. $\text{C}_1=4\mu\text{F},\ \text{C}_2=6\mu\text{F}$

In case of p & q, q = 0

$\therefore\text{C}_{\text{p}}=\frac{\text{C}_1}{2}=\frac{4}{2}=2\mu\text{F}$

$\text{C}_{\text{q}}=\frac{\text{C}_2}{2}=\frac{6}{2}=3\mu\text{F}$

& $\text{C}'=2+3=5\mu\text{F}$

$\text{C}\ \&\ \text{C}' =5\mu\text{F}$

$\therefore$ The equation of capacitor $\text{C}=\text{C}'+\text{C}''=5+5=10\mu\text{F}$

Initially when switch ‘s’ is closed

Total Initial Energy $=\Big(\frac{1}{2}\Big)\text{CV}^2+\Big(\frac{1}{2}\Big)\text{CV}^2=\text{CV}^2\ \dots(1)$

When switch is open the capacitance in each of capacitors varies, hence the energy also varies.

i.e. in case of ‘B’, the charge remains.

Same i.e. cv

$\text{C}_{\text{eff}}=3\text{C}$

$\text{E}=\frac{1}{2}\times\frac{\text{q}^2}{\text{c}}$

$=\frac{1}{2}\times\frac{\text{c}^2\text{v}^2}{3\text{c}}=\frac{\text{cv}^2}{6}$

In case of 'A'

$\text{C}_{\text{eff}}=3\text{c}$

$\text{E}=\frac{1}{2}\times\text{C}_{\text{eff}}\text{v}^2$

$=\frac{1}{2}\times\text{3c}\times\text{v}^2=\frac{3}{2}\text{cv}^2$

Total final energy $=\frac{\text{cv}^2}{6}+\frac{3\text{cv}^2}{2}=\frac{10\text{cv}^2}{6}$

Now, $\frac{\text{Initial Energy}}{\text{Final Energy}}=\frac{\text{cv}^2}{\frac{10\text{cv}^2}{6}}=3$

= C₅ and C₁ are in series

$\text{C}_{\text{eq}}=\frac{2\times2}{2+2}=1$

This is parallel to C₆ = 1 + 1 = 2

Which is series to $\text{C}_{\text{2}}=\frac{2\times2}{2+2}=1$

Which is parallel to C₇ = 1 + 1 = 2

Which is series to $\text{C}_{\text{3}}=\frac{2\times2}{2+2}=1$

Which is parallel to C₈ = 1 + 1 = 2

This is series to $\text{C}_{\text{4}}=\frac{2\times2}{2+2}=1$

A = 20cm × 20cm = 4 × 10^-2m

d = 1m = 1 × 10^-3m

k = 4

t = d

$\text{C}=\frac{\in_0\text{A}}{\text{d}-\text{t}+\frac{\text{t}}{\text{k}}}=\frac{\in_0\text{A}}{\text{d}-\text{d}+\frac{\text{d}}{\text{k}}}=\frac{\in_0\text{Ak}}{\text{d}}$

$=\frac{8.85\times10^{-12}\times4\times10^{-2}\times4}{1\times10^{-3}}$

$=141.6\times10^{-9}\text{F}=1.42\text{nf}$

The capacitance of the outer sphere $=2.2\mu\text{F}$

$\text{C}=2.2\mu\text{F}$

Potential, V = 10v

Let the charge given to individual cylinder = q.

$\text{C}=\frac{\text{q}}{\text{V}}$

$\Rightarrow\text{q}=\text{CV}=2.2\times10=22\mu\text{F}$

$\therefore$ The total charge given to the inner cylinder $=22+22=44\mu\text{F}$

1. Capacitor $=\frac{4\times8}{4+8}=\frac{8}{3}\mu\text{F}$

and $=\frac{6\times3}{6+3}=2\mu\text{F}$

1. The charge on the capacitance $\frac{8}{3}\mu\text{F}$

$\therefore\text{Q}=\frac{8}{3}\times50=\frac{400}{3}$

$\therefore$ The potential at $4\mu\text{F}=\frac{400}{3\times4}=\frac{100}{3}$

at $8\mu\text{F}=\frac{400}{3\times8}=\frac{100}{6}$

The Potential difference $=\frac{100}{3}-\frac{100}{6}=\frac{50}{3}\mu\text{V}$

2. Hence the effective charge at $2\mu\text{F}=50\times2=100\mu\text{F}$

$\therefore$ Potential at $3\mu\text{F}=\frac{100}{3};$ Potential at $6\mu\text{F}=\frac{100}{6}$

$\therefore$ Difference $=\frac{100}{3}-\frac{100}{6}=\frac{50}{3}\mu\text{V}$

$\therefore$ The potential at C & D is $\frac{50}{3}\mu\text{V}$

2. $\therefore\frac{\text{P}}{\text{Q}}=\frac{\text{R}}{\text{S}}=\frac{1}2{}=\frac{1}{2}=$ It is balanced. So, from it is cleared that the wheat star bridge balanced.

So, the potential at the point C & D are same. So, no current flow through the point C & D.

So, if we connect another capacitor at the point C & D the charge on the capacitor is zero.

1. When charge of $1\mu\text{C}$ is introduced to the B plate, we also get $0.5\mu\text{C}$ charge on the upper surface of Plate ‘A’.
2. Given $\text{C}=50\mu\text{F}=50\times10^{-9}\text{F}=5\times10^{-8}\text{F}$

Now charge $=0.5\times10^{-6}\text{C}$

$\text{V}=\frac{\text{q}}{\text{C}}=\frac{5\times10^{-7}\text{C}}{5\times10^{-8}\text{F}}=10\text{V}$

1. Before reconnection

$\text{C}=100\mu\text{f}$

$\text{V}=24\text{V}$

$\text{q}=\text{CV}=2400\mu\text{C}$ (Before reconnection)

After connection

When $\text{C}=100\mu\text{f}$

$\text{V}=12\text{V}$

$\text{q}=\text{CV}=1200\mu\text{C}$ (After connection)

2. C = 100, V = 12V

$\therefore$ q = CV = 1200v

3. We know $\text{V}=\frac{\text{W}}{\text{q}}$

W = vq = 12 × 1200 = 14400 J = 14.4 mJ

The work done on the battery.

4. Initial electrostatic field energy $\text{U}_{\text{i}}=\Big(\frac{1}{2}\Big)\text{CV}_1^2$

Final Electrostatic field energy $\text{U}_{\text{f}}=\Big(\frac{1}{2}\Big)\text{CV}_2^2$

$\therefore$ Decrease in Electrostatic

Field energy $=\Big(\frac{1}2{}\Big)\text{CV}_1^2-\Big(\frac{1}2{}\Big)\text{CV}_2^2$

$=\Big(\frac{1}{2}\Big)\text{C}(\text{V}_1^2-\text{V}_2^2)$

$=\Big(\frac{1}{2}\Big)\times100(576-144)=21600\text{J}$

$\therefore$ Energy = 21600j = 21.6mJ

5. After reconnection

$\text{C}=100\mu\text{c},\ \text{V}=12\text{v}$

$\therefore$ The energy appeared $=\Big(\frac{1}{2}\Big)\times100\times144$

$=7200\text{J}=7.2\text{mJ}$ 

This amount of energy is developed as heat when the charge flow through the capacitor.

The acceleration of electron $\text{a}_{\text{e}}=\frac{\text{qeme}}{\text{Me}}$

The acceleration of proton $=\frac{\text{qpe}}{\text{Mp}}=\text{ap}$

The distance travelled by proton $\text{X}=\frac{1}{2}\text{apt}^2\ \dots(1)$

The distance travelled by electron ...(2)

From (1) and (2)

$\Rightarrow2-\text{X}=\frac{1}{2}\text{a}_{\text{c}}\text{t}^2$

$\text{x}=\frac{1}{2}\text{a}_{\text{c}}\text{t}^2$

$\Rightarrow\frac{\text{x}}{2-\text{x}}=\frac{\text{a}_{\text{p}}}{\text{a}_{\text{c}}}=\frac{\big(\frac{\text{q}_{\text{p}\text{E}}}{\text{M}_{\text{p}}}\big)}{\big(\frac{\text{q}_{\text{c}\text{F}}}{\text{M}_{\text{c}}}\big)}$

$=\frac{\text{x}}{2-\text{x}}=\frac{\text{M}_{\text{c}}}{\text{M}_{\text{p}}}=\frac{9.1\times10^{-31}}{1.67\times10^{-27}}$

$=\frac{9.1}{1.67}\times10^{-4}=5.449\times10^{-4}$

$\Rightarrow\text{x}=10.898\times10^{-4}-5.449\times10^{-4}\text{x}$

$\Rightarrow\text{x}=\frac{10.898\times10^{-4}}{1.0005449}=0.001089226$

Let the capacitances be C₁ & C₂ net capacitance $\text{C}=\frac{\text{C}_1\text{C}_2}{\text{C}_1+\text{C}_2}$

Now, $\text{C}_1=\frac{\in_0\text{Ak}_1}{\text{d}_1}$

$\text{C}_2=\frac{\in_0\text{Ak}_2}{\text{d}_2}$

$\text{C}=\frac{\frac{\in_0\text{Ak}_1}{\text{d}_1}\times\frac{\in_0\text{Ak}_2}{\text{d}_2}}{\frac{\in_0\text{Ak}_1}{\text{d}_1}+\frac{\in_0\text{Ak}_2}{\text{d}_2}}$

$=\frac{\in_0\text{A}\Big(\frac{\text{k}_1\text{k}_2}{\text{d}_1\text{d}_2}\Big)}{\in_0\text{A}\Big(\frac{\text{k}_1\text{d}_2+\text{k}_2\text{d}_1}{\text{d}_1\text{d}_2}\Big)}$

$=\frac{8.85\times10^{-12}\times10^{-2}\times24}{6\times4\times10^{-3}+4\times6\times10^{-3}}$

$=4.425\times10^{-11}\text{C}=44.25\text{pc}.$

Let the equivalent capacitance be C. Since it is an infinite series. So, there will be negligible change if the arrangement is done an in Fig–II

$\text{C}_{\text{eq}}=\frac{2\times\text{C}}{2+\text{C}}+1$

$\Rightarrow\text{C}=\frac{2\text{C}+2+\text{C}}{2+\text{C}}$

$\Rightarrow(2+\text{C})\times\text{C}=3\text{C}+2$

$\Rightarrow\text{C}^2-\text{C}-2=0$

$\Rightarrow(\text{C}-2)(\text{C}+1)=0$

$\text{C}=-1$ (Impossible)

So, $\text{C}=2\mu\text{F}$

$\text{C}=\frac{\in_0\text{A}}{\text{d}-\text{t}+\frac{\text{t}}{\text{k}}}$

$=\frac{8.85\times10^{-12}\times4\times10^{-2}}{10^{-3}-5\times10^{-4}+\frac{5\times10^{-4}}{5}}$

$=\frac{35.4\times10^{-4}}{10^{-3}-0.5}$

C and $4\mu\text{F}$ are in series

So, $\text{C}_1=\frac{4\times\text{C}}{4+\text{C}}$

Then C₁ and $2\mu\text{F}$ are parallel

$\text{C}=\text{C}_1+2\mu\text{F}$

$\Rightarrow\frac{4\times\text{C}}{4+\text{C}}+2$

$\Rightarrow\frac{4\text{C}+8+2\text{C}}{4+\text{C}}=\text{C}$

$\Rightarrow4\text{C}+8+2\text{C}$

$=4\text{C}+\text{C}^2=\text{C}^2-2\text{C}-8=0$

$\text{C}=\frac{2\pm\sqrt{4+4\times1\times8}}{2}$

$=\frac{2\pm\sqrt{36}}{2}=\frac{2\pm6}{2}$

$\text{C}=\frac{2+6}{2}=4\mu\text{F}$

$\therefore$ The value of C is $4\mu\text{F}$

1. $\text{C}_1=5\mu\text{f}$

$\text{V}_1=24\text{V}$

$\text{q}_1=\text{C}_1\text{V}_1=5\times24=120\mu\text{c}$

and $\text{C}_2=6\mu\text{f}$

$\text{V}_2=\text{R}$

$\text{q}_2=\text{C}_2\text{V}_2=6\times12=72$

$\therefore$ Energy stored on first capacitor

$\text{E}_1=\frac{1}2{}\frac{\text{q}_1^2}{\text{C}_1}=\frac{1}{2}\times\frac{(120)^2}{2}=1440\text{J}=1.44\text{mJ}$

Energy stored on 2^nd capacitor

$\text{E}_2=\frac{1}2{}\frac{\text{q}_2^2}{\text{C}_1}=\frac{1}{2}\times\frac{(72)^2}{6}=432\text{J}=4.32\text{mJ}$

2. C₁V₁

C₂V₂

Let the effective potential = V

$\text{V}=\frac{\text{C}_1\text{V}_1-\text{C}_2\text{V}_2}{\text{C}_1+\text{C}_2}=\frac{120-72}{5+6}=4.36$

The new charge $\text{C}_1\text{V}=5\times4.36=21.8\mu\text{c}$

and $\text{C}_2\text{V}=6\times4.36=26.2\mu\text{c}$

3. $\text{U}_1=\Big(\frac{1}{2}\Big)\text{C}_1\text{V}^2$

$\text{U}_2=\Big(\frac{1}{2}\Big)\text{C}_2\text{V}^2$

$\text{U}_{\text{f}}=\Big(\frac{1}{2}\Big)\text{V}^2(\text{C}_1+\text{C}_2)$

$=\Big(\frac{1}{2}\Big)(4.36)^2(5+6)$

$=104.5\times10^{-6}\text{J}=0.1045\text{mJ}$

But $\text{U}_{\text{i}}=1.44+0.433=1.873$

$\therefore$ The loss in KE = 1.873 - 0.1045 = 1.7687 = 1.77mJ

Initial charge stored $=50\mu\text{c}$

Let the dielectric constant of the material induced be ‘k’.

Now, when the extra charge flown through battery is 100.

So, net charge stored in capacitor $=150\mu\text{c}$

Now, $\text{C}_1=\frac{\in_0\text{A}}{\text{d}}$ or $\frac{\text{q}_1}{\text{V}}=\frac{\in_0\text{A}}{\text{d}}\ \dots(1)$

$\text{C}_2=\frac{\in_0\text{Ak}}{\text{d}}$ or $\frac{\text{q}_2}{\text{V}}=\frac{\in_0\text{Ak}}{\text{d}}\ \dots(2)$

Dividing (1) and (2) we get $\frac{\text{q}_1}{\text{q}_2}=\frac{1}{\text{k}}$

$\Rightarrow\frac{50}{150}=\frac{1}{\text{k}}$

$\Rightarrow\text{k}=3$

$\therefore$ The equivalent capacity

$\text{C}=\frac{\text{C}_1\text{C}_2\text{C}_3}{\text{C}_2\text{C}_3+\text{C}_1\text{C}_3+\text{C}_1\text{C}_2}$

$=\frac{20\times30\times40}{30\times40+20\times40+20\times30}=\frac{24000}{2600}=9.23\mu\text{F}$

$\text{C}=\frac{\text{q}}{\text{V}}$

$\Rightarrow\text{q}=\text{C}\times\text{V}=9.23\times12=110\mu\text{C}$ on each.

$\text{C}_{\text{eff}}=\frac{\text{2C}}{3}\text{q}=\frac{\text{2C}}{3}\times50$

$=\frac{5}{2}\times10^{-4}=1.66\times10^{-4}\text{C}$

$\therefore\text{C}=2\mu\text{F}$

$\therefore$ In this system the capacitance are arranged in series. Then the capacitance is parallel to each other.

$\therefore$ The equation of capacitance in one row

$\text{C}=\frac{\text{C}}{3}$

Here,

Plate area = 100cm² = 10^-2m²

Separation d = 5cm = 5 × 10^-3m

Thickness of metal t = 4cm = 4 × 10^-3m

$\text{C}=\frac{\in_0\text{A}}{\text{d}-\text{t}+\frac{\text{t}}{\text{k}}}=\frac{\in_0\text{A}}{\text{d}-\text{t}}$

$=\frac{8.585\times10^{-12}\times10^{-2}}{(5-4)\times10^{-3}}=88\text{pf}$

Here the capacitance is independent of the position of metal. At any position the net separation is d - t.

As d is the separation and t is the thickness.

Plate area A = 25cm² = 2.5 × 10^-3m

Separation d = 2mm = 2 × 10^-3m

Potential v = 12v

1. We know

$\text{C}=\frac{\in_0\text{A}}{\text{d}}$

$=\frac{8.85\times10^{-12}\times2.5\times10^{-3}}{2\times10^{-3}}$

$=11.06\times10^{-12}\text{F}$

$\text{C}=\frac{\text{q}}{\text{v}}$

$\Rightarrow11.06\times10^{-12}=\frac{\text{q}}{12}$

$\Rightarrow\text{q}_1=1.32\times10^{-10}\text{C}.$

2. Then d = decreased to 1mm

$\therefore\ \text{d}=1\text{mm}=1\times10^{-3}\text{m}$

$\text{C}=\frac{\in_0\text{A}}{\text{d}}=\frac{\text{q}}{\text{v}}$

$=\frac{8.85\times10^{-12}\times2.5\times10^{-3}}{1\times10^{-3}}=\frac{2}{12}$

$\Rightarrow\text{q}_2=8.85\times2.5\times12\times10^{-12}$

$\Rightarrow\text{q}_2=2.65\times10^{-10}\text{C}.$

$\therefore$ The extra charge given to plate = (2.65 - 1.32) × 10^-10 = 1.33 × 10^-10C.

Consider an elemental capacitor of with dx our at a distance ‘x’ from one end. It is constituted of two capacitor elements of dielectric constants k₁ and k₂ with plate separation $\text{x}\tan\phi$ and $\text{d}-\text{x}\tan\phi$ respectively in series.

$\frac{1}{\text{dcR}}=\frac{1}{\text{dc}_1}+\frac{1}{\text{dc}_2}=\frac{\text{x}\tan\phi}{\in_0\text{k}_2(\text{bdx})}+\frac{\text{d}-\text{x}\tan\phi}{\in_0\text{k}_1(\text{bdx})}$

$\text{dcR}=\frac{\in_0\text{bdx}}{\frac{\text{x}\tan\phi}{\text{k}_2}+\frac{(\text{d}-\text{x}\tan\phi)}{\text{k}_1}}$

$\text{C}_{\text{R}}=\in_0\text{bk}_1\text{k}_2\int\frac{\text{dx}}{\text{k}_2\text{d}+(\text{k}_1+\text{k}_2)\text{x}\tan\phi}$

$=\frac{\in_0\text{bk}_1\text{k}_2}{\tan\phi(\text{k}_1-\text{k}_2)}[\log_{\text{e}}\text{k}_2\text{d}+(\text{k}_1+\text{k}_2)\text{x}\tan\phi]\text{a}$

$=\frac{\in_0\text{bk}_1\text{k}_2}{\tan\phi(\text{k}_1-\text{k}_2)}[\log_{\text{e}}\text{k}_2\text{d}+(\text{k}_1+\text{k}_2)\text{a}\tan\phi-\log_{\text{e}}\text{k}_2\text{d}]$

$\therefore\tan\phi=\frac{\text{c}}{\text{a}}$ and $\text{A}=\text{a}\times\text{a}$

$\text{C}_{\text{R}}=\frac{\in_0\text{ak}_1\text{k}_2}{\frac{\text{d}}{\text{a}}(\text{k}_1-\text{k}_2)}$ $\Big[\log_{\text{e}}\Big(\frac{\text{k}_1}{\text{k}_2}\Big)\Big]$

$\text{C}_{\text{R}}=\frac{\in_0\text{a}^2\text{k}_1\text{k}_2}{\text{d}(\text{k}_1-\text{k}_2)}$ $\Big[\log_{\text{e}}\Big(\frac{\text{k}_1}{\text{k}_2}\Big)\Big]$

$\text{C}_{\text{R}}=\frac{\in_0\text{a}^2\text{k}_1\text{k}_2}{\text{d}(\text{k}_1-\text{k}_2)}$ $\text{ln}\frac{\text{k}_1}{\text{k}_2}$

1. Area = A

Separation = d

$\text{C}_1=\frac{\in_0\text{Ak}_1}{\frac{\text{d}}{2}}$

$\text{C}_2=\frac{\in_0\text{Ak}_2}{\frac{\text{d}}{2}}$

$\text{C}=\frac{\text{C}_1\text{C}_2}{\text{C}_1+\text{C}_2}=\frac{\frac{2\in_0\text{Ak}_1}{\text{d}}\times\frac{2\in_0\text{Ak}_2}{\text{d}}}{\frac{2\in_0\text{Ak}_1}{\text{d}}+\frac{2\in_0\text{Ak}_2}{\text{d}}}$

$=\frac{\frac{(2\in_0\text{A})^2\text{k}_1\text{k}_2}{\text{d}^2}}{(2\in_0\text{A})\frac{\text{k}_1\text{d+}\text{k}_2\text{d}}{\text{d}^2}}=\frac{2\text{k}_1\text{k}_2\in_0\text{A}}{\text{d}(\text{k}_1+\text{k}_2)}$

2. $\frac{1}{\text{C}}=\frac{1}{\text{C}_1}+\frac{1}{\text{C}_2}+\frac{1}{\text{C}_3}$

$=\frac{1}{\frac{3\in_0\text{Ak}_1}{\text{d}}}+\frac{1}{\frac{3\in_0\text{Ak}_2}{\text{d}}}+\frac{1}{\frac{3\in_0\text{Ak}_3}{\text{d}}}$

$=\frac{\text{d}}{3\in_0\text{A}}\Big[\frac{1}{\text{k}_1}+\frac{1}{\text{k}_2}+\frac{1}{\text{k}_3}\Big]$

$=\frac{\text{d}}{3\in_0\text{A}}\Big[\frac{\text{k}_2\text{k}_3+\text{k}_1\text{k}_3+\text{k}_1\text{k}_2}{\text{k}_1\text{k}_2\text{k}_3}\Big]$

$\therefore\text{C}=\frac{3\in_0\text{A}\text{k}_1\text{k}_2\text{k}_3}{\text{d}(\text{k}_1\text{k}_2+\text{k}_2\text{k}_3+\text{k}_1\text{k}_3)}$

3. $\text{C}=\text{C}_1+\text{C}_2$

$=\frac{\in_0\frac{\text{A}}{2}\text{k}_1}{\text{d}}+\frac{\in_0\frac{\text{A}}{2}\text{k}_2}{\text{d}}=\frac{\in_0\text{A}}{2\text{d}}(\text{k}_1+\text{k}_2)$

$\therefore$ The capacitance $\text{C}=\frac{\in_0\text{A}}{\text{d}}=\frac{8.85\times10^{-12}\times10^{-2}}{10^{-2}}=8.85\times10^{-12}$

$\therefore$ The energy stored $\text{C}_1=\Big(\frac{1}{2}\Big)\text{CV}^2$

$=\Big(\frac{1}{2}\Big)\times10^{-12}\times(24)^2$

$=2548.8\times10^{-12}$

$\therefore$ The forced attraction between the plates $=\frac{\text{C}_1}{\text{d}}=\frac{2548.8\times10^{-12}}{10^{-2}}=2.54\times10^{-7}\text{N.}$

$\text{Cab}=\frac{4\pi\in_0\text{abk}}{(\text{b}-\text{a})}$

$\text{Cbc}=\frac{4\pi\in_0\text{bc}}{(\text{c}-\text{b})}$

$\frac{1}{\text{C}}=\frac{1}{\text{Cab}}+\frac{1}{\text{Cbc}}$

$=\frac{(\text{b}-\text{a})}{4\pi\in_0\text{abk}}+\frac{(\text{c}-\text{b})}{4\pi\in_0\text{bc}}$

$=\frac{\text{c}(\text{b}-\text{a})+\text{ka}(\text{c}-\text{b})}{\text{k}4\pi\in_0\text{abc}}$

$\text{C}=\frac{\text{k}4\pi\in_0\text{abc}}{\text{c}(\text{b}-\text{a})+\text{ka}(\text{c}-\text{b})}$

$\text{C}_1=\frac{\text{k}\in_0\text{A}}{\ell\text{d}}$

$\text{C}_2=\frac{\in_0(\ell-\text{a})\text{A}}{\ell\text{d}}$

$\therefore$ Net capacitance $\text{C}=\text{C}_1+\text{C}_2$

$=\frac{\in_0\text{A}}{\ell\text{d}}[\text{ka}+(\ell-\text{a})]$

$\text{C}=\frac{\in_0\text{A}}{\ell\text{d}}[\ell+\text{a}(\text{k}-1)]$

$\therefore$ dQ = (dc)E

$\therefore$ Work done by the force during the displacement, dx = fdx

$\therefore$ Increase in energy stored in the capacitor

$\Rightarrow\Big(\frac{1}{2}\Big)(\text{dc})\text{E}^2=(\text{dc})\text{E}^2-\text{fdx}$

$\Rightarrow\text{fdx}=\Big(\frac{1}2{}\Big)(\text{dc})\text{E}^2$

$\Rightarrow\text{f}=\frac{1}{2}\frac{{\text{E}^2\text{dc}}}{\text{dx}}$

$\text{C}=\frac{\in_0\text{A}}{\ell\text{d}}[\ell+\text{a}(\text{k}-1)]$ (Here x = a)

$\Rightarrow\frac{\text{dc}}{\text{da}}=\frac{-\text{d}}{\text{da}}\Big[\frac{\in_0\text{A}}{\ell\text{d}}\big\{\ell+\text{a}(\text{k}-1)\big\}\Big]$

$\Rightarrow\frac{\in_0\text{A}}{\ell\text{d}}(\text{k}-1)=\frac{\text{dc}}{\text{dx}}$

$\Rightarrow\text{f}=\frac{1}{2}\text{E}^2\frac{\text{dc}}{\text{dx}}=\frac{1}{2}\text{E}^2\Big\{\frac{\in_0\text{A}}{\ell\text{d}}(\text{k}-1)\Big\}$

$\therefore\text{a}_{\text{d}}=\frac{\text{f}}{\text{m}}=\frac{\text{E}^2\in_0\text{A}(\text{k}-1)}{2\ell\text{dm}}$ $\Big[\therefore(\ell-\text{a})=\frac{1}{2}\text{a}_{\text{d}}\text{t}^2\Big]$

$\Rightarrow\text{t}=\sqrt{\frac{2(\ell-\text{a})}{\text{a}_{\text{d}}}}=\sqrt{\frac{2(\ell-\text{a})2\ell\text{dm}}{\text{E}^2\in_0\text{A}(\text{k}-1)}}$

$=\sqrt{\frac{4\text{m}\ell\text{d}(\ell-\text{a})}{\in_0\text{AE}^2(\text{k}-1)}}$

$\therefore$ Time period $=\text{2t}=\sqrt{\frac{\text{8m}\ell\text{d}(\ell-\text{a})}{\in_0\text{AE}^2(\text{k}-1)}}$

$\text{C}_{\text{i}}=\frac{\in_0\times2\times10^{-3}}{10^{-3}}=2\in_0$

$\text{C}_{\text{f}}=\frac{\in_0\times2\times10^{-3}}{2\times10^{-3}}=\in_0$

$\text{q}_{\text{i}}=24\in_0$

$\text{q}_{\text{f}}=12\in_0$

1. So, q flown out $12\in_0.$ i.e., q_i - q_f

So, q = 12 × 8.85 × 10^-12 = 106.2 × 10^-12C = 1.06 × 10^-10C

2. Energy absorbed by battery during the process

= q × v = 1.06 × 10^-10C × 12

= 12.7 × 10^-10J

3. Before the process

$\text{E}_{\text{i}}=\Big(\frac{1}{2}\Big)\times\text{C}_{\text{i}}\times\text{V}^2$

$=\Big(\frac{1}{2}\Big)\times2\times8.85\times10^{-12}\times144$

$=12.7\times10^{-10}\text{J}$

After the force

$\text{E}_{\text{f}}=\Big(\frac{1}{2}\Big)\times\text{C}_{\text{f}}\times\text{V}^2$

$=\Big(\frac{1}{2}\Big)\times8.85\times10^{-12}\times144$

$=6.35\times10^{-10}\text{J}$

4. Workdone = Force × Distance

$=\frac{1}{2}\frac{\text{q}^2}{\in_0\text{A}}=1\times10^{3}$

$=\frac{1}{2}\times\frac{12\times12\times\in_0\times\in_0\times10^{-3}}{\in_0\times2\times10^{-3}}$

5. From (c) and (d) we have calculated, the energy loss by the separation of plates is equal to the work done by the man on plate. Hence no heat is produced in transformer.

$\text{C}_1=\frac{4\pi\in_0\text{ab}}{\text{b}-\text{a}}$ $(\therefore$ for a spherical capacitor formed by two spheres of radii R₂ > R₁$)$

$\text{C}=\frac{4\pi\in_0\text{R}_2\text{R}_1}{\text{R}_2-\text{R}_1}$

$\text{C}_2=\frac{4\pi\in_0\text{bc}}{\text{c}-\text{b}}$

$\text{C}_{\text{eff}}=\frac{\text{C}_1\text{C}_2}{\text{C}_1+\text{C}_2}$

$=\frac{\frac{(4\pi\in_0)^2\text{ab}^2\text{c}}{(\text{b}-\text{a})(\text{c}-\text{a})}}{4\pi\in_0\Big[\frac{\text{ab}(\text{c}-\text{b})+\text{bc}(\text{b}-\text{a})}{(\text{b}-\text{a})(\text{c}-\text{b})}\Big]}$

$=\frac{4\pi\in_0\text{ab}^2\text{c}}{\text{abc}-\text{ab}^2+\text{b}^2\text{c}-\text{abc}}$

$=\frac{4\pi\in_0\text{ab}^2\text{c}}{\text{b}^2(\text{c}-\text{a})}$

$=\frac{4\pi\in_0\text{a}\text{c}}{\text{c}-\text{a}}$

$\text{C}=\frac{\in_0\text{A}}{\text{d}}\text{C}$

$\text{Q}=\frac{\in_0\text{AV}}{\text{d}}\text{C}$

$\text{C}=\frac{\in_0\text{A}}{\frac{\text{d}}{\text{k}}}=\frac{\in_0\text{Ak}}{\text{d}}$

$\text{Q}_1=\frac{\in_0\text{Ak}}{\text{d}}\text{V}$

$\text{Q}=\text{Q}_1-\text{Q}$

$=\frac{\in_0\text{AkV}}{\text{d}}-\frac{\in_0\text{AV}}{\text{d}}=\frac{\in_0\text{AV}}{\text{d}}(\text{k}-1)$

$=\frac{1}2{}\frac{\text{q}^2}{\text{C}}=\frac{1}{2}\frac{\frac{\in_0^2\text{A}^2\text{V}^2}{\text{d}^2}(\text{k}-1)^2}{\frac{\in_0\text{A}}{\text{d}}(\text{k}-1)}$

$=\frac{\in_0\text{AV}^2}{2\text{d}}(\text{k}-1)$

1. Since the switch was open for a long time, hence the charge flown must be due to the both, when the switch is closed.

$\text{C}_\text{ef}=\frac{\text{C}}{2}$

so $\text{q}=\frac{\text{E}\times\text{C}}{2}$

2. Workdone 

$=\text{q}\times\text{v}=\frac{\text{EC}}{2}\times\text{E}=\frac{\text{E}^2\text{C}}{2}$

3. $\text{E}_{\text{i}}=\frac{1}{2}\times\frac{\text{C}}{2}\times\text{E}^2=\frac{\text{E}^2\text{C}}{4}$

$\text{E}_{\text{f}}=\frac{1}{2}\times\text{C}\times\text{E}^2=\frac{\text{E}^2\text{C}}{2}$

$\text{E}_{\text{i}}-\text{E}_{\text{j}}=\frac{\text{E}^2\text{C}}{4}$

4. The net charge in the energy is wasted as heat.

In the figure the three capacitors are arranged in parallel.

All have same surface area $=\text{a}=\frac{\text{A}}{3}$

First capacitance $\text{C}_1=\frac{\in_0\text{A}}{3\text{d}}$

Second capacitance $\text{C}_2=\frac{\in_0\text{A}}{3(\text{b}+\text{d})}$

Third capacitance $\text{C}_3=\frac{\in_0\text{A}}{3(2\text{b}+\text{d})}$

$\text{C}_{\text{eq}}=\text{C}_1+\text{C}_2+\text{C}_3$

$=\frac{\in_0\text{A}}{3\text{d}}+\frac{\in_0\text{A}}{3(\text{b}+\text{d})}+\frac{\in_0\text{A}}{3(2\text{b}+\text{d})}$

$=\frac{\in_0\text{A}}{3}\Big(\frac{1}{\text{d}}+\frac{1}{\text{b}+\text{d}}+\frac{2}{\text{2b}+\text{d}}\Big)$

$=\frac{\in_0\text{A}}{3}\Big(\frac{(\text{b}+\text{d})(2\text{b}+\text{d})+(2\text{b}+\text{d})\text{d}+(\text{b}+\text{d})\text{d}}{\text{d}(\text{b}+\text{d})(2\text{b}+\text{d})}\Big)$

$=\frac{\in_0\text{A}\big(3\text{d}^2+6\text{bd}+2\text{b}^2\big)}{3\text{d}(\text{b}+\text{d})(2\text{b}+\text{d})}$

$\text{Cac}=\frac{4\pi\in_0\text{ack}}{\text{k}(\text{c}-\text{a})}$

$\text{Cbc}=\frac{4\pi\in_0\text{bc}}{(\text{b}-\text{c})}$

$\frac{1}{\text{C}}=\frac{1}{\text{Cac}}+\frac{1}{\text{Cbc}}$

$=\frac{(\text{c}-\text{a})}{4\pi\in_0\text{ack}}+\frac{(\text{b}-\text{c})}{4\pi\in_0\text{bc}}$

$=\frac{\text{b}(\text{c}-\text{a})+\text{ka}(\text{b}-\text{c})}{\text{k}4\pi\in_0\text{abc}}$

$\text{C}=\frac{4\pi\in_0\text{kabc}}{\text{ka}(\text{b}-\text{c})+\text{b}(\text{c}-\text{a})}$