Capacitor

C_eq = ${6 \over {13}}$$\mu $F

Therefore three capacitors most be in parallel to get 6 in

${1 \over {{C_{eq}}}} = {1 \over {3C}} + {1 \over C} + {1 \over C} + {1 \over C} + {1 \over C}$

C_eq = ${{3C} \over {13}}$ = ${6 \over {13}}$$\mu $F

6$\mu $F & 4$\mu $F are in parallel & total charge on this combination is 30 $\mu $C

$ \therefore $  Charge on 6$\mu $F capacitor = ${6 \over {6 + 4}} \times 30$

= 18 $\mu $C

Since charge is asked on right plate therefore is +18$\mu $C

Initial energy of capacitor

U_i = ${1 \over 2}$ ${{{v^2}} \over c}$

= ${1 \over 2}$ $ \times $ ${{120 \times 120} \over {12}}$ = 600 J

Since battery is disconnected so charge remain same.

Final energy of capacitor

U_f = ${1 \over 2}{{{v^2}} \over c}$

= ${1 \over 2} \times {{120 \times 120} \over {12 \times 6.5}}$ = 92

W + U_f = U_i

W = 508 J

Let dielectric constant of material used be K.

$ \therefore $  ${{10{ \in _0}\,A/3} \over d} + {{12{ \in _0}\,A/3} \over d} + {{14{ \in _0}\,A/3} \over d} = {{K{ \in _0}A} \over d}$

$ \Rightarrow $    K $=$ 12

Here, ${C_1} = {{{K_1}{\varepsilon _0}.L.{L \over 2}} \over {{d \over 2}}} = {{{K_1}{\varepsilon _0}{L^2}} \over d}$

${C_2}{{{K_2}{\varepsilon _0}.L.{L \over 2}} \over {{d \over 2}}} = {{{K_2}{\varepsilon _0}{L^2}} \over d}$

${C_3} = {{{K_3}{\varepsilon _0}{L^2}} \over d}$

${C_4} = {{{K_4}{\varepsilon _0}{L^2}} \over d}$

Here  C₁ C₂ are in series and C₃, C₄ are in series.

Equivalent capacitance point A and B is,

C_eq $=$ ${{{C_1}{C_2}} \over {{C_1} + {C_2}}} + {{{C_3}{C_4}} \over {{C_3} + {C_4}}}$

$ = {C_{12}} + {C_{34}}$

${C_{12}}\,\, = \,\,{{{{{K_1}{\varepsilon _0}{L^2}} \over d} \times {{{K_2}{\varepsilon _0}{L^2}} \over d}} \over {{{{K_1}{\varepsilon _0}{L^2}} \over d} + {{{K_2}{\varepsilon _0}{L^2}} \over d}}}$

$ = \,\,{{{K_1}{K_2}} \over {{K_1} + {K_2}}} \times {{{\varepsilon _0}{L^2}} \over d}$

Similarly,

${C_{34}} = \,\,{{{K_3}{K_4}} \over {{K_3} + {K_4}}} \times {{{\varepsilon _0}{L^2}} \over d}$

$ \therefore $  C_eq $=$ $\left[ {{{{K_1}{K_2}} \over {{K_1} + {K_2}}} + {{{K_3}{K_4}} \over {{K_3} + {K_4}}}} \right]{{{\varepsilon _0}{L^2}} \over d}$

$ \Rightarrow $  ${{K{\varepsilon _0}{L^2}} \over d}$  $ = \left[ {{{{K_1}{K_2}} \over {{K_1} + {K_2}}} + {{{K_3}{K_4}} \over {{K_3} + {K_4}}}} \right]{{{\varepsilon _0}{L^2}} \over d}$

$ \Rightarrow $  $K = {{{K_1}{K_2}} \over {{K_1} + {K_2}}} + {{{K_3}{K_4}} \over {{K_3} + {K_4}}}$

Let the capacitance of the upper part of the strip of length dx is dC₁ and lower part of the strip is dC₂

$ \therefore $   dC₁ = ${{{\varepsilon _0}\,adx} \over {d - y}}$

and dC₂ = ${{K{\varepsilon _0}\,adx} \over y}$

Here dC₁ and dC₂ are in series.

So, equivalent capacitance,

${1 \over {dC}}$ = ${1 \over {d{C_1}}} + {1 \over {d{C_2}}}$

$ \Rightarrow $    ${1 \over {dC}}$ = ${{d - y} \over {{\varepsilon _0}a\,dx}} + {y \over {{\varepsilon _0}K\,adx}}$

$ \Rightarrow $   ${1 \over {dC}}$ = ${1 \over {{\varepsilon _0}\,adx}}$ ( d $-$ y + ${y \over K}$)

$ \Rightarrow $   dC = ${{{\varepsilon _0}\,adx} \over {\left( {d - y} \right) + {y \over K}}}$

We can divide entire parallel plate capacitor into similar part like the strip of length dx and all the strips will have common end A and B. So they are in parallel.

So equivalent capacitance is the sum of all the strips capacitance.

$ \therefore $   $\int {dC} $ = $\int\limits_0^a {{{{\varepsilon _0}\,adx} \over {\left( {d - y} \right) + {y \over K}}}} $

$ \Rightarrow $   C = $\int\limits_0^a {{{{\varepsilon _0}\,a\,dx} \over {\left( {d - y} \right) + {y \over K}}}} $

From above diagram you can find this $ \to $



tan$\theta $ = ${y \over x}$

Also tan$\theta $ = ${d \over a}$

$ \therefore $   ${y \over x}$ = ${d \over a}$

$ \Rightarrow $   y = ${d \over a}$ x

By putting this value of y in the integration we get,

C = $\int\limits_0^a {{{{\varepsilon _0}\,a\,dx} \over {d - {d \over a}x + {d \over {Ka}}x}}} $

= $\int\limits_0^a {{{{\varepsilon _0}\,a\,dx} \over {d + \left( {{1 \over K} - 1} \right){d \over a}x}}} $

= $\int\limits_0^a {{{{\varepsilon _0}\,{a^2}\,dx} \over {da + \left( {{{1 - K} \over K}} \right)xd}}} $

= $\int\limits_0^a {{{K{\varepsilon _0}\,{a^2}\,dx} \over {Kda + \left( {1 - K} \right)xd}}} $

= ${{K{\varepsilon _0}{a^2}} \over {d\left( {1 - k} \right)}}\left[ {\ln \left( {\left( {1 - K} \right)x + Ka} \right)} \right]_0^a$

= ${{K{\varepsilon _0}{a^2}} \over {d\left( {1 - k} \right)}}\left[ {\ln \left( a \right) - \ln \left( {Ka} \right)} \right]$

= ${{K{\varepsilon _0}\,{a^2}} \over {d\left( {1 - k} \right)}}\ln \left( {{a \over {Ka}}} \right)$

= ${{K{\varepsilon _0}\,{a^2}} \over {d\left( {1 - k} \right)}}\ln \left( {{1 \over K}} \right)$

= ${{K{\varepsilon _0}\,{a^2}} \over {d\left( {1 - k} \right)}}\ln \left( K \right)$

Charge on capacitor at time t,

q = q₀ [ 1 $-$ e^$-$t/RC_eq]

= C_eq E [ 1 $-$ e^$-$t/RC_eq]

[ as $\,\,\,\,\,\,$ q₀ = C_eq E]

this charge q will be on both capacitor C₁ and C₂, as both are in series.

Charge on Capacitor initially,

Q_i = CV

After inserting dielectric of dielectric constant = K,

new capacitance, Q_f = (KC) $ \vee $

$\therefore\,\,\,$ Induced charges on dielectric

= Q_f $-$ Q_i

= KCV $-$ CV

= (K $-$ ) CV

= $\left( {{5 \over 3} - 1} \right)$ $ \times $ 90 $ \times $ 10^$-$12 $ \times $ 20

= 1.2 $ \times $ 10^$-$9 C

= 1.2 nC

Given : C₁ = 1.0 $\mu$F; C₂ = 3.0 $\mu$F; C₃ = 6.0 $\mu$F

Capacitance C₂ and C₃ are in series, their equivalent capacitance is

${1 \over C} = {1 \over {{C_2}}} + {1 \over {{C_3}}} \Rightarrow {1 \over C} = {{{C_2} + {C_3}} \over {{C_2}{C_3}}}$

$ \Rightarrow C = {{{C_2}{C_3}} \over {{C_2} + {C_3}}} = {{3.0\mu F \times 6.0\mu F} \over {3.0\mu F + 6.0\mu F}} = 2\mu F$

Charge on C₁ when battery is connected to switch (1) is

$ = {C_1} \times 60$ ..... (1)

Charge when battery is connected to switch (2) is

$ = ({C_1} + C)V'$ ...... (2)

Equating Eq. (1) and (2), we get

$({C_1} + C)V' = 60{C_1} \Rightarrow V' = {{60{C_1}} \over {{C_1} + C}}$

$ \Rightarrow V' = {{60 \times 1\mu F} \over {1\mu F + 2\mu F}} = {{60 \times 1\mu F} \over {3\mu F}} = 20\,V$

Final charge on C₂ and C₃ is $Q = CV' = 2\mu F \times 20 = 40\mu C$

Therefore, $Q = 40\mu C$

Given area of Parallel plate capacitor, A = 200 cm²

Separation between the plates, d = 1.5 cm

Force of attraction between the plates, F = 25 × 10^–6 N

F = QE

$ \Rightarrow $ F = ${{{Q^2}} \over {2A{ \in _0}}}$

[ As E due to parallel plate = ${\sigma \over {2{ \in _0}}} = {Q \over {2A{ \in _0}}}$]

Also we know, Q = CV = ${{{ \in _0}AV} \over d}$

$ \therefore $ F = ${{{{\left( {{ \in _0}AV} \right)}^2}} \over {{d^2} \times 2A{ \in _0}}}$

$ \Rightarrow $ V = d$\sqrt {{{2F} \over {{ \in _0}A}}} $

$ \Rightarrow $ V = $1.5 \times {10^{ - 2}}\sqrt {{{2 \times 25 \times {{10}^{ - 6}}} \over {8.85 \times {{10}^{ - 12}} \times 2 \times {{10}^{ - 2}}}}} $

= $1.5 \times {10^{ - 2}}\sqrt {{{25} \over {8.85}}} $ = 250 V

Given circuit :


Simplified circuit.


The equivalent capacitance between C and D = 2 + 5 + 5 = 12 $\mu $F

Equivalent capacitance between E and B = 4 + 2 = 6 $\mu $F

Now equivalent capacitance between A and B

${1 \over {{C_{eq}}}}$ = ${1 \over 6}$ + ${1 \over 12}$ + ${1 \over 6}$ = ${5 \over 12}$

$ \Rightarrow $$\,\,\,\,$ C_eq = ${12 \over 5}$ = 2.4 $\mu $F

Note :

(1) When capacitors C₁, C₂, . . . .C_n are in parallel then equivalent capacitance

C_eq = C₁ + C₂ + . . . . + C_n

(2) When capacitor C₁, C₂ . . . . . . C_n are in series the equivalent capacitance

${1 \over {{C_{eq}}}}$ = ${1 \over {{C_1}}}$ + ${1 \over {{C_2}}}$ + . . . . . + ${1 \over {{C_n}}}$

Before introducing the slab between the plates of all three capacitors, we have

${C_1} = {{{\varepsilon _0}A} \over 3}$

After introducing the slab between the plates of all three capacitors, we have

${C_1} = K{{{\varepsilon _0}A} \over 3} + {{{\varepsilon _0}A} \over {2.4}}$

As the capacitors are in series, we get

${{{\varepsilon _0}A} \over 3} = {{\left( {{{K{\varepsilon _0}A} \over 3}} \right)\,.\,\left( {{{{\varepsilon _0}A} \over {2.4}}} \right)} \over {\left( {{{K{\varepsilon _0}A} \over 3}} \right) + \left( {{{{\varepsilon _0}A} \over {2.4}}} \right)}}$

Therefore, the dielectric constant of the slab is

$ \Rightarrow 3K = 2.4K + 3$

$ \Rightarrow 0.6K = 3$

$ \Rightarrow K = {3 \over {0.6}} = {{30} \over 6} = 5$

Energy of sphere = ${{{Q^2}} \over {2C}}$

$\therefore\,\,\,$ ${{16 \times {{10}^{ - 12}}} \over {2C}}$ = 4.5

$ \Rightarrow $$\,\,\,$ C = ${{16 \times {{10}^{ - 12}}} \over 9}$

We know capacity of spherical conductor,

C = 4$\pi $$\varepsilon $₀R

$\therefore\,\,\,$ 4$\pi $$\varepsilon $₀R = ${{16 \times {{10}^{ - 12}}} \over 9}$

$ \Rightarrow $$\,\,\,$ R = ${1 \over {4\pi {\varepsilon _0}}} \times {{16 \times {{10}^{ - 12}}} \over 9}$

= 9 $ \times $ 10⁹ $ \times $ ${{16 \times {{10}^{ - 12}}} \over 9}$

= 16 mm

To get a capacitance of 2 μF arrangement of capacitors of capacitance 1μF as shown in figure 8 capacitors of 1μF in parallel with four such branches in series i.e., 32 such capacitors are required.

${1 \over {{C_{eq}}}} = {1 \over 8} + {1 \over 8} + {1 \over 8} + {1 \over 8}$

$ \Rightarrow $ ${1 \over {{C_{eq}}}} = {1 \over 2}$

$ \Rightarrow $ ${{C_{eq}} = 2}$

In steady state, flow of current through capacitor will be zero.

Current through the circuit,

i = ${E \over {r + {r_2}}}$

Potential difference through capacitor

V_c = ${Q \over C}$ = E - ir = E - $\left( {{E \over {r + {r_2}}}} \right)r$

$ \therefore $ Q = $CE{{{r_2}} \over {(r + {r_2})}}$

Energy supplied to the circuit is CV$_0^2$

Energy stored in capacitor, E_C = ${1\over 2}$CV$_0^2$

Therefore,

Energy dissipated E₀ = Energy supplied $-$ Energy stored

= CV$_0^2$ $-$ ${1\over 2}$CV$_0^2$ = ${1\over 2}$CV$_0^2$

$\Rightarrow$ E_C = E_D

Process 1 :

Voltage is set to V₀/3

Charge supplied = ${{C{V_0}} \over 3}$

Energy supplied = ${{{V_0}} \over 3} \times {{C{V_0}} \over 3} = {{CV_0^2} \over 9}$

Process 2 :

Voltage is raised to 2V₀/3

additional charge supplied = ${{2{V_0}C} \over 3} - {{{V_0}C} \over 3} = {{C{V_0}} \over 3}$

Energy supplied = ${{2{V_0}} \over 3} \times {{C{V_0}} \over 3} = {{2CV_0^2} \over 9}$

Process 3 :

Voltage is raised to V₀

additional charge supplied = ${V_0}C - {{2{V_0}C} \over 3} = {{C{V_0}} \over 3}$

Energy supplied = ${V_0} \times {{C{V_0}} \over 3} = {{CV_0^2} \over 3}$

Total energy supplied to circuit = ${{CV_0^2} \over 9} + {{2CV_0^2} \over 9} + {{CV_0^2} \over 3}$

$ = {6 \over 9}CV_0^2 = {2 \over 3}CV_0^2$

Final energy stored in capacitor = ${1 \over 2}CV_0^2$

Therefore,

Energy dissipated E_D = Energy supplied $-$ Energy stored

$ = {2 \over 3}CV_0^2 - {1 \over 2}CV_0^2$

$ = {{4CV_0^2 - 3CV_0^2} \over 6} = {1 \over 6}CV_0^2$

$ = {1 \over 3}\left( {{1 \over 2}CV_0^2} \right)$

Equivalent capacitance of 6 $\mu $F and 12 $\mu $F is = ${{6 \times 12} \over {12 + 6}}$ = 4 $\mu $F

Equivalent capacitance of 4$\mu $F and 4$\mu $F

is = 4 + 4 = 8 $\mu $F

New circuit is $ \to $



equivalent capacitance of 1 $\mu $F and 8 $\mu $F is

   = ${{1 \times 8} \over {8 + 1}}$ = ${8 \over 9}$ $\mu $F

Equivalent capacitance of 8 $\mu $F and 4 $\mu $F is

   = ${{8 \times 4} \over {12}}$ = ${8 \over 3}$ $\mu $F

New circuit is $ \to $



Equation capacitance of ${8 \over 9}$ $\mu $F and ${8 \over 3}$ $\mu $F is

   = ${8 \over 9}$ + ${8 \over 3}$ = ${32 \over 9}$ $\mu $F

$ \therefore $   Equivalent capacitance of AB is

C_AB = ${{C \times {{32} \over 9}} \over {C + {{32} \over 9}}}$

Given that,

C_AB = 1 $\mu $F

$ \therefore $   1 = ${{C \times {{32} \over 9}} \over {C + {{32} \over 9}}}$

$ \Rightarrow $   C + ${{{32} \over 9}}$ = ${{32C} \over 9}$

$ \Rightarrow $   ${{23C} \over 9} = {{32} \over 9}$

$ \Rightarrow $   C = ${{32} \over {23}}$ $\mu $F

(a)   ${1 \over {{C_{eq}}}} = {1 \over 4} + {1 \over 4} + {1 \over 4} = {3 \over 4}$

$ \Rightarrow $   ${C_{eq}} = {4 \over 3}\,\mu F$

(b)   ${C_{eq}} = {{4 \times 4} \over {4 + 4}} + 4 = 6\mu F$

(c)   ${C_{eq}} = 4 + 4 + 4 = 12\,\mu F$

(d)   ${C_{eq}} = {{\left( {4 + 4} \right) \times 4} \over {\left( {4 + 4} \right) + 4}} = {8 \over 3}\mu F$

Charge on ${C_1}$ is ${q_1} = \left[ {\left( {{{12} \over {4 + 12}}} \right) \times 8} \right] \times 4 = 24\mu c$

The voltage across ${C_P}$ is ${V_P} = {4 \over {4 + 12}} \times 8 = 2v$

$\therefore$ Voltage across $9\mu F$ is also $2V$

$\therefore$ Charge on $9\mu F$ capacitor $ = 9 \times 2 = 18\mu C$

$\therefore$ Total charge on $4\mu F$ and $9\mu F$ $ = 42\mu c$

$\therefore$ $E = {{KQ} \over {{r^2}}} = 9 \times {10^9} \times {{42 \times {{10}^{ - 6}}} \over {30 \times 30}} = 420N{c^{ - 1}}$

From figure, ${Q_2} = {2 \over {2 + 1}}Q = {2 \over 3}Q$

$Q = E\left( {{{C \times 3} \over {C + 3}}} \right)$

$\therefore$ ${Q_2} = {2 \over 3}\left( {{{3CE} \over {C + 3}}} \right) = {{2CE} \over {C + 3}}$

Therefore graph $d$ correctly dipicts.

We can think of this configuration to be made up of three parts:

(a) Capacitor of plate area s/2, plate separation d, and filled with a dielectric of relative permittivity $\in$₁ = 2 (lower half). The capacitance of this part is ${C_a} = { \in _1}{ \in _0}{{s/2} \over d} = { \in _0}{s \over d} = {C_1}$.

(b) Capacitor of plate area s/2, plate separation d/2, and filled with a dielectric of relative permittivity $\in$₁ = 2 (left upper half). The capacitance of this part is ${C_b} = { \in _1}{ \in _0}{{s/2} \over {d/2}} = 2{ \in _0}{s \over d} = 2{C_1}$.

(c) Capacitor of plate area s/2, plate separation d/2, and filled with a dielectric of relative permittivity $\in$₂ = 4 (right upper half). The capacitance of this part is ${C_c} = { \in _2}{ \in _0}{{s/2} \over {d/2}} = 4{ \in _0}{s \over d} = 4{C_1}$.

The capacitors C_b and C_c are connected in series with equivalent capacitance

${C_{bc}} = {{{C_b}{C_c}} \over {{C_b} + {C_c}}} = {{(2{C_1})(4{C_1})} \over {(2{C_1}) + (4{C_1})}} = {4 \over 3}{C_1}$.

The capacitor C_bc is connected in parallel with C_a. The equivalent capacitance is

$\left. {{C_2} = {C_{abc}} = {C_a}} \right\|{C_{bc}} = {C_1} + {4 \over 3}{C_1} = {7 \over 3}{C_1}$.

Electric field in presence of dielectric between the two plates of a parallel plate capacitor is given by,

$E = {\sigma \over {K{\varepsilon _0}}}$

Then, charge density

$\sigma = K{\varepsilon _0}E$

$ = 2.2 \times 8.85 \times {10^{ - 12}} \times 3 \times {10^4} \approx 6 \times {10^{ - 7}}\,\,C/{m^2}$

Let A be area of each plate and d is the distance between the plates.

The given capacitor is equivalent to two capacitors in parallel with capacitances

${C_1} = {{K{\varepsilon _0}(A/3)} \over d} = {{K{\varepsilon _0}A} \over {3d}}$

${C_2} = {{{\varepsilon _0}(2A/3)} \over d} = {{2{\varepsilon _0}A} \over {3d}}$

$\therefore$ $C = {C_1} + {C_2}$

$ = {{K{\varepsilon _0}A} \over {3d}} + {{2{\varepsilon _0}A} \over {3d}} = {{{\varepsilon _0}A} \over {3d}}(K + 2)$

$\therefore$ ${C \over {{C_1}}} = {{K + 2} \over K}$

Hence, option (d) is correct.

Let V be potential difference between the plates. Then

${E_1} = {V \over d}$ and ${E_2} = {V \over d}$

$\therefore$ ${{{E_1}} \over {{E_2}}} = 1$

Hence, option (a) is correct and option (b) is incorrect.

${Q_1} = {C_1}V = {{K{\varepsilon _0}A} \over {3d}}V$ and ${Q_2} = {C_2}V = {{2{\varepsilon _0}A} \over {3d}}V$

$\therefore$ ${{{Q_1}} \over {{Q_2}}} = {K \over 2}$

Hence, option (c) is incorrect.

For potential to be made zero, after connection

$120{C_1} = 200{C_2}$

$\left[ {\,\,} \right.$ as $\left. {C = {q \over v}\,\,} \right]$

$ \Rightarrow 3{C_1} = 5{C_2}$

$ \mathrm{C}_1=\mathrm{C}_2=\mathrm{C} $

* When $\mathrm{S}_1$ is closed. charge on $\mathrm{C}_1=+2 \mathrm{VC}$ on upper plate and $-2 \mathrm{Vc}$ on lower plate.

* When $S_1$ is released and $S_2$ is closed. charge on $C_1$ and $C_2=\mathrm{VC}$

* When $\mathrm{S}_2$ is released and $\mathrm{C}_3$ is closed. charge on $\mathrm{C}_2=\mathrm{CV}$

* When $\mathrm{S}_2$ is closed ( $\mathrm{S}_1$ open)

As $\mathrm{C}_1=\mathrm{C}_2$ the charge gets distributed equal. The upper plates of $C_1$ and $C_2$ now take charge to $\mathrm{CV}_0$ each and lower plate $-\mathrm{CV}_0$ each.

During discharging of a capacitor, V = V₀${e^{ - {t \over \tau }}}$

At t = $\tau $,

V = ${{{V_0}} \over e}$ = 0.37V₀

From the graph, t = 0, V₀ = 25 V

$ \therefore $ V = 0.37 $ \times $ 25 = 9.25 V

This voltage will occur at time between 100 sec and 150 sec. Hence, time constant $\tau $ of this circuit lies between 100 sec and 150 sec.

Let the charges on the $ 3 \mu \mathrm{F} $ and $ 2 \mu \mathrm{F} $ capacitors be $ q $ and $ q' $ respectively. The charge on the lower plate of the $ 4 \mu \mathrm{F} $ capacitor is $ -80 \mu \mathrm{C} $. The lower plate of the $ 4 \mu \mathrm{F} $ capacitor and the upper plates of the $ 2 \mu \mathrm{F} $ and $ 3 \mu \mathrm{F} $ capacitors form an isolated system. Therefore, the net charge on this system must be zero, which can be represented as :

$ q + q' - 80 \mu \mathrm{C} = 0 \quad \text{..........(i)} $

The potentials ($ V = q / C $) across these two capacitors are equal, which gives :

$ \frac{q}{3} = \frac{q'}{2} \quad \text{..........(ii)} $

By substituting $ q' $ from equation (ii) into equation (i), we can solve for $ q $ and find :

$ q = 48 \mu \mathrm{C} $

When switch S is connected to terminal 1, the potential difference across the 2 $\mu$F capacitor is V volt. Therefore, energy stored in the system is

${U_1} = {1 \over 2}{C_1}{V^2} = {1 \over 2} \times 2 \times {V^2}$

$ = {V^2}\,\mu J$

When switch S is turned to terminal 2, the charge will flow from 2 $\mu$F capacitor to 8 $\mu$F capacitor until their potentials are equalized. The common potential is

${V^2} = {q \over {{C_1} + {C_2}}} = {{{C_1}V} \over {{C_1} + {C_2}}}$

$ = {{2V} \over {(2 + 8)}} = {V \over 5}$ volt

$\therefore$ Energy stored in the system now will be

${U_2} = {1 \over 2}({C_1} + {C_2})V_2^2$

$ = {1 \over 2}(2 + 8) \times {\left( {{V \over 5}} \right)^2} = {{{V^2}} \over 5}\mu J$

$\therefore$ Percentage loss of energy is

${{{U_1} - {U_2}} \over {{U_1}}} \times 100 = {{\left( {{V^2} - {{{V^2}} \over 5}} \right)} \over {{V^2}}} \times 100 = 80\% $

Initial energy of capacitor, ${E_1} = {{q_1^2} \over {2C}}$

Final energy of capacitor, ${E_2} = {1 \over 2}{E_1} = {{q_1^2} \over {4C}} = {\left( {{{{{{q_1}} \over {\sqrt 2 }}} \over {2C}}} \right)^2}$

$\therefore$ ${t_1}=$ time for the charge to reduce to ${1 \over {\sqrt 2 }}$ of its initial value

and ${t_2} = $ time for the charge to reduce to ${1 \over 4}$ of its initial value

We have, ${q_2} = {q_1}{e^{ - t/CR}}$

$ \Rightarrow \ln \left( {{{{q_2}} \over {{q_1}}}} \right) = - {t \over {CR}}$

$\therefore$$\ln \left( {{1 \over {\sqrt 2 }}} \right) = {{ - {t_1}} \over {CR}}...\left( 1 \right)$

and $\ln \left( {{1 \over 4}} \right) = {{ - {t_2}} \over {CR}}\,\,...\left( 2 \right)$

By $(1)$ and $(2),$ ${{{t_1}} \over {{t_2}}} = {{\ln \left( {{1 \over {\sqrt 2 }}} \right)} \over {\ln \left( {{1 \over 4}} \right)}}$

$ = {1 \over 2}{{\ln \left( {{1 \over 2}} \right)} \over {2\ln \left( {{1 \over 2}} \right)}} = {1 \over 4}$

The given capacitance is equal to two capacitances connected in series where

${C_1} = {{{k_1}{ \in _0}A} \over {d/3}} = {{3{k_1}{ \in _0}A} \over d}$

$ = {{3 \times 3{ \in _0}A} \over d} = {{9{ \in _0}A} \over d}$

and

${C_2} = {{{k_2}{ \in _0}A} \over {2d/3}} = {{3{k_2}{ \in _0}A} \over {2d}}$

$ = {{3 \times 6{ \in _0}A} \over {2d}} = {{9{ \in _0}A} \over d}$

The equivalent capacitance ${C_{eq}}$ is

${1 \over {C{}_{eq}}} = {1 \over {{C_1}}} + {1 \over {{C_2}}}$

$ = {d \over {9{ \in _0}A}} + {d \over {9{ \in _0}A}}$

$ = {{2d} \over {9{ \in _0}A}}$

$\therefore$ ${C_{eq}} = {9 \over 2}{{{\varepsilon _0}A} \over d} = {9 \over 2} \times 9pF = 40.5pF$

Required ratio

$ = {{Energy\,\,\,stored\,\,in\,\,capacitor} \over {Workdone\,\,by\,\,the\,\,battery}} = {{{1 \over 2}C{V^2}} \over {C{e^2}}}$

where $C=$ Capacitance of capacitor

$V=$ Potential difference,

$e=$ $emf$ of battery

$ = {{{1 \over 2}C{e^2}} \over {C{e^2}}} = {1 \over 2}$

$\left( \, \right.$ as $V=e$ $\left. \, \right)$

First, let's consider the capacitor with the dielectric between its plates. The charge on the capacitor is Q, its capacitance is C (which includes the effect of the dielectric), and the potential difference (voltage) across its plates is V. According to the formula for the energy stored in a capacitor :

$U = \frac{1}{2} CV^2$

we find that the energy is equal to half of the product of the capacitance and the square of the voltage.

Now, consider the process where the dielectric slab is removed from the capacitor. When the dielectric is removed, the capacitance of the capacitor decreases. However, because the capacitor is not connected to anything that can supply or absorb charge, the charge Q on the capacitor stays the same. Since the charge stays the same but the capacitance decreases, the voltage V across the capacitor must increase to keep Q = CV true.

The energy of the capacitor without the dielectric is still given by :

$U = \frac{1}{2} CV^2$

but now C is smaller and V is larger. However, because both C and V² change in such a way that their product remains constant, the energy of the capacitor doesn't change when the dielectric is removed.

Therefore, the energy of the capacitor before the dielectric is removed is the same as the energy of the capacitor after the dielectric is removed. When the dielectric is reinserted, the process is just reversed, so again no net work is done.

So the total work done by the system in the process of removing the dielectric and then reinserting it is zero. This is because work is the transfer of energy, and in this case, the energy of the system (the charged capacitor) does not change. Therefore, no energy is transferred, so no work is done.

Step 1: Initial State (Switch Open)

Let $q_1$ and $V_1$ be the charge and voltage on capacitor $C_1 = 3\,\mu\mathrm{F}$. Let $q_2$ and $V_2$ be the charge and voltage on capacitor $C_2 = 6\,\mu\mathrm{F}$.

When the switch $S$ is open, the two capacitors are connected in series. This means that the charge stored on each capacitor is the same. So, $q_1 = q_2$.

The voltages on the capacitors add up to the battery voltage: $V_1 + V_2 = 9\,\mathrm{V}$

The charges and voltages on capacitors in series also follow this equation: $C_1 V_1 = C_2 V_2$ This means the charge on each capacitor is the same since $q_1 = C_1 V_1$ and $q_2 = C_2 V_2$.

Solving these two equations gives:

• $V_1 = 6\,\mathrm{V}$, $q_1 = 3\,\mu\mathrm{F} \times 6\,\mathrm{V} = 18\,\mu\mathrm{C}$
• $V_2 = 3\,\mathrm{V}$, $q_2 = 6\,\mu\mathrm{F} \times 3\,\mathrm{V} = 18\,\mu\mathrm{C}$

Total charge on the side connected to $X$ is $-q_1 + q_2 = 0$. No net charge has moved yet.

Step 2: After Closing the Switch

When the switch is closed, the resistors $R_1 = 3\,\Omega$ and $R_2 = 6\,\Omega$ are connected in series with the capacitors and battery.

Total resistance is $R_1 + R_2 = 9\,\Omega$.

The current from the battery is: $I = \\frac{9\,\mathrm{V}}{9\,\Omega} = 1\,\mathrm{A}$

The voltage drop across each resistor is:

• $V_1' = I \times R_1 = 1\,\mathrm{A} \times 3\,\Omega = 3\,\mathrm{V}$
• $V_2' = I \times R_2 = 1\,\mathrm{A} \times 6\,\Omega = 6\,\mathrm{V}$

The new charges on the capacitors are:

• $q_1' = C_1 \times V_1' = 3\,\mu\mathrm{F} \times 3\,\mathrm{V} = 9\,\mu\mathrm{C}$
• $q_2' = C_2 \times V_2' = 6\,\mu\mathrm{F} \times 6\,\mathrm{V} = 36\,\mu\mathrm{C}$

Final Calculation

The total charge on the plates connected to $X$ is now $-q_1' + q_2' = -9 + 36 = 27\,\mu\mathrm{C}$.

This means that $27\,\mu\mathrm{C}$ of charge flows from $Y$ to $X$ when the switch is closed.

(A) When capacitor is shorted, current will begin to flow and capacitor will start discharging. Because of discharging, magnitude of current in circuit and potential across the two ends will reduce continually until becomes zero. Thermal energy will be generated due to flow of charges.

(B) When wire is moved perpendicularly in magnetic field, due to change in flux associated with wire, a potential difference will be developed across its ends, but not current will flow as circuit is not complete and no heat will be generated.

(C) Since wire is placed in electric field, electrons inside the wire will align in such a way that net electric field inside the metal is zero. Net charges will be developed across the two ends of wire, creating a potential difference. Here, circuit is incomplete, so no current will flow and no heat will be generated.

(D) When a battery is connected, across two ends of wire, there will be constant potential difference. In this case circuit is complete, so current will start flowing and heat will be generated.

For (i) : The net capacitance

$ \begin{aligned} \mathrm{C} & =\frac{C_1 C_2}{C_1+C_2} \\ & =\frac{2 \times 4}{2+4}=\frac{8}{6}=\frac{4}{3} \mu \mathrm{~F} \end{aligned} $

And net resistance, $R=\frac{R_1 R_2}{R_1+R_2}$

$ =\frac{1 \times 2}{1+2}=\frac{2}{3} \Omega $

Thus, time constant, $\mathrm{T}=\mathrm{C} . \mathrm{R}=\frac{4}{3} \times \frac{2}{3}=\frac{8}{9} \mu \mathrm{~s}$

For (ii) :

$ \begin{aligned} \mathrm{C} & =\mathrm{C}_1+\mathrm{C}_2 \\ & =2+4=6 \mu \mathrm{~F} \text { and } \\ \mathrm{R} & =\mathrm{R}_1+\mathrm{R}_2 \\ & =1+2=3 \Omega \end{aligned} $

Time constant T = C.R

$ =6 \times 3=18 \mu \mathrm{~s} $

$ \begin{aligned} &\text { For (iii) : }\\ &\begin{aligned} \mathrm{C} & =\mathrm{C}_1+\mathrm{C}_2 \\ & =2+4=6 \mu \mathrm{~F} \text { and } \\ \mathrm{R} & =\frac{\mathrm{R}_1 \mathrm{R}_2}{\mathrm{R}_1+\mathrm{R}_2} \\ & =\frac{1 \times 2}{1+2}=\frac{2}{3} \Omega \\ \mathrm{~T} & =\mathrm{C} . \mathrm{R}=6 \times \frac{2}{3} \\ & =4 \mu \mathrm{~s} \end{aligned} \end{aligned} $

Applying conservation of energy,

${1 \over 2}C{V^2} = m.s\Delta T;\,\,\,$

$V = \sqrt {{{2m.s.\Delta T} \over C}} $

As $n$ plates are joined, it means $(n-1)$ capacitor joined in parallel.

$\therefore$ resultant capacitance $=(n-1)C$

Given, $Q=Q_{0}\left(1-e^{-\alpha t}\right)$

Where, Q$_0$ is the steady state charge stored in the capacitor.

Q$_0$ = C[PD across capacitor in steady state]

$\mathrm{Q}_{0}=C\left(\frac{\mathrm{V}}{\mathrm{R}_{1}+\mathrm{R}_{2}}\right) \mathrm{R}_{2}$

$\therefore \quad \mathrm{Q}_{0}=\frac{\mathrm{CVR}_{2}}{\mathrm{R}_{1}+\mathrm{R}_{2}}$

From, $Q = {Q_0}(1 - {e^{ - \alpha t}})$

$\alpha$ is $\frac{1}{\tau}$ or $\frac{1}{\mathrm{CR}_{e q}}$

Here, $\mathrm{R}_{e q}$ is equivalent resistance across capacitor after short circuiting the battery.

$\begin{aligned} \therefore \quad R_{e q} & =\frac{\mathrm{R}_{1} R_{2}}{\mathrm{R}_{1}+\mathrm{R}_{2}} \\ \alpha & =\frac{1}{C\left(\frac{R_{1} R_{2}}{R_{1}+R_{2}}\right)}=\frac{R_{1}+R_{2}}{C_{1} R_{2}} \end{aligned}$

The work done is stored as the potential energy. The potential energy stored in a capacitor is given by

$U = {1 \over 2}{{{Q^2}} \over C}$

$ = {1 \over 2} \times {{{{\left( {8 \times {{10}^{ - 18}}} \right)}^2}} \over {100 \times {{10}^{ - 6}}}}$

$ = 32 \times {10^{ - 32}}J$

The capacitancce of parallel plate capacitor in which a metal plate of thickness $t$ is inserted is given by

$C = {{{\varepsilon _0}A} \over {d - t}}.\,\,\,\,\,$

Here $t \to 0\,\,\,\,\,\,$ $\therefore$ $C = {{{\varepsilon _0}A} \over d}$

The equivalent capacitance of $n$ identical capacitors of capacitance $C$ is equal to $nC.$ Energy stored in this capacitor

$E = {1 \over 2}\left( {nC} \right){V^2} = {1 \over 2}nC{V^2}$

For an isolated sphere, the capacitance is given by

$C = 4\pi \,{ \in _0}\,r$

$ = {1 \over {9 \times {{10}^9}}} \times 1$

$ = 1.1 \times {10^{ - 10}}F$