Capacitor

When connected in parallel

C_eq = C₁ + C₂

When in series

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

${C_1} + {C_2} = {{15} \over 4}\left( {{{{C_1}{C_2}} \over {{C_1} + {C_2}}}} \right)$

$4{({C_1} + {C_2})^2} = 15{C_1}{C_2}$

$4{C_1}^2 + 4{C_2}^2 - 7{C_1}{C_2} = 0$

dividing by ${C_1}^2$

$4{\left( {{{{C_2}} \over {{C_1}}}} \right)^2} - {{7{C_2}} \over {{C_1}}} + 4 = 0$

Let ${{{C_2}} \over {{C_1}}} = x$

$4{x^2} - 7x + 4 = 0$

${b^2} - 4ac = 49 - 64 < 0$

$ \therefore $ No solution exixts.

$ \because $ ${v_1}\cos \alpha = {v_2}\cos \beta $

${{{v_1}} \over {{v_2}}} = {{\cos \beta } \over {\cos \alpha }}$

Then the ratio of kinetic energies

${{{k_1}} \over {{k_2}}} = {{{1 \over 2}m{v_1}^2} \over {{1 \over 2}m{v_2}^2}} = {\left( {{{{v_1}} \over {{v_2}}}} \right)^2} = {\left( {{{\cos \beta } \over {\cos \alpha }}} \right)^2}$

$ \Rightarrow $ ${{{k_1}} \over {{k_2}}} = {{{{\cos }^2}\beta } \over {{{\cos }^2}\alpha }}$

Given, C₁ = C₂ = C

When both capacitors are connected in series, their equivalent capacitance will be

${1 \over {{C_s}}} = {1 \over C} + {1 \over C} = {2 \over C}$

$ \Rightarrow {C_s} = {C \over 2}$

When both capacitors are connected in parallel, their equivalent capacitance will be

C_p = C + C = 2C

$\therefore$ The ratio of equivalent capacitance in series and parallel combination is

${{{C_s}} \over {{C_p}}} = {{C/2} \over {2C}} = {1 \over 4}$

$\therefore$ C_s : C_p = 1 : 4

$\tau $ = RC = 10$\mu $s

For 0 < t < 5 $\mu $s, capacitor is charging.

For 5 < t < 10 $\mu $s potential is constant so capacitor is discharging but during discharging not possible to discharge 100%.

After that capacitor again gets charged.

Let potential of point O v₀ = 0.

Now, using junction analysis

We can say, q₁ + q₂ + q₃ = 0

$ \Rightarrow $ 2(x – 6) + 4(x – 6) + 5(x) = 0

$ \Rightarrow $ x = ${{36} \over {11}}$

q₃ = CV = ${{36} \over {11}} \times 5$ = 16.36 $\mu $C

C_i = ${{{\varepsilon _0}A} \over d} = {{{\varepsilon _0}lw} \over d}$

U_i = ${1 \over 2}{C_i}{V^2}$ = ${1 \over 2}{{{\varepsilon _0}lw} \over d}{V^2}$

C_f = C₁ + C₂

= ${{K{\varepsilon _0}{A_1}} \over d} + {{{\varepsilon _0}{A_2}} \over d}$

= ${{K{\varepsilon _0}wx} \over d} + {{{\varepsilon _0}w\left( {l - x} \right)} \over d}$

= ${{{\varepsilon _0}w} \over d}\left[ {Kx + l - x} \right]$

$ \therefore $ U_f = ${1 \over 2}{C_f}{V^2}$

= ${1 \over 2}{{{\varepsilon _0}w} \over d}\left[ {Kx + l - x} \right]{V^2}$

Given U_f = 2U_i

$ \Rightarrow $ ${1 \over 2}{{{\varepsilon _0}w} \over d}\left[ {Kx + l - x} \right]{V^2}$ = 2 $ \times $ ${1 \over 2}{{{\varepsilon _0}lw} \over d}{V^2}$

$ \Rightarrow $ kx + l – x = 2l

$ \Rightarrow $ 4x – x = l

$ \Rightarrow $ 3x = l

$ \Rightarrow $ x = ${l \over 3}$

By conservation of charge,

${q_1} + {q_2} = {q_1}' + {q_2}'$

$ \Rightarrow $ $ - CV + (2C)(2V) = (C + 2C)V'$

$V' = {{3CV} \over {3C}} = V$

${U_f} = {1 \over 2}C{v^2} + {1 \over 2}(2C){V^2}$

${U_f} = {3 \over 2}C{V^2}$

Heat loss = ${1 \over 2}\left( {{{{C_1}{C_2}} \over {{C_1} + {C_2}}}} \right)V_0^2$

= ${1 \over 2}\left( {{{C \times {C \over 2}} \over {C + {C \over 2}}}} \right)V_0^2$

= ${1 \over 6}CV_0^2$

Q₂ = CV = 8 $ \times $ 20 = 160 $\mu $C

Q₁ = 750 - 160 = 590 $\mu $C

Initially,

Charge on capacitor 10 μF

Q = CV = (10 μF) (50V)

Q = 500 μC

Final Charge on 10 μF capacitor

Q = CV = (10 μF) (20V)

Q = 200 μC

From charge conservation,

Charge on unknown capacitor

Q = 500 μC – 200 μC = 300 μC

$ \Rightarrow $ Capacitance (C) = ${Q \over V}$ = ${{300} \over {20}}$ = 15 $\mu $F

dC = ${{{\varepsilon _0}adx} \over {d + \alpha x}}$

$ \therefore $ C = $\int\limits_0^a {{{{\varepsilon _0}adx} \over {d + \alpha x}}} $

= ${{{\varepsilon _0}a} \over \alpha }\left[ {\ln \left( {d + \alpha } \right)} \right]_0^a$

= ${{{\varepsilon _0}a} \over \alpha }\ln \left[ {{{d + \alpha a} \over d}} \right]$

= ${{{\varepsilon _0}a} \over \alpha }\ln \left[ {1 + {{\alpha a} \over d}} \right]$

= ${{{\varepsilon _0}a} \over \alpha }\left( {{{\alpha a} \over d} - {{{\alpha ^2}{a^2}} \over {{d^2}}}} \right)$

= ${{{\varepsilon _0}{a^2}} \over d}\left( {1 - {{\alpha a} \over {2d}}} \right)$

Note : ln(1 + x) = $x - {{{x^2}} \over 2} + {{{x^3}} \over 3} - ...$

C₁ + C₂ = 10 ......(1)

${1 \over 2}{C_2}{V^2} = 4 \times {1 \over 2}{C_1}{V^2}$

$ \Rightarrow $ C₂ = 4C₁ .....(2)

From (1) ans (2),

C₁ = 2 and C₂ = 8

For series combination

C_eq = ${{{C_1}{C_2}} \over {{C_1} + {C_2}}}$ = ${{8 \times 2} \over {8 + 2}}$ = 1.6 $\mu $F

k(x) = K(1 + $\alpha $x)
Capacitance of element dC₁ = ${{K{\varepsilon _0}A} \over {dx}}$ = ${{K\left( {1 + \alpha x} \right){\varepsilon _0}A} \over {dx}}$

${1 \over {{C_{eq}}}}$ = $\int {{1 \over {d{C_1}}}} $

= $\int\limits_0^d {{{dx} \over {KA\left( {1 + \alpha x} \right){\varepsilon _0}}}} $

= ${{1 \over {KA\alpha {\varepsilon _0}}}\ln \left( {1 + \alpha x} \right)}$

Using expansion of ln (1 + x) keeping x $ \ll $ 1

$ \Rightarrow $ ${1 \over {{C_{eq}}}}$ = ${{1 \over {KA\alpha {\varepsilon _0}}}\left( {\alpha d - {{{\alpha ^2}{d^2}} \over 2}} \right)}$

$ \Rightarrow $ C_eq = ${{{{\varepsilon _0}KA} \over d}{{\left( {1 - {{\alpha d} \over 2}} \right)}^{ - 1}}}$

$ \Rightarrow $ C_eq = ${{{{\varepsilon _0}KA} \over d}\left( {1 + {{\alpha d} \over 2}} \right)}$

As V = ${q \over C}$

$ \therefore $ V $ \propto $ ${1 \over C}$

So ${{{V_1}} \over {{V_2}}} = {{{C_2}} \over {{C_1}}}$

$ \Rightarrow $ ${{{V_1}} \over {{V_2}}} = {6 \over 4} = {3 \over 2}$

Also V₁ + V₂ = 10

$ \therefore $ V₁ + ${2 \over 3}$V₁ = 10

$ \Rightarrow $ ${{5{V_1}} \over 3}$ = 10

$ \Rightarrow $ V₁ = 6 V

The charge on 4 $\mu $F capacitor is = C₁V₁ = 4 $ \times $ 6 = 24 $\mu $C

${1 \over {{C_1}}} = {d \over {3A{\varepsilon _0}}}\left( {{1 \over {{K_1}}} + {1 \over {{K_2}}} + {1 \over {{K_3}}}} \right)$

${C_1} = {{3A{\varepsilon _0}\left( {{K_1}{K_2}{K_3}} \right)} \over {d\left( {{K_1}{K_2} + {K_2}{K_3} + {K_3}{K_1}} \right)}}$

${C_2} = {{A{\varepsilon _0}} \over {3d}}\left( {{K_1} + {K_2} + {K_3}} \right)$

${{{E_1}} \over {{E_2}}} = {{{C_1}} \over {{C_2}}} = {{3{K_1}{K_2}{K_3}} \over {\left( {{K_1}{K_2} + {K_2}{K_3} + {K_3}{K_1}} \right)}} \times {3 \over {\left( {{K_1} + {K_2} + {K_3}} \right)}}$

$ \Rightarrow {{{E_1}} \over {{E_2}}} = {{9{K_1}{K_2}{K_3}} \over {\left( {{K_1} + {K_2} + {K_3}} \right)\left( {{K_1}{K_2} + {K_2}{K_3} + {K_3}{K_1}} \right)}}$

$t = 2\pi \sqrt {{L \over {{g_{eff}}}}} $

$ \Rightarrow {g_{eff}} = \sqrt {{g^2} + {{\left( {{{gE} \over m}} \right)}^2}} $

$ \Rightarrow t = 2\pi \sqrt {{L \over {\sqrt {{g^2} + {{\left( {{{qE} \over m}} \right)}^2}} }}} $

As q = CV
Hence slope of graph will give capacitance. Slope will be more in parallel combination. Hence capacitance in parallel should be 50 $\mu $F and in series combination must be 8 $\mu $F. Only in option 40 $\mu $f and 10 $\mu $F.

C_parallel = 40 + 10 = 50 $\mu $F

C_series = ${{40 \times 10} \over {40 + 10}} = 8\mu F$

Initially Q = CV(1 + n)

$ \therefore $ C_eq = (K + n)C

$ \therefore $ $V = {{CV\left( {1 + n} \right)} \over {\left( {K + n} \right)C}} = {{V\left( {1 + n} \right)} \over {\left( {K + n} \right)}}$

Work done = $\Delta $U

= U_f – U_i

$ = {{{q^2}} \over {2{C_r}}} - {{{q^2}} \over {2{C_i}}}$

$ = {{{{\left( {5 \times {{10}^{ - 6}}} \right)}^2}} \over 2}.\left( {{1 \over {2 \times {{10}^{ - 6}}}} - {1 \over {5 \times {{10}^{ - 6}}}}} \right)$

$ = {{15} \over 4} \times {10^{ - 6}} = 3.75{\rm{ }} \times {\rm{ }}{10^{-6}}{\rm{ }}J$

Assuming potential x and y at the top junctions A and B respectively and zero potential at lower junctions.

At the junction A applying KCL,

i₁ + i₂ + i₃ = 0

${{x - 72} \over 6} + {x \over 4} + {{x - y} \over 2} = 0$

$ \Rightarrow $ ${{2x - 144 + 6x - 6y + 3x} \over {12}} = 0$

$ \Rightarrow $ 11x - 6y = 144 ......(1)

At the junction B applying KCL,

i₄ + i₅ = 0

${{y - x} \over 2} + {y \over {10}} = 0$

$ \Rightarrow $ ${{5y - 5x + y} \over {10}}$ = 0

$ \Rightarrow $ 6y = 5x ......(2)

Solving (1) and (2)

11x - 5x = 144

$ \Rightarrow $ x = 24 V

$ \therefore $ y = 20 V

The charge on the capacitor, q = CV

$ \Rightarrow $ q = 10$\mu $F × 20 = 200 $\mu $C

Charges at inner plates are 1 $\mu $C and –1 $\mu $C.

$ \therefore $ Potential difference across capacitor
= ${q \over c} = {{1\mu C} \over {1\mu F}} = {{1 \times {{10}^{ - 6}}C} \over {1 \times {{10}^{ - 6}}Farad}} = 1V$

A = 10^–4 m²

E_max = 10⁶ V/m

C = 15 $\mu $F

$C = {{k{\varepsilon _0}A} \over d};{{Cd} \over {{\varepsilon _0}A}} = k$

$k = {{15 \times {{10}^{ - 12}} \times 500 \times {{10}^{ - 6}}} \over {8.86 \times {{10}^{ - 12}} \times {{10}^4}}}$

= ${{15 \times 5} \over {8.86}} = 8.465$

k $ \approx $ 8.5

From equs.

${{{{7C} \over 3}} \over {{7 \over 3} + C}} = {1 \over 2}$

$ \Rightarrow $  14 C $=$ 7 + 3 C

$ \Rightarrow $  C $=$ ${7 \over {11}}$

As we know, Current, $I = {{dq} \over {dt}}$

= Slope of q versus t graph

= Zero at t = 4s; (as graph is a line parallel to time axis at t = 4s)

$E = {\sigma \over {{ \in _0}}} = {Q \over {A\,{ \in _0}}}$

Q = AE$ \in $₀

Q = (1) (100) (8.85 $ \times $ 10^$-$12)

Q = 8.85 $ \times $ 10^$-$10C

V_i = ${1 \over 2}$CE²

V_f = ${{{{\left( {CE} \right)}^2}} \over {2 \times 4c}}$ = ${1 \over 2}{{C{E^2}} \over 4}$

$\Delta $E = ${1 \over 2}$CE² $ \times $ ${3 \over 4}$ = ${3 \over 8}$ CE²

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