Capacitor

Given data

• Dielectric constant $ K = 3 $

• Electric field $ E = 15\pi \, \mathrm{N/C} $

• We are to find electric displacement $ D $.

Step 1. Recall the relation

$ D = \varepsilon_0 E + P = \varepsilon_0 K E $

(using $ K = \frac{D}{\varepsilon_0 E} $).

So,

$ D = \varepsilon_0 K E $

Step 2. Substitute values

$ \varepsilon_0 = 8.854 \times 10^{-12} \, \mathrm{F/m} $

$ D = (8.854 \times 10^{-12}) \times 3 \times (15\pi) $

Step 3. Compute

$ D = 8.854 \times 10^{-12} \times 45\pi $

$ 45\pi \approx 141.37 $

$ D = 8.854 \times 10^{-12} \times 141.37 $

$ D \approx 1.252 \times 10^{-9} \, \mathrm{C/m^2} $

Step 4. Express in given unit format

$ 1.252 \times 10^{-9} \, \mathrm{C/m^2} = 1250 \times 10^{-12} \, \mathrm{C/m^2} $

✅ Final Answer:

$ \boxed{D = 1250 \times 10^{-12} \, \mathrm{C/m^2}} $

Correct Option: A

$ C_1: C_2: C_3: C_4=1: 2: 3: 4 $

$ \therefore C_1=C, C_2=2 C, C_3=3 C, C_4=4 C $

$C_1, C_2$ and $C_3$ are in series.

$ \begin{aligned} & \therefore \frac{1}{C^{\prime}}=\frac{1}{C_1}+\frac{1}{C_2}+\frac{1}{C_3}=\frac{1}{C}+\frac{1}{2 C}+\frac{1}{3 C} \\ &\,\,\,\,\,\,\,\,\,\,\,\,\,\, =\frac{11}{6 C} \\ & \therefore C^{\prime}=\frac{6 C}{11} \end{aligned} $

$C^{\prime}$ and $C_4$ are in parallel.

∴ equivalent capacitance

$ C^{\prime \prime}=C^{\prime}+C_4=\frac{6 C}{11}+4 C=\frac{50 C}{11} $

Charge on $C_4, q_4=C_4 V=4 C V$

Total charge drawn from battery,

$ q=C^{\prime \prime} \cdot V=\frac{50 C V}{11} $

∴ Charge on $C_2$,

$ q_2=q-q_4=\frac{50 C V}{11}-4 C V=\frac{6 C V}{11} $

Thus, $\frac{q_2}{q_4}=\frac{\frac{6 C V}{11}}{4 C V}=\frac{6}{44}=\frac{3}{22}$

$ \Rightarrow q_2: q_4=3: 22 $

Step 1: Find the charge on the first capacitor.

The $2~\mu\text{F}$ capacitor is charged using the $60~\text{V}$ battery, so its charge is: $q_1 = C \times V = 2 \times 10^{-6} \times 60 = 120 \times 10^{-6}~\text{C}$

Step 2: Find the charge on the second capacitor.

The $1~\mu\text{F}$ capacitor is not charged yet, so $q_2 = 0~\text{C}$

Step 3: Find total charge after connection.

The total charge when we join them is the sum of both charges: $q_{\text{total}} = 120 \times 10^{-6} + 0 = 120 \times 10^{-6}~\text{C}$

Step 4: Find total capacitance when connected in parallel.

When capacitors are connected in parallel, their capacitances add: $C_{\text{total}} = 2~\mu\text{F} + 1~\mu\text{F} = 3~\mu\text{F} = 3 \times 10^{-6}~\text{F}$

Step 5: Find the new (final) voltage across the capacitors.

The new voltage is the total charge divided by the total capacitance: $V_{\text{final}} = \frac{q_{\text{total}}}{C_{\text{total}}} = \frac{120 \times 10^{-6}}{3 \times 10^{-6}} = 40~\text{V}$

$ \begin{aligned} &\text { Energy stored in a capacitor is, }\\ &\begin{aligned} U & =\frac{1}{2} C V^2 \\ \Rightarrow \quad U & =\frac{1}{2} \times 10 \times 10^{-6} \times\left(6 \times 10^3\right)^2 \end{aligned} \end{aligned} $

$ \begin{aligned} & =\frac{1}{2} \times 10 \times 10^{-6} \times 36 \times 10^6 \\ & =180 \mathrm{~J} \end{aligned} $

Here dielectric fills the space between plates completely so,

$ \begin{aligned} C & =\frac{\varepsilon_0 K A}{d} \\ & =\frac{8.85 \times 10^{-12} \times 4.5 \times 0.4 \pi}{0.5 \times 10^{-3}} \\ & \approx 100 \mathrm{nF} \end{aligned} $

In given circuit capacitor does not conducts once it is charged. So in steady state we have following circuit.

$\Rightarrow$ Now, let $I=$ current drawn from cell. Then we have above current distribution.

Using KVL in loop $A B C D E F G A$; we have;

$ \begin{array}{ll} & -6 I-1 \frac{I}{2}-\frac{3 I}{2}-I+9=0 \\ \Rightarrow \quad & 9 I=9 \text { or } I=1 \mathrm{~A} \end{array} $

Now, potential difference between $B$ and $D$ (using KVL);

$ \begin{aligned} & V_B-6 \times 1-1 \times \frac{1}{2}=V_D \\ \Rightarrow & V_B-V_D=6+\frac{1}{2}=\frac{13}{2}=6.5 \mathrm{~V} . \end{aligned} $

$ \begin{aligned} U_i & =\frac{1}{2} C V_1^2+\frac{1}{2} C V_2^2 \\ & =\frac{1}{2} C\left(V_1^2+V_2^2\right) \end{aligned} $

Common potential,

$ \begin{aligned} V & =\frac{Q_1+Q_2}{C_1+C_2}=\frac{C V_1+C V_2}{C+C} \\ & =\frac{V_1+V_2}{2} \\ \Rightarrow \quad V & =\frac{V_1+V_2}{2} \\ \therefore \quad U_f & =\frac{1}{2} C^1 V^2 \\ & =\frac{1}{2}(2 C)\left(\frac{V_1+V_2}{2}\right)^2=C\left(\frac{V_1+V_2}{2}\right)^2 \\ \therefore \quad \Delta U & =U_i-U_f \\ = & \frac{1}{2} C\left(V_1^2+V_2^2\right)-C\left(\frac{V_1+V_2}{2}\right)^2 \\ = & \frac{C}{2}\left(V_1^2+V_2^2\right)-\frac{C}{4}\left(V_1^2+V_2^2+2 V_1 V_2\right) \\ = & \frac{C}{4}\left[2 V_1^2+2 V_2^2-V_1^2-V_2^2-2 V_1 V_2\right] \\ = & \frac{C}{4}\left[V_1^2+V_2^2-2 V_1 V_2\right]=\frac{C}{4}\left(V_1-V_2\right)^2 \end{aligned} $

Capacitance of small conducting sphere

$ C=4 \pi \varepsilon_0 r=10 \mu \mathrm{~F} $

Since, $27 \times \frac{4}{3} \pi r^3=\frac{4}{3} \pi R^3 \Rightarrow R=3 r$

$\therefore$ Capacitance of big sphere

$ \begin{aligned} C^{\prime} & =4 \pi \varepsilon_0 R=4 \pi \varepsilon_0(3 r) \\ & =3\left(4 \pi \varepsilon_0 r\right)=3 \times 10=30 \mu \mathrm{~F} \end{aligned} $

$ \begin{aligned} &\text { } b-a=1 \mathrm{~cm}=1 \times 10^{-2} \mathrm{~m} \text {, }\\ &\begin{gathered} b=\left(a+1 \times 10^{-2}\right) \mathrm{m} \\ C=4 \pi \varepsilon_0\left(\frac{a b}{b-a}\right) \\ \Rightarrow 100 \times 10^{-12}=\frac{1}{9 \times 10^9} \times \frac{a\left(a+1 \times 10^{-2}\right)}{1 \times 10^{-2}} \\ 9 \times 10^{-3}=a^2+10^{-2} a \\ \Rightarrow 0.9=100 a^2+a \\ \Rightarrow 100 a^2+a-0.9=0 \\ \Rightarrow a=\frac{-1 \pm \sqrt{1^2-4 \times 100(-0.9)}}{2 \times 100} \\ =-1 \pm \frac{\sqrt{361}}{200}=\frac{-1 \pm 19}{200} \\ =\frac{-1+19}{200}=\frac{18}{200} \mathrm{~m} \\ =0.09 \mathrm{~m}=9 \mathrm{~cm} \end{gathered} \end{aligned} $

$ \begin{aligned} & W=\frac{Q^2}{2 C} \\ & W^{\prime}=\frac{(2 Q)^2}{2 C}=4\left(\frac{Q^2}{2 C}\right)=4 W \end{aligned} $

$\therefore$ Additional work done,

$ W^{\prime}-W=4 W-W=3 W $

$ \begin{aligned} \tau_{\max } & =M B \\ \tau_{\max } & =I A B \end{aligned} $

Since,  $ 2 \pi r=10 $

$ \Rightarrow \quad r=\frac{10}{2 \pi} $

$\therefore$ Area of the loop

$ \begin{aligned} A & =\pi r^2=\pi\left(\frac{10}{2 \pi}\right)^2 \\ & =\frac{100}{4 \pi}=\frac{25}{\pi} \mathrm{~m}^2 \\ \therefore \tau_{\max } & =L A B=1 \times \frac{25}{\pi} \times 2 \pi \times 10^{-4} \\ & =50 \times 10^{-4} \mathrm{~N}-\mathrm{m} \end{aligned} $

$ \text { } \begin{aligned} & C=4 \pi \varepsilon_0 k \frac{r_1 r_2}{r_2-r_1} \\ = & 4 \pi\left(8.85 \times 10^{-12}\right) \times 5 \times \frac{(0.08)(0.09)}{0.09-0.08} \\ & \simeq 400 \times 10^{-12} \mathrm{~F} \\ & \simeq 400 \mathrm{pF} \end{aligned} $

Initial stored energy

$ E_i=\frac{1}{2} C V^2 $

Final stored energy,

$ \begin{aligned} E_f & =E_i-0.25 E_i \\ E_f & =0.75 E_i \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\\ \therefore \quad E_f & =\frac{1}{2} C^{\prime} V^{\prime 2} \\ & =\frac{1}{2}(K C)\left(\frac{V}{k}\right)^2 \\ E_f & =\frac{E_i}{K} \\ \Rightarrow 0.75 E_i & =\frac{E_i}{K} \,\,\,\,\text { [from eq. (i)] }\\ \Rightarrow \quad \frac{1}{K} & =0.75 \\ \Rightarrow \quad K & =\frac{1}{0.75}=\frac{4}{3} \end{aligned} $

The capacitance of a parallel plate capacitor is given by the formula:

$ C = \frac{A \varepsilon_0}{d} $

where:

$ A $ is the area of the plates,

$ \varepsilon_0 $ is the permittivity of free space,

$ d $ is the distance between the plates.

In this scenario, the distance between the plates is halved, and a dielectric material with a relative permittivity ($ k $) of 10 is introduced. The new capacitance ($ C' $) can be calculated as follows:

$ C' = \frac{k A \varepsilon_0}{d'} = \frac{10 \cdot A \varepsilon_0}{d/2} $

Simplifying gives:

$ C' = \frac{10 \cdot A \varepsilon_0}{d/2} = \frac{20 \cdot A \varepsilon_0}{d} = 20C $

Thus, the ratio of the final capacitance to the initial capacitance is:

$ \frac{C'}{C} = 20 $

$ \text { Given capacitance, } C=2 \mu \mathrm{~F} $

Equivalent circuit, Thus, $C_{\mathrm{eq}}=\frac{C}{2}+\frac{C}{2}+2 C+2 C$

$ \begin{aligned} & =C+4 C \\\ \quad C_{\mathrm{eq}} & =5 C \\ \therefore C_{\mathrm{eq}} & =5 \times 2=10 \mu \mathrm{~F} \end{aligned} $

Let initial thickness $=d$

Thickness of dielectrics $=2 \mathrm{~mm}$

It is given that final potential is same.

Let initial capacitance $=C_1$

Final capacitance $=C_2$

$ \begin{array}{ll} \therefore & C_1=C_2 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\\ C_1=\frac{A \varepsilon_0}{d} \text { and } C_2 & =\frac{A \varepsilon_0}{d^{\prime}-t+\frac{t}{k}} \end{array} $

where, $d^{\prime}=d+1.6 \mathrm{~mm}$

$ \begin{aligned} & \Rightarrow \quad \frac{A \varepsilon_0}{d}=\frac{A \varepsilon_0}{(d+1.6)-t\left(1-\frac{1}{k}\right)} \\ & \Rightarrow \quad \frac{1}{d}=\frac{1}{(d+1.6)-2\left(1-\frac{1}{k}\right)} \\ & \Rightarrow \quad d+1.6-2+\frac{2}{k}=d \\ & \Rightarrow \quad k=5 \end{aligned} $

Given,

Radius of isolated sphere $=r_1$

Radius of earthed sphere $=r_2$

Capacitance of isolated sphere,

$ C_1=4 \pi \varepsilon_0 r_1 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)$

Capacitance of isolated sphere enclosed in earthed sphere,

$ C_2=\frac{4 \pi \varepsilon_0 r_1 r_2}{r_2-r_1} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)$

Now, capacitance is 5 times

$ \begin{aligned} & \Rightarrow \quad 5 C_1=C_2 \\ & \Rightarrow 5\left(4 \pi \varepsilon_0 r_1\right)=\frac{4 \pi \varepsilon_0 r_1 r_2}{r_2-r_1} \Rightarrow 5=\frac{r_2}{r_2-r_1} \end{aligned} $

$ \begin{aligned} & \Rightarrow 5 r_2-5 r_1=r_2 \Rightarrow-5 r_1=-4 r_2 \\ & \Rightarrow \quad \frac{r_1}{r_2}=\frac{4}{5} \end{aligned} $

$ \text { For fig (A), } $

$ \begin{aligned} &\begin{aligned} & R^{\prime}=\frac{6 \times 12}{6+12}+4=4+4 \\ & R^{\prime}=8 \end{aligned}\\ &\therefore \text { Time constant for circuit }(A)\\ &\tau_A=\frac{L}{R^{\prime}}=\frac{2}{8}=\frac{1}{4} \mathrm{~s} \end{aligned} $

Similar for figure (B)

$ R^{\prime \prime}=\frac{10 \times 10}{10+10}=5 \Omega $

and equivalent capacitance,

$ C_{\mathrm{eq}}=0.5+0.5=1 \mathrm{~F} $

$\therefore$ Time constant, $\tau_B=R C$

$ =5 \times 1=5 \mathrm{~s} $

Given the capacitances:

$ C_1 = 4 \, \mu \mathrm{F} $

$ C_2 = 6 \, \mu \mathrm{F} $

$ C_3 = 12 \, \mu \mathrm{F} $

When connected in series:

To find the equivalent capacitance for capacitors in series, we use the formula:

$ C_{\text{eq}} = \frac{C_1 C_2 C_3}{C_1 C_2 + C_2 C_3 + C_3 C_1} $

Substituting the given values:

$ C_{\text{eq}} = \frac{4 \times 6 \times 12}{4 \times 6 + 6 \times 12 + 12 \times 4} = \frac{288}{144} = 2 \, \mu \mathrm{F} $

When connected in parallel:

For capacitors in parallel, the equivalent capacitance is the sum of the capacitances:

$ C_{\text{eq}} = C_1 + C_2 + C_3 $

Substituting the given values:

$ C_{\text{eq}} = 4 + 6 + 12 = 22 \, \mu \mathrm{F} $

Calculating the ratio:

The ratio of the effective capacitances when connected in series to when connected in parallel is:

$ \frac{2}{22} = \frac{1}{11} $

Thus, the ratio is $ 1:11 $.

Let the potential of point $D$ be $V$.

In the above figure, capacitors $C_1$ and $C_2$ are in series combination.

Potential difference across both the capacitors are,

$ \begin{aligned} & \Delta V_1=V_1-V \\ & \Delta V_2=V-V_2 \end{aligned} $

We knows that in series arrangement, the charge on capacitor will be equal.

$ \begin{array}{ll} \Rightarrow & Q_1=Q_2 \\ & C_1\left(V_1-V\right)=C_2\left(V-V_2\right) \\ \Rightarrow & C_1 V_1-C_1 V=C_2 V-C_2 V_2 \\ \Rightarrow & C_1 V+C_2 V=C_2 V_2+C_1 V_1 \\ \Rightarrow & V\left(C_1+C_2\right)=C_1 V_1+C_2 V_2 \\ \Rightarrow & V=\frac{C_1 V_1+C_2 V_2}{\left(C_1+C_2\right)} \end{array} $

The given arrangement of capacitor are represented as

$\therefore C_{A B}=8+8+8+8=32 \mu \mathrm{~F}$

According to figure,

$4 \mu \mathrm{~F}, 8 \mu \mathrm{~F}, 4 \mu \mathrm{~F}$ are in parallel combination. So,

$ C^{\prime}=4+8+4=16 \mu \mathrm{~F} \quad \ldots \text { (i) } $

Now, $C^{\prime}$ and $5 \mu \mathrm{~F}$ are in series combination. So, total capacitance in the circuit is,

$ \frac{1}{C_{\text {total }}}=\frac{1}{5}+\frac{1}{C^{\prime}} $

From Eq. (i), $C_{\text {total }}=\frac{16 \times 5}{16+5}=\frac{80}{21} \mu \mathrm{~F}$

Total charge in the given circuit,

$ \begin{aligned} q_{\text {total }} & =C_{\text {total }} \cdot V \\ & =\frac{80}{21} \times 63=240 \mu \mathrm{C} \end{aligned} $

Charge remains same in series combination. So, $5 \mu \mathrm{~F}$ also have $240 \mu \mathrm{C}$ charge.

The potential difference across $5 \mu \mathrm{~F}$ capacitor is,

$ \begin{aligned} & V^{\prime}=\frac{Q}{C}=\frac{240}{5} \\ & V^{\prime}=48 \mathrm{~V} \end{aligned} $

The given circuit diagram can be represented as

Since, $\frac{C_{\mathrm{r}}}{C_2}=\frac{C_3}{C_4}=\frac{1}{3}$

Hence, given circuit follows the condition of balance wheatstone bridge.

Hence, capacitor $C_5$ is ineffective.

Now, the circuit diagram is redrawn as

$\begin{aligned} C_{A B} & =\frac{C_1 C_2}{C_1+C_2}+\frac{C_3 C_4}{C_3+C_4}=\frac{1 \times 3}{1+3}+\frac{2 \times 6}{2+6} \\ & =\frac{3}{4}+\frac{12}{8}=\frac{9}{4} \mu \mathrm{F} \end{aligned}$

Given, $C_1=1 \mu \mathrm{F}, V=9 \mathrm{~V}$,

$C_2=2 \mu \mathrm{F}, C_3=3 \mu \mathrm{F}$

Equivalent capacitance of $C_2$ and $C_3$

$C^{\prime}=C_2+C_3=2+3=5 \mu \mathrm{F}$

Common potential difference when $C_1$ and $C^{\prime}$ are combined, is given as

$\begin{aligned} & V^{\prime}=\frac{\text { total charge }(Q)}{\text { total copacitance }\left(C_1+C^{\prime}\right)} \\ & V^{\prime}=\frac{Q}{C_1+C^{\prime}} \end{aligned}$

$\begin{aligned} & \text { But } Q=C_1 V \\ & \therefore V^{\prime}=\frac{C_1 V}{C_1+C^{\prime}}=\frac{1 \times 9}{1+5}=\frac{9}{6}=1.5 \mathrm{~V} \end{aligned}$

Charge on capacitor $C_3$ is given as

$\begin{aligned} Q_3 & =C_3 V^{\prime} \\ & =3 \times 10^{-6} \times 1.5=4.5 \times 10^{-6} \mathrm{C} \end{aligned}$

Given,

Capacitance of parallel plate capacitor, $C_0=60 \propto \mathrm{F}$

$=60 \times 10^{-6} \mathrm{~F}$

Separation between plates, $d=6 \mathrm{~mm}$

$=6 \times 10^{-3} \mathrm{~m}$

Electric potential, $V_0=250 \mathrm{~V}$

Dielectic constant of slab, $K=5$

and thickness of slab, $t=3 \mathrm{~mm}$

$=3 \times 10^{-3} \mathrm{~m}$

Since, $C=\frac{q}{V}=\frac{\varepsilon_0 A}{d}=\frac{\varepsilon_0 A}{d-t\left(1-\frac{1}{K}\right)}$

where, $q$ is charge stored, $V$ is potential, $\varepsilon_0$ is free space permittivity $=8.854 \times 10^{-12} \mathrm{Nm}^2 \mathrm{C}^{-2}$ and $A$ is plate area

$\begin{aligned} \therefore \quad q_0 & =C_0 V_0 \\ & =60 \times 10^{-6} \times 250=15 \times 10^{-3} \mathrm{C}\end{aligned}$

$\text { and new capacitance, } C^{\prime}=\frac{\varepsilon_0 A}{\left[6-3\left(1-\frac{1}{5}\right)\right] \times 10^{-3}} .... \text{(i)}$

and $\varepsilon_0 A=C_0 d$

$\begin{aligned} \Rightarrow \quad \varepsilon_0 A & =60 \times 10^{-6} \times 6 \times 10^{-3} \\ & =360 \times 10^{-9} \end{aligned}$

Put this value in Eq. (i), we get

$C^{\prime}=\frac{360 \times 10^{-9}}{3.6 \times 10^{-3}}=10^{-4} \mathrm{~F}$

Final potential stored in capacitor $=\frac{q_0}{C^{\prime}}=\frac{15 \times 10^{-3}}{10^{-4}}=150 \mathrm{~V}$

Hence, voltage difference $=250-150=100 \mathrm{~V}$

Given, $C_1,C_2,C_3$ and $C_4$ are 1 $\mu$F, 1.5 $\mu$F, 2.5 $\mu$F and 0.5 $\mu$F.

Supply voltage = 30 V

Let $V_{a b}$ is potential across $a$ and $b$,

$Q$ is charge in circuit and $Q_1, Q_2$ are branch charges.

Since, series equivalent capacitance $\frac{1}{C_s}=\frac{1}{C_1}+\frac{1}{C_2}$

$\Rightarrow \quad C_s=\frac{C_1 C_2}{C_1+C_2} $

and in parallel connection voltages remain same

$\begin{aligned} \therefore \quad Q_1 & =\frac{C_1 C_2}{C_1+C_2} V \Rightarrow Q_1=\frac{1 \times 1.5}{1+1.5} \times 30 \\\\ & =\frac{1.5}{2.5} \times 30=18 \mu \mathrm{C} \end{aligned}$

Similarly, $\quad Q_2=\frac{2.5 \times 0.5}{2.5+0.5} \times 30=12.5 \mu \mathrm{C}$

$\begin{aligned} \therefore \quad V_{a b} & =V_a-V_b=\left(V-V_b\right)-\left(V-V_a\right) \\\\ & =\frac{Q_2}{C_3}-\frac{Q_1}{C_1}=\frac{12.5}{2.5}-\frac{18}{1}=-13 \mathrm{~V} \\\\ \therefore \quad V_{b a} & =13 \mathrm{~V} \end{aligned}$

According to given circuit diagram,

Voltage across $A$ and $B, V_{A B}=80 \mathrm{~V}$

Let charge in $9 \propto \mathrm{F}, 10 \propto \mathrm{F}$ and $5 \propto \mathrm{F}$ be $Q_1, Q_2$ and $Q_3$.

Net charge, $Q=Q_{9 \propto \mathrm{F}}+Q_{10 \propto \mathrm{F}}+Q_{5 \propto \mathrm{F}}$

As we know that,

Parallel equivalent capacitance,

$\begin{aligned} C_{\text {eq }}^{\prime} & =C_1+C_2+C_2 \\ \therefore C_{\text {eq }}^{\prime} & =9+10+5=24 \propto \mathrm{F} \end{aligned}$

and series equivalent capacitance,

$\begin{aligned} \quad \frac{1}{C_{\mathrm{eq}}} & =\frac{1}{C_a}+\frac{1}{C_b}+\frac{1}{C_c} \\ \Rightarrow \quad \frac{1}{C_{\mathrm{eq}}} & =\frac{1}{12}+\frac{1}{24}+\frac{1}{8} \Rightarrow \frac{1}{C_{\mathrm{eq}}}=\frac{2+1+3}{24} \\ \Rightarrow \quad C_{\mathrm{eq}} & =\frac{24}{6} \propto \mathrm{F}=4 \propto \mathrm{F} \end{aligned}$

Since, $C=\frac{Q}{V}$

$\begin{aligned} \therefore \quad Q & =C V \\ & =4 \times 10^{-6} \times 80=320 \propto \mathrm{C} \end{aligned}$

Since, in series connection,

$\begin{aligned} & V=V_{12 \propto \mathrm{F}}+V_{10 \propto \mathrm{F}}+V_{8 \propto \mathrm{F}} \\ & \Rightarrow \quad 80=\frac{320}{12}+V_{10 \propto \mathrm{F}}+\frac{320}{8} \\ \end{aligned}$

$ \Rightarrow {V_{10 \propto F}} = 13.33\,V$

$\therefore$ ${Q_{10 \propto F}} = {V_{10 \propto F}} \times 10$

$ \simeq 133 \propto C$