Capacitor

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$

In a uniform electric field, potential difference across two parallel plates of capacitor,

$V = Ed$

where, E is electric field and d is separation between plates.

$ \Rightarrow V = 2000 \times (1 \times {10^{ - 3}}) = 2V$

$ \text { The given circuit is } $

The two capacitors of $5 \mu \mathrm{~F}$ each are in parallel. So, their equivalent capacitance is $(5+5) 10 \mu \mathrm{~F}$. Similarly,

two $2 \mu \mathrm{~F}$ capacitors are also in parallel. So, their equivalent capacitance is $(2+2) 4 \mu \mathrm{~F}$. The circuit is now reduced to as shown

In steady state, when all the capacitors are charged, no current flow in the circuit. So, potential across resistors is zero.

As in parallel combination, potential is same in both wires $C D$ and $E G$, i.e. 100 V .

In series combination, the potential is divided between the capacitor in the inverse ratio of their capacitance, i.e.

$ \begin{aligned} & \frac{V_{A F}}{V_{F B}}=\frac{4}{10}=\frac{2}{5} \text { or } 2: 5 \\ \therefore & V_{A F}=\frac{2}{7} \times 100=28.5 \mathrm{~V} \\ \text { and } & V_{F B}=\frac{5}{7} \times 100=71.4 \mathrm{~V} \end{aligned} $

Capacitance of an oil drop of radius $r$ is

$ C=4 \pi \varepsilon r $

when $n$ drops combine to form a large drop of radius $R$, then

$n \times$ volume of 1 small drop $=$ volume of 1 large drop

$ \begin{aligned} n \times \frac{4}{3} \pi r^3=\frac{4}{3} \pi R^3 \Rightarrow & \\ R=n^{1 / 3} \cdot r & \end{aligned} $

So, capacitance of large drop is

$ C^{\prime}=4 \pi \varepsilon R=4 \pi \varepsilon \cdot n^{\frac{1}{3}} \cdot r=n^{\frac{1}{3}} C $

Growth of charge on capacitor in a $C R$ circuit occurs as

$ Q=Q_0\left(1-e^{-t / R C}\right) $

If product $C R$ is small, then time constant for $C R$ circuit is small and growth of charge will be more rapid.

Since capacitors C₁ and C₂ are parallel, so their equivalent capacitance will be (C₁ + C₂). Now, (C₁ + C₂) and C₃ are in series, so the net equivalent capacitance of circuit will be,

${1 \over C} = {1 \over {{C_3}}} + {1 \over {{C_1} + {C_2}}} = {{{C_1} + {C_2} + {C_3}} \over {({C_1} + {C_2}){C_3}}}$

or, $C = {{({C_1} + {C_2}){C_3}} \over {{C_1} + {C_2} + {C_3}}}$

Since V is the voltage of the battery, so charge on this system,

$q = CV \Rightarrow q = {{({C_1} + {C_2}){C_3}V} \over {({C_1} + {C_2} + {C_3})}}$

If the capacitance C₃ breaks down, then total equivalent capacitance will become

$C' = {C_1} + {C_2}$

$\therefore$ New charge stored,

$q' = C'V$

$ \Rightarrow q' = ({C_1} + {C_2})V$

Now, change in total charge,

$\Delta q = q' - q$ [$\because$ $q' > q$]

$ = ({C_1} + {C_2})V - {{({C_1} + {C_2}){C_3}V} \over {{C_1} + {C_2} + {C_2}}}$

or, $\Delta q = ({C_1} + {C_2})V\left[ {1 - \left( {{{{C_3}} \over {{C_1} + {C_2} + {C_3}}}} \right)} \right]$

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)