Capacitor

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

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.

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}$

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}$

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.

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.

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 $=(n-1)C$

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$