Capacitor

$ \text { From figure I, } $

$ \begin{aligned} &\begin{aligned} Q & =C V \\ & =10 \times 220 \\ & =2200 \mu \mathrm{C} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\end{aligned}\\ &\text { From figure II, } \end{aligned} $

$ \text { } \begin{aligned}and\,\,\, Q_1+Q_2 & =2200 \mu \mathrm{C} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)\\ \frac{Q_1}{C_1} & =\frac{Q_2}{C_2} \end{aligned} $

$ \begin{aligned} \frac{Q_1}{10} & =\frac{Q_2}{12} \Rightarrow 6 Q_1=5 Q_2 \\ \text { or } \quad Q_2 & =\frac{6}{5} Q_1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(iii) \end{aligned} $

Putting Eq. (iii) in Eq. (ii)

$ \begin{array}{ll} & Q_1+\frac{6}{5} Q_1=2200 \\ \Rightarrow \quad & Q_1=\frac{5 \times 2200}{11}=1000 \mu \mathrm{C} \\ \text { and } & Q_2=\frac{6}{5} \times 1000=1200 \mu \mathrm{C} \end{array} $

Now, initial energy,

$ \Sigma_i=\frac{1}{2} \frac{Q^2}{C}=\frac{1}{2} \times \frac{(2200)^2 \times 10^{-6}}{10} $

Final energy,

$ \Sigma_f=\frac{1}{2}\left[\frac{(1000)^2}{10}+\frac{(1200)^2}{12}\right] \times 10^{-6} $

Loss of energy, $\Delta E=E_i-E_f$

$ \begin{aligned} & =\frac{1}{2} \times 10^{-6}\left[\frac{(2200)^2}{10}-\frac{(1000)^2}{10}-\frac{(1200)^2}{12}\right] \\ & =132 \mathrm{~mJ} \end{aligned} $

Displacement current

$ \begin{aligned} & I_d=\varepsilon_0 \frac{d \phi}{d t}=\varepsilon_0 \frac{d}{d t}\left(E^{\prime} A\right) \\ & I_d=\varepsilon_0 A \frac{d E^{\prime}}{d t} \end{aligned} $

Here, $\frac{d E^{\prime}}{d t}=E$

$ \begin{array}{lll} \therefore & I_d & =\varepsilon_0 A E \\ \Rightarrow & A & =\frac{I_d}{\varepsilon_0 E}=\frac{I}{\varepsilon_0 E} \quad\left[\because I_d=I\right] \end{array} $

Step 1: Find Initial Capacitance

The original capacitor has air and its capacitance is $4~\mu\mathrm{F}$. When a dielectric with constant 5 fills the gap, the new capacitance becomes $5$ times bigger:

$C_i = 5 \times 4~\mu\mathrm{F} = 20~\mu\mathrm{F}$

Step 2: Find Charge on the Capacitor

The capacitor is charged to $100~\mathrm{V}$. The charge stored is:

$Q = C_i \times V = 20 \times 10^{-6}~\mathrm{F} \times 100~\mathrm{V} = 2 \times 10^{-3}~\mathrm{C}$

Step 3: Work Out Initial Energy With Dielectric

The energy stored to start with (with dielectric in place) is:

$U_i = \frac{Q^2}{2C_i} = \frac{(2 \times 10^{-3})^2}{2 \times 20 \times 10^{-6}} = \frac{4 \times 10^{-6}}{40 \times 10^{-6}} = 0.1~\mathrm{J}$

Step 4: Find Final Capacitance After Removing Dielectric

When the dielectric is taken out, the capacitance returns to air value: $C_f = 4~\mu\mathrm{F}$.

Step 5: Calculate Final Energy With Dielectric Gone

The charge on the plates stays the same because the capacitor is disconnected from the battery. The energy now is:

$U_f = \frac{Q^2}{2C_f} = \frac{(2 \times 10^{-3})^2}{2 \times 4 \times 10^{-6}} = \frac{4 \times 10^{-6}}{8 \times 10^{-6}} = 0.5~\mathrm{J}$

Step 6: Work Done In Removing the Dielectric

Work done is the increase in energy:

$W = U_f - U_i = 0.5 - 0.1 = 0.4~\mathrm{J}$

$ \begin{aligned} &\text { Displacement current }\\ &\begin{aligned} I_D & =\varepsilon_0 \frac{d \phi}{d t} \\ & =8.854 \times 10^{-12} \times 9 \pi \times 10^3 \\ & =0.25 \times 10^{-6} \mathrm{~A} \\ & =0.25 \mu \mathrm{~A} \end{aligned} \end{aligned} $

When dielectric is filled, we model this as two capacitors in series.

For capacitor with dielectric,

$ C_1=\frac{\varepsilon_0 A}{\frac{d}{2 k}}=\frac{2 K A \varepsilon_0}{d} $

For capacitor without dielectric,

$ C_2=\frac{\varepsilon_0 A}{\frac{d}{2}}=\frac{2 \varepsilon_0 A}{d} $

For series combination,

$ \begin{aligned} & C=\frac{C_1 \times C_2}{C_1+C_2} \\ & C=\frac{\frac{2 K A \varepsilon_0}{d} \times \frac{2 \varepsilon_0 A}{d}}{\frac{2 K A \varepsilon_0}{d}+\frac{2 \varepsilon_0 A}{d}} \\ & C=\frac{2 \varepsilon_0 A}{d\left(1+\frac{1}{K}\right)} \end{aligned} $

Charge stored initially,

$ Q=C V=\frac{2 \varepsilon_0 A}{d\left(1+\frac{1}{K}\right)} \cdot V $

Now, the entire capacitor just vacuum,

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

So, final voltage, $V^{\prime}=\frac{Q}{C^{\prime}}$

$ \begin{aligned} & =\frac{2 \varepsilon_0 A V}{d\left(1+\frac{1}{K}\right)} \times \frac{d}{\varepsilon_0 A} \\ V^{\prime} & =\frac{2 V}{\left(1+\frac{1}{K}\right)} \end{aligned} $

When a capacitor of capacitancec is charged to a' potential $ V $, then initial charge on capacitor is,

$ Q=C V $

When a dielectric material with dielectric constant $ K $ is inserted between the plates then new capacitance,

$ C^{\prime}=K C $

Charge on the capacitor remains the same because it is disconnected from the battery.

$ Q^{\prime}=Q=C V $

Final potential on the capacitor,

$ \begin{array}{l} V^{\prime}=\frac{Q^{\prime}}{C^{\prime}}=\frac{C V}{K C} \\\\ V^{\prime}=\frac{V}{K} \end{array} $

Given,

Current, $ i_{C}=150 \mu \mathrm{~A} $

Capacitance of capacitor, $ C=30 \mu \mathrm{~F} $

Rate of change of potential, $ \frac{d V}{d t}= $ ?

We know that,

$ \begin{array}{l} C=\frac{q}{V} \\\\ q=C V \end{array} $

Differentiate the Eq. (i) w.r.t. to time $ t $, we have

$ \frac{d q}{d t}=C \frac{d V}{d t} $

As we know the rate to change of charge per unit of time is equal to the current, then

$ i_{C}=C \cdot \frac{d V}{d t} $

Putting the value,

$ \begin{aligned} 150 \times 10^{-6} & =30 \times 10^{-6} \times \frac{d V}{d t} \\\\ \frac{d V}{d t} & =\frac{150}{30} \\\\ \frac{d V}{d t} & =5 \mathrm{Vs}^{-1} \end{aligned} $

A capacitor has a capacitance of $ (4.0 \pm 0.2) \, \mu \mathrm{F} $ and is charged to a potential of $ (10.0 \pm 0.1) \, \mathrm{V} $.

Given:

Capacitance, $ C = (4.0 \pm 0.2) \, \mu \mathrm{F} $

Voltage, $ V = (10.0 \pm 0.1) \, \mathrm{V} $

The charge $ Q $ on the capacitor can be calculated using the formula:

$ Q = C \times V $

So, the absolute charge value is:

$ Q = 4.0 \, \mu \mathrm{F} \times 10.0 \, \mathrm{V} = 40 \, \mu \mathrm{C} $

To find the error in the charge, we use the formula for the combined relative error:

$ \frac{\Delta Q}{Q} = \frac{\Delta C}{C} + \frac{\Delta V}{V} $

Substitute the given values:

$ \frac{\Delta Q}{Q} = \left(\frac{0.2}{4.0}\right) + \left(\frac{0.1}{10.0}\right) = 0.05 + 0.01 = 0.06 $

Therefore, the percentage error in the charge $ Q $ is:

$ \frac{\Delta Q}{Q} \times 100 = 0.06 \times 100\% = 6\% $

The final charge on the capacitor, including the percentage error, is:

$ Q = (40 \, \mu \mathrm{C} \pm 6\%) $

Capacitor, $ C_{1}=1 \mu \mathrm{~F} $

$ C_{2}=2 \mu \mathrm{~F} $

According to question, maximum potential with stand by $ C_{1} $ is 6 kV .

$ \Rightarrow V_{1 \max }=\frac{Q_{\max }}{C_{1}} $

$ \begin{array}{l} Q_{\max }=C_{1} V_{1 \max } \\\\ \Rightarrow 1 \times 10^{-6} \times 6 \times 10^{3} \\\\ Q_{\max }=6 \times 10^{-3} \end{array} $

On first capacitor

When $ C_{1} $ and $ C_{2} $ are connected in series, the charge on each capacitor is equal and is equal to the maximum charge the combination can withstand.

Since $ C_{1} V_{1 \text { max }} < C_{2} V_{2 \text { max }} $

The resultant capacitance of combination is

$ \begin{aligned} C_{R} & =\frac{C_{2} C_{1}}{C_{1}+C_{2}}=\frac{2 \times 1}{2+1} \\\\ C_{R} & =\frac{2}{3} \times 10^{-6} \mathrm{~F} \\\\ V_{\max } & =\frac{Q_{\max }}{C_{R}}=\frac{6 \times 10^{-3}}{\frac{2}{3} \times 10^{-6}} \\\\ V_{\max } & =9 \times 10^{3} \mathrm{~V} \\\\ V_{\max } & =9 \mathrm{kV} \end{aligned} $

Capacitor $ C_{1}=10 \mu \mathrm{~F} $

Charging voltage, $ V=100 \mathrm{~V} $

Initial stored energy of capacitor

$ \begin{aligned} U_{\text {initial }} & =\frac{1}{2} C_{1} V^{2} \\\\ & =\frac{1}{2} \times 10^{-6} \times 100 \times 100 \\\\ & =0.5 \times 10^{-2} \mathrm{~J} \end{aligned} $

Energy lost when connect to uncharged capacitor

$ \begin{array}{l} \Delta U=\frac{1}{2} \frac{\left(C_{1} C_{2}\right)}{C_{1}+C_{2}}\left(V_{1}-V_{2}\right)^{2} \\\\ V_{2}=0 \\\\ \Delta U=\frac{1}{2} \times \frac{10 \times 30 \times 10^{-6}}{40} \times 100 \times 100 \\\\ \Delta U=\frac{3}{80} \times 10^{-2} \times 100=0.0375 \end{array} $

or $ \Delta U=3.75 \times 10^{-2} \mathrm{~J} $

Let us assume that the area of the plates is $A$ and the separation between them is $d$

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

The capacitance with the dielectric of thickness, $t=\frac{d}{2}$,

$ \begin{aligned} C_1 & =\frac{\varepsilon_0 A}{d-t+\frac{t}{K}}=\frac{\varepsilon_0 A}{d-\frac{d}{2}+\frac{d}{2 K}} \\\\ & =\frac{2 \varepsilon_0 A K}{d(K+1)}=\frac{2 K C}{(K+1)} \\\\ & =\frac{2 \times 4 \times C}{(5)}=\frac{8 C}{5} \end{aligned} $

Again, when capacitor is filled upto $t=$ ?

$ \begin{aligned} \therefore \quad C_2 & =\frac{\varepsilon_0 A}{d-t+\frac{t}{K}}=\frac{\varepsilon_0 A}{d-\frac{d}{3}+\frac{d}{3 K}} \\\\ & =\frac{\varepsilon_0 A}{\frac{2 d}{3}+\frac{d}{3 K}}=\frac{\varepsilon_0 A}{d\left(\frac{2 K+1}{3 K}\right)} \\\\ \Rightarrow \quad \frac{\varepsilon_0 A}{d} \frac{3 K}{(2 K+1)} & =C \times \frac{3 K}{(2 K+1)} \\\\ C_2 & =\frac{C \times 3 \times 4}{9}=\frac{4}{3} C \end{aligned} $

Now, $\frac{C_1}{C_2}=\frac{\frac{8}{5} C}{\frac{4}{3} C} \Rightarrow \frac{8}{5} \times \frac{3}{4}=\frac{6}{5}$

Natural frequency of $L-C$ circuit

$ 2 \pi v=\frac{1}{\sqrt{L C}} $

When capacitor is filled with dielectric,

$ C^{\prime}=K C $

Thus $\omega_1=\frac{1}{\sqrt{L C_1}}$ and $\omega_2=\frac{1}{\sqrt{L C_2}}$

According to question, $2 \pi \times 120=\frac{1}{\sqrt{L C}}$

$ \begin{array}{l} \text { and } 2 \pi \times 100=\frac{1}{\sqrt{L K C}} \\\\ \Rightarrow \frac{2 \pi \times 120}{2 \pi \times 100}=\frac{1}{\sqrt{L C}} \times \frac{\sqrt{L K C}}{1} \\\\ \frac{120}{100}=\sqrt{K} \\\\ 1.2=\sqrt{K} \\\\\Rightarrow \quad K=(1.2)^2=1.44 \end{array} $

When capacitor are connected in series, the total capacitance $C_{\text {Toal }}$,

$ \frac{1}{C_{\text {Total }}}=\frac{1}{C_1}+\frac{1}{C_2}+\frac{1}{C_3} $

Given,

$ \begin{aligned} & C_1=10 \mu \mathrm{~F}, \quad C_2=5 \mu \mathrm{~F}, \quad C_3=20 \mu \mathrm{~F} \\\\ & \frac{1}{C_{\text {Toul }}} =\frac{1}{10}+\frac{1}{5}+\frac{1}{20} \\\\ & =\frac{2}{20}+\frac{4}{20}+\frac{1}{20}=\frac{7}{20} \end{aligned} $

For, $5 \mu \mathrm{C}$,

$ \begin{aligned} Q & =C_{\text {Total }} \times V \\\\ & =\frac{20}{7} \times 14=\frac{280}{7}=40 \mu \mathrm{C} \end{aligned} $

Given,

Capacitance of $A=C$

Capacitance of $B=2 C$

$A$ and $B$ are in series, series capacitance is given by

$ \frac{1}{C_{\mathrm{eq}}}=\frac{1}{C}+\frac{1}{2 C} \Rightarrow C_{\mathrm{eq}}=\frac{2}{3} C $

Charge on each cell before removing battery

$ Q=C_{\mathrm{eq}} E=\frac{2 C E}{3} $

Energy before disconnecting battery.

$ E=\frac{1}{2} C_{\mathrm{eq}} V^2=\frac{1}{2} \times \frac{2}{3} \times C E^2=\frac{1}{3} C E^2 $

When battery is disconnected and plates are joined in opposite polarity

$ C=\frac{Q-Q}{C+2 C}=0 $

So, all energy is lost $=\frac{1}{3} C E^2$

Hence, statement (ii) and (iii) are correct and correct option is (ii).

$ \text { Given circuit } $

Each capacitor has $C$ capacitance. Draw the circuit in simplified manner.

Now, From (1)

Capacitor are in parallel connect

$ C_{\mathrm{eq}_1}=C_1+C_2=2 C $

From (3) [both capacitor are parallely connected]

$ C_{\mathrm{eq}_2}=C_1+C_2=2 C $

The circuit will resolves as

$C_{\mathrm{eq}}$ and $C_{\mathrm{eq}}$ are in series connection

$ \begin{array}{ll} \therefore & C_{\mathrm{eq}_3}=\frac{1}{C_{\mathrm{eq}_1}}+\frac{1}{C_{\mathrm{eq}_2}} \\ \Rightarrow & C_{\mathrm{eq}_3}=\frac{1}{2 C}+\frac{1}{2 C}=C \end{array} $

Now, $C_{\mathrm{eq}_3}$ and $C$ are parallely connected. So, net capacitance is $=C+C=2 C$

The given circuit diagram is given as,

$ \therefore \quad C_{A B}=\frac{C_1 C_2}{C_1+C_2}+\frac{C_3 C_4}{C_3+C_4} $

$ \begin{aligned} & =\frac{20 \times 20}{20+20}+\frac{10 \times 10}{10+10} \\ & =10+5=15 \mu \mathrm{~F} \end{aligned} $

We know that, displacement current

$ \begin{array}{rlr} I_D & =\varepsilon_0 \frac{d \phi}{d t} & \\ & =\varepsilon_0 \frac{d}{d t}(E A) & (\because \phi=E A) \\ & =\varepsilon_0 A \frac{d E}{d t} & \\ & =\varepsilon_0 A \frac{d}{d t}\left(\frac{V}{d}\right) & \left(\because E=\frac{V}{d}\right) \\ & =\frac{\varepsilon_0 A}{d} \frac{d V}{d t} & \\ I_D & =C \frac{d V}{d t} & \left(\because C=\frac{\varepsilon_0 A}{d}\right) \\ \frac{d V}{d t} & =\frac{I_D}{C} & \end{array} $

Here, $I_D=150 \mu \mathrm{~A}=1.5 \times 10^{-4} \mathrm{~A}$

$ \begin{aligned} & C =30 \mu \mathrm{~F}=3 \times 10^{-5} \mathrm{~F} \\ \therefore & \frac{d V}{d t} =\frac{1.5 \times 10^{-4}}{3 \times 10^{-4}}=5 \mathrm{Vs}^{-1} \end{aligned} $

Given:

Capacitance $ C = 100 \, \mu\text{F} $

Charging rate $ q = 100 \, \mu\text{C/s} $

Desired potential difference $ V = 100 \, \text{V} $

To find the time $ t $ required to achieve a potential difference of 100 V, use the relationship between charge, capacitance, and potential difference:

$ Q = C \cdot V $

Since charge $ Q $ is also given by:

$ Q = q \cdot t $

Equate the two expressions:

$ C \cdot V = q \cdot t $

Rearrange to solve for time $ t $:

$ t = \frac{C \cdot V}{q} $

Insert the given values:

$ t = \frac{100 \times 10^{-6} \, \text{F} \times 100 \, \text{V}}{100 \times 10^{-6} \, \text{C/s}} $

$ t = 100 \, \text{s} $

Therefore, the time required to produce a potential difference of 100 V between the capacitor plates is 100 seconds.

Given, plate area, $A=10 \mathrm{~cm}^2$

$ =10^{-3} \mathrm{~m}^2 $

Plate separation, $d=3 \mathrm{~mm}=3 \times 10^{-3} \mathrm{~m}$

Potential difference, $V=12 \mathrm{~V}$

Capacitance of parallel plate capacitor,

$ \begin{aligned} & C=\frac{\varepsilon_0 A}{d}=\frac{\varepsilon_0 \times 10^{-3}}{3 \times 10^{-3}}=\frac{\varepsilon_0}{3} \\ \Rightarrow \quad & C=\frac{\varepsilon_0}{3} \end{aligned} $

Charge stored in capacitor,

$ q=C V=\frac{\varepsilon_0}{3} \times 12=4 \varepsilon_0 C $

∵ Work done, $W=q V$

$ =4 \varepsilon_0 12=48 \varepsilon_0 $

When dielectric material (with $K=3$ ) is introduced between the parallel plate capacitor, then new value of capacitance,

$ C^{\prime}=\frac{K \varepsilon_0 A}{d}=K C=3 \times \frac{\varepsilon_0}{3}=\varepsilon_0 $

Charge stored $q^{\prime}=C^{\prime} V^{\prime}$

$ \Rightarrow \quad q^{\prime}=\varepsilon_0 \cdot \frac{V}{K}=\varepsilon_0 \frac{12}{3}=4 \varepsilon_0 $

Work done $W=q^{\prime} V^{\prime}$

$ \begin{aligned} & =4 \varepsilon_0 \cdot \frac{V}{K}=4 \varepsilon_0 \frac{12}{3}=16 \varepsilon_0 \\ & =\alpha \varepsilon_0\,\,\,\,\text { (Given) } \\ \alpha & =16 \end{aligned} $

The given circuit diagram is redrawn as,

Equivalent capacitance between point $A$ And $B$ is given as,

$ \begin{aligned} C_{A B} & =\frac{\left(C_2+C_3\right) \times C_1}{\left(C_2+C_3\right)+C_1} \\ & =\frac{(3 C+2 C) \times C}{3 C+2 C+C} \\ & =\frac{5 C \times C}{6 C}=\frac{5 C}{6} \end{aligned} $

Given, $C_1=C_2=C_3=1 \mu \mathrm{~F}=10^{-6} \mathrm{~F}$

$ V=9 \mathrm{~V} $

When switch is thrown to the right side, then left side of switch is open. Hence, capacitors $C_2$ and $C_3$ are uneffective. Therefore, only capacitor $C_2$ is in valid connection.

∴ Charge on capacitor $C_1$,

$ Q_1=C_1 V=1 \times 10^{-6} \times 9=9 \times 10^{-6} \mathrm{C}=9 \mu \mathrm{C} $

Now, when switch is thrown to left, then capacitors $C_1, C_2$ and $C_3$ are connected in series and charge of charged capacitor $C_1$ is flowing towards uncharged capacitors $C_2$ and $C_3$.

Since, $C_1=C_2=C_3$, hence $V_{C_1}=V_{C_2}=V_{C_3}$

Therefore, $Q_2=Q_3=\frac{Q_1}{3}=\frac{9}{3}=3 \mu \mathrm{C}$

$ \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.