Capacitor

To solve the problem, let's start by considering the energy stored in each capacitor before they are connected together.

The energy stored in a capacitor is given by the formula:

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

Where $E$ is the energy, $C$ is the capacitance, and $V$ is the potential.

For the first capacitor charged to the potential $V$, the energy stored is:

$E_1 = \frac{1}{2} C V^2$

For the second capacitor charged to the potential $2V$, the energy stored is:

$E_2 = \frac{1}{2} C (2V)^2 = \frac{1}{2} C \cdot 4V^2 = 2 CV^2$

The total initial energy stored in the system is the sum of $E_1$ and $E_2$:

$E_{\text{initial}} = E_1 + E_2 = \frac{1}{2} CV^2 + 2 CV^2 = \frac{5}{2} CV^2$

When the two capacitors are connected together, their potentials will become equal because they are identical capacitors. Let's denote this final potential as $V_f$. The total charge before and after the connection remains constant because charge is conserved. Therefore, we can write:

$C \cdot V + C \cdot 2V = 2C \cdot V_f$

Simplifying this equation gives us the final potential:

$V + 2V = 2 V_f$

$3V = 2 V_f$

$V_f = \frac{3}{2} V$

The final energy stored in the system when the capacitors are connected is now the total energy stored across both capacitors at the final potential $V_f$:

$E_{\text{final}} = 2 \cdot \frac{1}{2} C V_f^2 = 2 \cdot \frac{1}{2} C \left(\frac{3}{2} V\right)^2 = C \cdot \frac{9}{4} V^2$

The decrease in energy $ \Delta E $ of the combined system is the initial energy minus the final energy:

$\Delta E = E_{\text{initial}} - E_{\text{final}}$

$\Delta E = \frac{5}{2} CV^2 - \frac{9}{4} CV^2$

$\Delta E = \frac{10}{4} CV^2 - \frac{9}{4} CV^2$

$\Delta E = \frac{1}{4} CV^2$

The correct answer is:

Option A: $\frac{1}{4} CV^2$

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} $

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 electric field inside a parallel plate capacitor is uniform and given by $\mathbf{E}=\frac{V}{d} \hat{\mathbf{j}}$ where $d$ is the separation between the plates.

The electric flux through a closed surface enclosing only the positive plate of the capacitor is given by $\Phi_E = \oint_S \mathbf{E}\cdot d\mathbf{A}$.

Since the electric field is perpendicular to the surface of the plate, the flux through the surface will be constant and given by $\Phi_E = E A$, where $A$ is the area of the plate.

The area of the positive plate is $A = \frac{Q}{\varepsilon_0 V}$, where $Q = C V$ is the charge on the positive plate. Therefore, the electric flux through the surface is given by:

$\Phi_E = \frac{Q}{\varepsilon_0 V} \frac{V}{d} = \frac{Q}{\varepsilon_0 d} = \frac{C V}{\varepsilon_0 d}$

Thus, the answer is $\frac{C V}{\varepsilon_0}$.

To determine the ratio $\mathrm{E}_{2} / \mathrm{E}_{1}$, we will follow these steps:

1. Calculate the initial energy stored in the original capacitor $\mathrm{E}_{1}$.

2. Determine the energy stored in the system when two capacitors are connected in parallel, which is $\mathrm{E}_{2}$.

3. Find the ratio $\mathrm{E}_{2} / \mathrm{E}_{1}$.

Step 1: Calculate the initial energy stored in the original capacitor $\mathrm{E}_{1}$.

The energy stored in a capacitor is given by the formula:

$E = \frac{1}{2} CV^2$

Given that the capacitance $C$ is $2 \mathrm{~F}$, and the potential difference is $\mathrm{V}$, the initial energy $\mathrm{E}_{1}$ is:

$E_{1} = \frac{1}{2} \cdot 2 \mathrm{~F} \cdot V^2$

$E_{1} = V^2 \mathrm{~J}$

Step 2: Determine the energy stored when two capacitors are connected in parallel

When the charged capacitor (capacitor 1) is connected to an identical uncharged capacitor (capacitor 2), the charge will redistribute between the two capacitors. The total capacitance of the parallel combination is:

$C_{\text{total}} = 2 \mathrm{~F} + 2 \mathrm{~F} = 4 \mathrm{~F}$

The initial charge on capacitor 1 is:

$Q_1 = CV = 2 \mathrm{~F} \cdot V = 2V \mathrm{~C}$

After connection, this charge will be shared equally by the two capacitors because they are identical. Therefore, the voltage across each capacitor in the parallel combination will be:

$V_{\text{across each capacitor}} = \frac{\text{Total charge}}{\text{Total capacitance}} = \frac{2V}{4 \mathrm{~F}} = \frac{V}{2}$

The energy stored in the parallel combination is:

$E_{2} = \frac{1}{2} \cdot 4 \mathrm{~F} \cdot \left(\frac{V}{2}\right)^2$

$E_{2} = \frac{1}{2} \cdot 4 \mathrm{~F} \cdot \frac{V^2}{4}$

$E_{2} = V^2 \cdot \frac{1}{2} \mathrm{~J}$

Step 3: Find the ratio $\mathrm{E}_{2} / \mathrm{E}_{1}$

We know that:

$E_{1} = V^2 \mathrm{~J}$

$E_{2} = \frac{1}{2} V^2 \mathrm{~J}$

The ratio is then:

$\frac{E_{2}}{E_{1}} = \frac{\frac{1}{2} V^2}{V^2} = \frac{1}{2} = 1 : 2$

Therefore, the correct answer is:

Option A: 1 : 2

When a metal sheet of thickness ($\frac{2d}{3}$) is introduced between the plates of a capacitor, it divides the capacitor into two separate capacitors. The metal sheet acts as a new plate in each capacitor, and because the metal is a conductor, it is at the same potential as the plates on either side.

The capacitance of the original capacitor with air between the plates is given by:

$\mathrm{C}_1 = \frac{\epsilon_0 \mathrm{A}}{\mathrm{d}}$

where ($\epsilon_0$) is the permittivity of free space, ($\mathrm{A}$) is the area of one of the plates, and ($\mathrm{d}$) is the distance between the plates.

When the metal sheet of thickness ($\frac{2d}{3}$) is introduced, the distance between the plates is reduced by ($\frac{2d}{3}$), so the capacitance of each of the new capacitors is:

$\mathrm{C}' = \frac{\epsilon_0 \mathrm{A}}{\mathrm{d}-\frac{2d}{3}}$

Simplifying, we get:

$\mathrm{C}' = \frac{3\epsilon_0 \mathrm{A}}{\mathrm{d}}$

Since the two capacitors are in parallel with each other, the total capacitance is:

$\mathrm{C}_2 = 2\mathrm{C}' = 2 \times \frac{3\epsilon_0 \mathrm{A}}{\mathrm{d}} = 3\mathrm{C}_1$

So the ratio of the capacitances is:

$\frac{\mathrm{C}_2}{\mathrm{C}_1} = 3:1$