Capacitor

In a series combination, the effective capacitance ( $\mathrm{C}_{\mathrm{eff}}$ ) is given by :

$ \frac{1}{\mathrm{C}_{\mathrm{eff}}}=\frac{1}{\mathrm{C}_1}+\frac{1}{\mathrm{C}_2}+\cdots+\frac{1}{\mathrm{C}_{\mathrm{n}}} $

Mathematically, the sum of reciprocals always results in a total value that is less than any individual component. Hence, statement (A) is correct.

While dielectrics are insulators and do not allow the flow of current (conduction), they absolutely allow the displacement of charges. Under an external electric field, the molecules in a dielectric polarize, i.e., their positive and negative charges shift slightly in opposite directions. This is known as dielectric polarization.

Hence, the statement (B) is incorrect.

The capacitance $(C)$ of a parallel plate capacitor is given by:

$ C=\frac{\kappa \epsilon_0 A}{d} $

Where A is the area and d is the distance (thickness of dielectric).

If A increases, C increases $(\mathrm{C} \propto \mathrm{A})$.

If d decreases, C increases $\left(\mathrm{C} \propto \frac{1}{\mathrm{~d}}\right)$.

Hence, the statement (C) is correct.

The potential ( $V$ ) due to a point charge is $V=\frac{\mathrm{kq}}{\mathrm{r}}$. This means the potential depends only on the distance ( $r$ ) from the charge. For a fixed r , the potential is constant. All points at a constant distance r from a center form a spherical shell. Therefore, these concentric shells are equipotential surfaces.

Hence, the statement (D) is correct.

Statements A, C, and D are true, while statement B is false. Hence, the correct option is (A).

The capacitance of a parallel plate capacitor is $C=\frac{k \epsilon_0 A}{d}$, where $k$ is the dielectric constant of the material, $A$ is the area of the plate and d is the separation between the plates.

When capacitors are in parallel then the equivalent capacitance is given as

$ C_{e q}=C_1+C_2+\cdots+C_n $

When capacitors are in series then the equivalent capacitance is given as

$ \frac{1}{\mathrm{C}_{\mathrm{eq}}}=\frac{1}{\mathrm{C}_1}+\frac{1}{\mathrm{C}_2}+\cdots+\frac{1}{\mathrm{C}_n} $

Let $\mathrm{C}_0=\frac{\epsilon_0 \mathrm{~A}}{\mathrm{~d}}$.

For Capacitor (A)

The capacitance, $C_{A 1}=\frac{\mathrm{k}_1 \epsilon_0 \mathrm{~A}}{\mathrm{~d} / 2}=\frac{2 \mathrm{k}_1 \epsilon_0 \mathrm{~A}}{\mathrm{~d}}=2 \mathrm{k}_1 \mathrm{C}_0$

The capacitance, $C_{A 2}=\frac{k_1 \epsilon_0\left(\frac{A}{2}\right)}{d / 2}=\frac{k_1 \epsilon_0 A}{d}=k_1 C_0$

The capacitance, $\mathrm{C}_{\mathrm{A} 3}=\frac{\mathrm{k}_2 \epsilon_0 \mathrm{~A} / 2}{\mathrm{~d} / 2}=\frac{\mathrm{k}_2 \epsilon_0 \mathrm{~A}}{\mathrm{~d}}=\mathrm{k}_2 \mathrm{C}_0$

$\mathrm{C}_{\mathrm{A} 2}$ and $\mathrm{C}_{\mathrm{A} 3}$ are in parallel,

$ \mathrm{C}_{\mathrm{A}(2,3)}=\mathrm{k}_1 \mathrm{C}_0+\mathrm{k}_2 \mathrm{C}_0=\left(\mathrm{k}_1+\mathrm{k}_2\right) \mathrm{C}_0 $

Now, $C_{A 1}$ and $C_{A(2,3)}$ are in series, so the equivalent capacitance of capacitor $A$ is

$ \frac{1}{C_A}=\frac{1}{C_{A 1}}+\frac{1}{C_{A(2,3)}}=\frac{1}{2 k_1 C_0}+\frac{1}{\left(k_1+k_2\right) C_0} $

$\Rightarrow $ $ \frac{1}{\mathrm{C}_{\mathrm{A}}}=\frac{\mathrm{k}_1+\mathrm{k}_2+2 \mathrm{k}_1}{2 \mathrm{k}_1\left(\mathrm{k}_1+\mathrm{k}_2\right) \mathrm{C}_0}=\frac{3 \mathrm{k}_1+\mathrm{k}_2}{2 \mathrm{k}_1\left(\mathrm{k}_1+\mathrm{k}_2\right) \mathrm{C}_0} $

So, $C_A=\frac{2 \mathrm{k}_1\left(\mathrm{k}_1+\mathrm{k}_2\right) \mathrm{C}_0}{3 \mathrm{k}_1+\mathrm{k}_2}$.

For Capacitor (B)

The capacitance, $C_{B 1}=\frac{\mathrm{k}_2 \epsilon_0 \mathrm{~A}}{\mathrm{~d} / 2}=\frac{2 \mathrm{k}_2 \epsilon_0 \mathrm{~A}}{\mathrm{~d}}=2 \mathrm{k}_2 \mathrm{C}_0$

The capacitance, $C_{B 2}=\frac{\mathrm{k}_1 \epsilon_0\left(\frac{\mathrm{~A}}{2}\right)}{\mathrm{d} / 2}=\frac{\mathrm{k}_1 \epsilon_0 \mathrm{~A}}{\mathrm{~d}}=\mathrm{k}_1 \mathrm{C}_0$

The capacitance, $C_{B 3}=\frac{\mathrm{k}_2 \epsilon_0 \mathrm{~A} / 2}{\mathrm{~d} / 2}=\frac{\mathrm{k}_2 \epsilon_0 \mathrm{~A}}{\mathrm{~d}}=\mathrm{k}_2 \mathrm{C}_0$

$\mathrm{C}_{\mathrm{B} 2}$ and $\mathrm{C}_{\mathrm{B} 3}$ are in parallel,

$ \mathrm{C}_{\mathrm{B}(2,3)}=\mathrm{k}_1 \mathrm{C}_0+\mathrm{k}_2 \mathrm{C}_0=\left(\mathrm{k}_1+\mathrm{k}_2\right) \mathrm{C}_0 $

Now, $C_{B 1}$ and $C_{B(2,3)}$ are in series, so the equivalent capacitance of capacitor $B$ is

$ \frac{1}{C_B}=\frac{1}{C_{B 1}}+\frac{1}{C_{B(2,3)}}=\frac{1}{2 \mathrm{k}_2 \mathrm{C}_0}+\frac{1}{\left(\mathrm{k}_1+\mathrm{k}_2\right) \mathrm{C}_0} $

$\Rightarrow $ $ \frac{1}{\mathrm{C}_{\mathrm{A}}}=\frac{\mathrm{k}_1+\mathrm{k}_2+2 \mathrm{k}_2}{2 \mathrm{k}_2\left(\mathrm{k}_1+\mathrm{k}_2\right) \mathrm{C}_0}=\frac{\mathrm{k}_1+3 \mathrm{k}_2}{2 \mathrm{k}_2\left(\mathrm{k}_1+\mathrm{k}_2\right) \mathrm{C}_0} $

So, $C_B=\frac{2 \mathrm{k}_2\left(\mathrm{k}_1+\mathrm{k}_2\right) \mathrm{C}_0}{\mathrm{k}_1+3 \mathrm{k}_2}$.

For Capacitor (C)

The capacitance, $C_{C 1}=\frac{\mathrm{k}_1 \epsilon_0 \mathrm{~A} / 2}{\mathrm{~d} / 2}=\frac{\mathrm{k}_1 \epsilon_0 \mathrm{~A}}{\mathrm{~d}}=\mathrm{k}_1 \mathrm{C}_0$

The capacitance, $C_{C 2}=\frac{\mathrm{k}_2 \epsilon_0 \mathrm{~A} / 2}{\mathrm{~d} / 2}=\frac{\mathrm{k}_2 \epsilon_0 \mathrm{~A}}{\mathrm{~d}}=\mathrm{k}_2 \mathrm{C}_0$

The capacitance, $C_{C 3}=\frac{\mathrm{k}_2 \epsilon_0 \mathrm{~A} / 2}{\mathrm{~d} / 2}=\frac{\mathrm{k}_2 \epsilon_0 \mathrm{~A}}{\mathrm{~d}}=\mathrm{k}_2 \mathrm{C}_0$

The capacitance, $\mathrm{C}_{\mathrm{C} 4}=\frac{\mathrm{k}_1 \epsilon_0 \mathrm{~A} / 2}{\mathrm{~d} / 2}=\frac{\mathrm{k}_1 \epsilon_0 \mathrm{~A}}{\mathrm{~d}}=\mathrm{k}_1 \mathrm{C}_0$

$\mathrm{C}_{\mathrm{C} 1}$ and $\mathrm{C}_{\mathrm{C} 2}$ are in parallel,

$ \mathrm{C}_{\mathrm{C}(1,2)}=\mathrm{k}_1 \mathrm{C}_0+\mathrm{k}_2 \mathrm{C}_0=\left(\mathrm{k}_1+\mathrm{k}_2\right) \mathrm{C}_0 $

Now, $\mathrm{C}_{\mathrm{C} 3}$ and $\mathrm{C}_{\mathrm{C} 4}$ are in parallel,

$ \mathrm{C}_{\mathrm{C}(3,4)}=\mathrm{k}_2 \mathrm{C}_0+\mathrm{k}_1 \mathrm{C}_0=\left(\mathrm{k}_1+\mathrm{k}_2\right) \mathrm{C}_0 $

Now, $\mathrm{C}_{\mathrm{C}(1,2)}$ and $\mathrm{C}_{\mathrm{C}(3,4)}$ are in series, so the equivalent capacitance of capacitor C is

$ \frac{1}{\mathrm{C}_{\mathrm{C}}}=\frac{1}{\mathrm{C}_{\mathrm{C}(1,2)}}+\frac{1}{\mathrm{C}_{\mathrm{C}(3,4)}}=\frac{1}{\left(\mathrm{k}_1+\mathrm{k}_2\right) \mathrm{C}_0}+\frac{1}{\left(\mathrm{k}_1+\mathrm{k}_2\right) \mathrm{C}_0} $

$\Rightarrow $ $\frac{1}{\mathrm{C}_{\mathrm{C}}}=\frac{2}{\left(\mathrm{k}_1+\mathrm{k}_2\right) \mathrm{C}_0}$

So, $\mathrm{C}_{\mathrm{C}}=\frac{\left(\mathrm{k}_1+\mathrm{k}_2\right) \mathrm{C}_0}{2}$.

It is given that $\mathrm{k}_1>\mathrm{k}_2$, let $\mathrm{k}_1=2$ and $\mathrm{k}_2=1$

$ C_A=\frac{2 \times 2 \times 3}{7} C_0=\frac{12}{7} C_0 \approx 1.71 C_0 $

$C_B=\frac{2 \times 1 \times 3}{5} C_0=\frac{6}{5} C_0=1.2 C_0$

$\mathrm{C}_{\mathrm{C}}=2 \times 2 \times \frac{1}{3} \mathrm{C}_0=\frac{4}{3} \mathrm{C}_0 \approx 1.33 \mathrm{C}_0$

$ \Rightarrow \mathrm{C}_{\mathrm{A}}>\mathrm{C}_{\mathrm{C}}>\mathrm{C}_{\mathrm{B}} $

Hence, the correct option is (C).

Initially, the capacitor has air between the plates with separation $\mathrm{d}=5 \mathrm{~mm}=5 \times 10^{-3} \mathrm{~m}$.

The initial capacitance ( $\mathrm{C}_0$ ) is:

$ C_0=\frac{\epsilon_0 A}{d}=\frac{\epsilon_0 A}{5} $

Where A is the area of the plates and d is in mm ).

A mica sheet of thickness $\mathrm{t}=2 \mathrm{~mm}$ and dielectric constant K is introduced. The new capacitance $\left(\mathrm{C}^{\prime}\right)$ of a capacitor partially filled with a dielectric is given by:

$ C^{\prime}=\frac{\epsilon_0 A}{d-t+\frac{t}{K}} $

Substituting the values $\mathrm{d}=5$ and $\mathrm{t}=2$ :

$ C^{\prime}=\frac{\epsilon_0 A}{5-2+\frac{2}{K}}=\frac{\epsilon_0 A}{3+\frac{2}{K}} $

The battery remains connected, meaning the potential difference (V) stays constant.

Initial charge: $\mathrm{Q}_0=\mathrm{C}_0 \mathrm{~V}$

Final charge: $Q^{\prime}=C^{\prime} V$

The capacitor draws $25 \%$ more charge :

$ \mathrm{Q}^{\prime}=\mathrm{Q}_0+0.25 \mathrm{Q}_0=1.25 \mathrm{Q}_0 $

$\Rightarrow $ $\mathrm{C}^{\prime} \mathrm{V}=1.25 \mathrm{C}_0 \mathrm{~V}$

$\Rightarrow C^{\prime}=\frac{5}{4} C_0$

$\Rightarrow $ $\frac{\epsilon_0 \mathrm{~A}}{3+\frac{2}{\mathrm{~K}}}=\frac{5}{4}\left(\frac{\epsilon_0 \mathrm{~A}}{5}\right)$

$\Rightarrow $ $\frac{1}{3+\frac{2}{K}}=\frac{1}{4}$

$\Rightarrow $ $3+\frac{2}{K}=4$

$\Rightarrow $ $\frac{2}{\mathrm{~K}}=1$

$\Rightarrow $ $ K=2 $

The dielectric constant K is 2.0 .

Hence, the correct option is (D).

For a parallel plate capacitor with plate area $A$, separation $d$, and vacuum between the plates, the capacitance $C$ is given by:

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

A dielectric sheet of thickness $t=\frac{d}{3}$ and relative permittivity (dielectric constant) $K$ is inserted.

So, the system is made of two capacitors connected in series:

• Capacitor $1\left(\mathrm{C}_1\right)$ : A capacitor filled with the dielectric material of thickness $\mathrm{t}=\frac{\mathrm{d}}{3}$.

• Capacitor $2\left(\mathrm{C}_2\right)$ : A capacitor with vacuum filling the remaining space of thickness $(\mathrm{d}-\mathrm{t})=\mathrm{d}- \frac{\mathrm{d}}{3}=\frac{2 \mathrm{~d}}{3}$.

For the capacitor $\left(\mathrm{C}_1\right)$ :

$ C_1=\frac{K \epsilon_0 A}{t}=\frac{K \epsilon_0 A}{d / 3}=\frac{3 K \epsilon_0 A}{d}=3 K C $

For the capacitor $\left(\mathrm{C}_2\right)$ :

$ C_2=\frac{\epsilon_0 A}{d-t}=\frac{\epsilon_0 A}{2 d / 3}=\frac{3 \epsilon_0 A}{2 d}=\frac{3}{2} C $

Since they are in series, the equivalent capacitance is :

$ \frac{1}{\mathrm{C}_{\mathrm{eq}}}=\frac{1}{\mathrm{C}_1}+\frac{1}{\mathrm{C}_2} $

Substituting the values :

$ \frac{1}{\mathrm{C}_{\mathrm{eq}}}=\frac{1}{3 \mathrm{KC}}+\frac{1}{3 / 2 \mathrm{C}} $

$\Rightarrow $ $\frac{1}{C_{e q}}=\frac{1}{3 K C}+\frac{2}{3 C}$

$\Rightarrow $ $\frac{1}{\mathrm{C}_{\mathrm{eq}}}=\frac{1+2 \mathrm{~K}}{3 \mathrm{KC}}$

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

Therefore, the new capacitance of the system is $\frac{3 \mathrm{KC}}{2 \mathrm{~K}+1}$.

Hence, the correct option is (D).

The work done in charging a capacitor is stored as electrostatic potential energy. If a charge dq is moved to a capacitor of capacitance C at potential $\mathrm{V}=\mathrm{q} / \mathrm{C}$, the work dW is:

$ \mathrm{dW}=\mathrm{Vdq}=\frac{\mathrm{q}}{\mathrm{C}} \mathrm{dq} $

Integrating from $\mathrm{q}=0$ to total charge $\mathrm{q}=\mathrm{Q}$ :

$ \mathrm{U}=\int_0^{\mathrm{Q}} \frac{\mathrm{q}}{\mathrm{C}} \mathrm{dq}=\frac{1}{\mathrm{c}} \int_0^{\mathrm{Q}} \mathrm{qdq}=\frac{\mathrm{Q}^2}{2 \mathrm{C}} $

Using $\mathrm{Q}=\mathrm{CV}$,

$ \mathrm{U}=\frac{1}{2} \mathrm{CV}^2 $

Initially, only the first sphere ( $\mathrm{C}_1=100 \mathrm{pF}$ ) is charged to $\mathrm{V}_1=100 \mathrm{~V}$.

$ U_i=\frac{1}{2} C_1 V_1^2 $

$\Rightarrow $ $U_i=\frac{1}{2} \times\left(100 \times 10^{-12}\right) \times(100)^2$

$\Rightarrow $ $ \mathrm{U}_{\mathrm{i}}=\frac{1}{2} \times 10^{-10} \times 10^4=0.5 \times 10^{-6} \mathrm{~J}=5 \times 10^{-7} \mathrm{~J} $

Finally, when the spheres touch, they share charge until they reach a common potential $\mathrm{V}_{\mathrm{c}}$.

Since the spheres are identical ( $\mathrm{C}_1=\mathrm{C}_2=100 \mathrm{pF}$ ), the charge Q is shared equally, and

$ \mathrm{V}_{\mathrm{c}}=\frac{\text { Total Charge }}{\text { Total Capacitance }}=\frac{\mathrm{C}_1 \mathrm{~V}_1+0}{\mathrm{C}_1+\mathrm{C}_2}=\frac{\mathrm{V}_1}{2}=50 \mathrm{~V} . $

So, the final energy is,

$ U_f=\frac{1}{2}\left(C_1+C_2\right) V_c^2 $

$\Rightarrow $ $U_f=\frac{1}{2} \times\left(200 \times 10^{-12}\right) \times(50)^2$

$\Rightarrow $ $ U_f=100 \times 10^{-12} \times 2500=2.5 \times 10^{-7} J $

So, the change in energy is,

$ \Delta U=U_i-U_f $

$\Rightarrow $ $\Delta \mathrm{U}=5 \times 10^{-7} \mathrm{~J}-2.5 \times 10^{-7} \mathrm{~J}=2.5 \times 10^{-7} \mathrm{~J}=\alpha \times 10^{-7} \mathrm{~J}$

$\Rightarrow \alpha=2.5=\frac{5}{2}$

Update Capacitance of $ C_1 $:

The initial capacitance of $ C_1 $ is $ 5 \, \mu \text{F} $. Adding a dielectric with a dielectric constant of 4 increases $ C_1 $ to:

$ C_1 = 4 \times 5 = 20 \, \mu \text{F} $

Identify Capacitances of $ C_2 $ and $ C_3 $:

Both $ C_2 $ and $ C_3 $ remain at $ 5 \, \mu \text{F} $ since no dielectric is added to them.

$ C_2 = 5 \, \mu \text{F}, \quad C_3 = 5 \, \mu \text{F} $

Calculate Effective Capacitance:

Capacitors $ C_1 $ and $ C_2 $ are in series, and this series combination is parallel to $ C_3 $.

Series Combination of $ C_1 $ and $ C_2 $:

The formula for capacitors in series is:

$ C_{\text{series}} = \frac{C_1 \times C_2}{C_1 + C_2} = \frac{20 \times 5}{20 + 5} = \frac{100}{25} = 4 \, \mu \text{F} $

Parallel Combination with $ C_3 $:

The formula for capacitors in parallel is:

$ C_{\text{eq}} = C_{\text{series}} + C_3 = 4 + 5 = 9 \, \mu \text{F} $

Thus, the effective capacitance between points $ A $ and $ B $ is $ 9 \, \mu \text{F} $.

At first, a capacitor with capacitance $ C_0 = 100 $ pF is charged to a voltage $ V_0 = 60 $ V. This means it stores a charge $ Q = C_0 V_0 $.

Then, the battery is removed, so the total charge stays the same. Now, a second uncharged capacitor, $ C $, is connected in parallel. The two capacitors will now share the charge.

The final voltage across both capacitors is given in the question as 20 V. The charge is now shared between both capacitors:
Total charge = $ (C_0 + C) \times 20 $

But total charge stays the same as before, so:
$ (C_0 + C) \times 20 = C_0 \times 60 $

Divide both sides by 20:
$ C_0 + C = 3 C_0 $
So, $ C = 2 C_0 $

Since $ C_0 = 100 $ pF, the second capacitor $ C $ is $ 2 \times 100 = 200 $ pF.

Area of plate is A. then

$\begin{aligned} & \mathrm{C}=\frac{\varepsilon_2 \varepsilon_0 \mathrm{~A}}{\mathrm{~d} / 2}=\frac{2 \varepsilon_2 \varepsilon_0 \mathrm{~A}}{\mathrm{~d}} \\ & \mathrm{C}^{\prime}=\frac{\varepsilon_1 \varepsilon_0 \mathrm{~A}}{\mathrm{~d} / 2}=\frac{2 \varepsilon_1 \varepsilon_0 \mathrm{~A}}{\mathrm{~d}} \end{aligned}$

$\begin{aligned} &\text { Let } \mathrm{C}_0=\frac{\varepsilon_0 \mathrm{~A}}{\mathrm{~d}}\\ &\mathrm{C}=2 \varepsilon_2 \mathrm{C}_0\\ &\mathrm{C}^{\prime}=2 \varepsilon_1 \mathrm{C}_0 \end{aligned}$

C & C' are in series

$\begin{aligned} & \mathrm{C}_1=\frac{\mathrm{CC}^{\prime}}{\mathrm{C}+\mathrm{C}^{\prime}}=\frac{4 \varepsilon_2 \varepsilon_1 \mathrm{C}_0^2}{2 \mathrm{C}_0\left(\varepsilon_2+\varepsilon_1\right)} \\ & =\frac{2 \varepsilon_2 \varepsilon_1 \mathrm{C}_0}{\left(\varepsilon_2+\varepsilon_1\right)} \end{aligned}$

Here $C=\frac{\varepsilon_1 \varepsilon_0 A}{2 d}=\frac{\varepsilon_1 C_0}{2}$

$C^{\prime}=\frac{\varepsilon_2 C_0}{2}$

$\mathrm{C} \& \mathrm{C}^{\prime}$ are inparallel

$\mathrm{C}_2=\mathrm{C}^{\prime}+\mathrm{C}=\left(\varepsilon_1+\varepsilon_2\right) \frac{\mathrm{C}_0}{2}$

Thus $\frac{\mathrm{C}_1}{\mathrm{C}_2}=\frac{2 \varepsilon_2 \varepsilon_1 \mathrm{C}_0}{\left(\varepsilon_2+\varepsilon_1\right)} \times \frac{2}{\left(\varepsilon_1+\varepsilon_2\right) \mathrm{C}_0}$

$=\frac{4 \varepsilon_2 \varepsilon_1}{\left(\varepsilon_2+\varepsilon_1\right)^2}$

Given, ${c_1} = 6\mu F,{c_2} = 12\mu F,v = 5v$

We know, for a capacitor $Q = CV$

Initial charge on ${c_1}\,.\,Q = {c_1}v$

$ \Rightarrow Q = 6 \times 5$

$ \Rightarrow Q = 30\mu C$ .... (1)

when switch 'S' is closed, let q charge flows from C$_1$ to C$_2$

So final charges are $Q-q$ and q on C$_1$ and C$_2$ respectively.

Now, when equilibrium condition is reached potential on C$_1$ and C$_2$ becomes equal.

Hence, ${{Q - q} \over {{c_1}}} = {q \over {{c_2}}}$ (As $v = {Q \over c}$)

$ \Rightarrow {Q \over {{c_1}}} = q\left( {{{{c_1} + {c_2}} \over {{c_1}{c_2}}}} \right)$

$ \Rightarrow q = \left( {{{{c_2}} \over {{c_1} + {c_2}}}} \right)Q$

$ \Rightarrow q = \left( {{{12} \over {6 + 12}}} \right) \times 30\mu C$ (From (1))

$ \Rightarrow q = \left( {{{12} \over {18}}} \right) \times 30 = 20\mu c$

and $Q - q = 30 - 20 = 10\mu c$

Hence, $q = 10\mu c$ and ${q_2} = 20\mu c$

The energy density ($u$) between the plates of a capacitor is given by the formula:

$ u = \frac{1}{2} \varepsilon_0 E^2 $

where $\varepsilon_0$ is the permittivity of free space ($8.85 \times 10^{-12} \, \text{F/m}$) and $E$ is the electric field between the plates. The electric field $E$ is related to the potential difference $V$ and the separation $d$ between the plates by:

$ E = \frac{V}{d} $

Given:

Capacitance ($C$) = 1 µF = $1 \times 10^{-6} \, \text{F}$

Potential Difference ($V$) = 20 V

Distance ($d$) = 1 µm = $1 \times 10^{-6} \, \text{m}$

First, calculate the electric field $E$:

$ E = \frac{V}{d} = \frac{20 \, \text{V}}{1 \times 10^{-6} \, \text{m}} = 2 \times 10^7 \, \text{V/m} $

Now plug this into the formula for energy density:

$ u = \frac{1}{2} \times 8.85 \times 10^{-12} \, \text{F/m} \times (2 \times 10^7 \, \text{V/m})^2 $

Calculate $u$:

$ u = \frac{1}{2} \times 8.85 \times 10^{-12} \times 4 \times 10^{14} $

$ u = 2 \times 8.85 \times 10^2 $

$ u = 1770 \, \text{J/m}^3 $

This value is approximately $1.8 \times 10^3 \, \text{J/m}^3$. Therefore, the correct option is:

Option A: $1.8 \times 10^3 \, \text{J/m}^3$

Since parallel combination, so v is same for both

we know, $q=cv$

$\Rightarrow q \propto C$ (for constant v)

we can see in the graph,

at final state, $q_2 > q_1$,

so $c_2 > c_1$

Now, as $u = {1 \over 2}c{v^2} \Rightarrow u \propto c$ (for constant v)

$ \Rightarrow {u_2} > {u_1}$

Hence, option 1 is correct.

The capacitance of a parallel plate capacitor is given by

$ C = \frac{\epsilon_0 A}{d}, $

where:

$\epsilon_0$ is the permittivity of free space,

$A$ is the area of a plate, and

$d$ is the separation between the plates.

A change in the capacitor’s dimensions will change its capacitance by a factor of

$ \frac{C_{\text{new}}}{C_{\text{initial}}} = \frac{A_{\text{new}}}{A_{\text{initial}}} \cdot \frac{d_{\text{initial}}}{d_{\text{new}}}. $

The initial dimensions are:

Length: $l = 3\,\text{cm} = 0.03\,\text{m},$

Breadth: $b = 1\,\text{cm} = 0.01\,\text{m},$

Hence, the original area is

$ A_{\text{initial}} = 0.03 \times 0.01 = 3 \times 10^{-4}\,\text{m}^2, $

Plate separation: $d_{\text{initial}} = 3\,\mu\text{m} = 3 \times 10^{-6}\,\text{m}.$

For each modification, we compare the new factor:

Case C:

New dimensions:

$l = 6\,\text{cm} = 0.06\,\text{m}, \quad b = 5\,\text{cm} = 0.05\,\text{m}.$

New area:

$ A_{\text{new}} = 0.06 \times 0.05 = 3 \times 10^{-3}\,\text{m}^2. $

Ratio of areas:

$ \frac{A_{\text{new}}}{A_{\text{initial}}} = \frac{3 \times 10^{-3}}{3 \times 10^{-4}} = 10. $

The plate separation is unchanged:

$ \frac{d_{\text{initial}}}{d_{\text{new}}} = \frac{3 \times 10^{-6}}{3 \times 10^{-6}} = 1. $

Overall factor:

$ 10 \times 1 = 10. $

Case E:

New dimensions:

$l = 5\,\text{cm} = 0.05\,\text{m}, \quad b = 2\,\text{cm} = 0.02\,\text{m}.$

New area:

$ A_{\text{new}} = 0.05 \times 0.02 = 1 \times 10^{-3}\,\text{m}^2. $

Ratio of areas:

$ \frac{A_{\text{new}}}{A_{\text{initial}}} = \frac{1 \times 10^{-3}}{3 \times 10^{-4}} \approx 3.33. $

New plate separation:

$d_{\text{new}} = 1\,\mu\text{m} = 1 \times 10^{-6}\,\text{m},$

so the ratio of separations is

$ \frac{d_{\text{initial}}}{d_{\text{new}}} = \frac{3 \times 10^{-6}}{1 \times 10^{-6}} = 3. $

Overall factor:

$ 3.33 \times 3 \approx 10. $

Both Case C and Case E yield a capacitance that is 10 times the original value.

Thus, the correct modifications are those in Case C and Case E.

Initially, when the key in the circuit is closed, the capacitor behaves like a short circuit. This means:

Maximum Current: The current flowing through the circuit will be at its maximum since there is effectively no resistance offered by the capacitor, as it behaves like a wire.

Current Through R: Since the current is flowing through the circuit, including any resistors in series, the statement that there is no current through resistor $ R $ is incorrect.

Charge on the Capacitor: The charge stored on the capacitor plates is initially zero because it behaves as a short circuit.

Potential Difference: The potential difference across the capacitor is also zero at this instant because it is fully discharged.

Therefore, the valid statements at this instant are that the current in the circuit is at a maximum, the charge on the capacitor is zero, and the potential difference across the capacitor is zero.

The capacitance (in farads) is defined as the ratio of charge to potential difference. Let's go through the steps:

Capacitance is given by:

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

where:

$ Q $ is the charge with dimensional symbol $ C $.

$ V $ is the potential difference.

Voltage (potential difference) is defined as energy per unit charge:

$V = \frac{W}{Q}$

where:

$ W $ is energy with dimensions:

$[W] = [M L^2 T^{-2}]$

Therefore, the dimensions of voltage are:

$[V] = \frac{[ML^2T^{-2}]}{[C]}$

Now, substitute this back into the expression for capacitance:

$[C] = \frac{[Q]}{[ML^2T^{-2}]/[C]} = \frac{[C]^2}{ML^2T^{-2}}$

Simplifying gives:

$[C] = [C^2 M^{-1}L^{-2}T^2]$

Comparing with the given options, we see that Option C is:

$[F] = [C^2 M^{-1}L^{-2}T^2]$

Thus, the correct dimensional formula for the capacitance in farads is given by Option C.

No force in horizontal direction, i.e., $F_x=0$

$\Rightarrow a_x=0$

So, $v_x$ = constant

hence, $t=\frac{l}{v_x}$ ..... (1)

Now, for y-direction, using Ist equation of motion,

${v_y} = {u_y} + {a_y}t$

$ \Rightarrow {v_y} = 0 + {{eE} \over m}\left( {{l \over {{v_x}}}} \right)$ (from (1) and usiung $F=ma$ and $a=\frac{F}{m}$)

$ \Rightarrow {v_y} = {{1.6 \times {{10}^{ - 19}} \times 9.1 \times {{10}^2} \times 0.1} \over {9.1 \times {{10}^{ - 31}} \times {{10}^6}}}$

$ = 1.6 \times {10^7}$ m/s

$ \Rightarrow {v_y} = 16 \times {10^6}$ m/s.

Given, $K = 2,\,C = 40\mu F,\,V = 100\,V$

$\Delta q = (KC)V - CV$ (As $q = CV$)

$ = (K - 1)CV$

$ = (2 - 1) \times 40 \times {10^{ - 6}} \times 100 = 4 \times {10^{ - 3}}C$

$\Delta q = 4\,mC$

$\Delta u = {1 \over 2}C'{V^2} - {1 \over 2}C{V^2}$

$ = {1 \over 2}{V^2}(C' - C) = {1 \over 2}{V^2}(KC - C)$

$ = {1 \over 2}{V^2}C(K - 1) = {1 \over 2} \times {10^4} \times 40 \times {10^{ - 6}}(2 - 1)$

$ = \Delta u = 2 \times {10^{ - 1}} = 0.2\,J$

$ \Rightarrow \Delta u = 0.2\,J$

$ \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} $

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} $

$\begin{aligned} & C=C_1+C_2+C_3 \\ & =\frac{A \epsilon_0}{3 d}+\frac{A}{6} \frac{\epsilon_0}{d}+\frac{A \epsilon_0}{g D} \\ & =\frac{A \epsilon_0}{d}\left(\frac{1}{3}+\frac{1}{6}+\frac{1}{9}\right) \\ & =\frac{11}{18} \frac{A \epsilon_0}{d} \end{aligned}$

$\begin{aligned} & \frac{\in_0 A}{d}=\frac{\epsilon_0 A}{\frac{d}{k}+\frac{0.2}{1}} \\ & \Rightarrow \quad 0.6-0.2=\frac{0.6}{k} \\ & \quad k=\frac{0.6}{0.4}=1.5 \end{aligned}$

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} $