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).

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$

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

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

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

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:

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

Charge on the left surface of plate $\mathrm{A} = {{\mathrm{Total\,charge}} \over 2}$

$ = {{{q_1} + {q_2}} \over 2}$

Let right surface of plate A has charge $= x$

And total charge on plate $A = {q_1}$

$\therefore$ $q_1$ = Charge on left surface of plate A + Charge on right surface of plate A

$ = {{{q_1} + {q_2}} \over 2} + x$

$ \Rightarrow x = {q_1} - {{{q_1} + {q_2}} \over 2}$

$ = {{2{q_1} - {q_1} - {q_2}} \over 2}$

$ = {{{q_1} - {q_2}} \over 2}$

Let potential difference between two plates $= V$

For capacitor we know,

$q = CV$

$\therefore$ ${{{q_1} - {q_2}} \over 2} = CV$

$ \Rightarrow V = {{{q_1} - {q_2}} \over {2C}}$

${C_0} = {{{\varepsilon _0}A} \over d}$

$C = {{{\varepsilon _0}A} \over {d - {{3d} \over 4} + {{3d} \over {4K}}}} = {{4{\varepsilon _0}AK} \over {3d + Kd}}$

$ = {{4K{C_0}} \over {3 + K}}$

${{40K \times 40} \over {40K + 40}} = 24$

$40K = 24(K + 1)$

$40K = 24K + 24$

$16K = 24$

$K = {{24} \over {16}} = {3 \over 2} = 1.5$

${E_1} = {1 \over 2}(2C){V^2}$

$ \Rightarrow {E_1} = C{V^2}$ ...... (i)

${E_2} = {1 \over 2}(5C){V^2} + {1 \over 2}{{{{(CV)}^2}} \over {5C}}$

$ = {{13} \over 5}C{V^2}$

$ \Rightarrow {{{E_1}} \over {{E_2}}} = {5 \over {13}}$

Equivalent $C = \sum {{C_i}} $

$ = 10\,\mu F$

$\Rightarrow$ Charge $Q = CV$

$ = 200\,\mu C$

Initially $ = {C_0} = 4\pi {\varepsilon _0}{R_1}$

Finally ${{4\pi {\varepsilon _0}{R_1}{R_2}} \over {{R_2} - {R_1}}} = n{C_0} = 4\pi {\varepsilon _0}n{R_1}$

$ \Rightarrow $ ${{{R_2}} \over {{R_2} - {R_1}}} = n$

$ \Rightarrow $$1 - {{{R_1}} \over {{R_2}}} = {1 \over n}$

$ \Rightarrow $ ${{{R_1}} \over {{R_2}}} = {{n - 1} \over n}$

$ \Rightarrow $ ${{{R_2}} \over {{R_1}}} = {n \over {n - 1}}$

Q = CV

As capacitance is constant so, Q $\propto$ V

So graph between V and Q will be a straight line.

Initially voltage, ${V_i} = {{{Q_i}} \over C} = {0 \over {2 \times {{10}^{ - 6}}}} = 0 V$

and ${V_f} = {{{Q_f}} \over C} = {5 \over {2 \times {{10}^{ - 6}}}} = 2.5 \times {10^6}\,V$

So correct graph will be A.

Capacitance of Fig. 1, ${C_0} = {{{\varepsilon _0}A} \over d}$

In Fig. 2, C₁ and C₂ are parallel with each other,

$\therefore$ Equivalent capacitance,

${C_{eq}} = {C_1} + {C_2}$

$ = {{{\varepsilon _0}A/2} \over d} + {{{\varepsilon _0}{\varepsilon _r}A/2} \over d}$

$ = {{{\varepsilon _0}A} \over {2d}} + {{{\varepsilon _0}{\varepsilon _r}A} \over {2d}}$

$ = {{A{\varepsilon _0}} \over {2d}}\left( {1 + {\varepsilon _r}} \right) = {{{C_0}} \over 2}\left( {1 + {\varepsilon _r}} \right)$

For a discharging capacitor when energy reduces to half the charge would become ${1 \over {\sqrt 2 }}$ times the initial value.

$ \Rightarrow {\left( {{1 \over 2}} \right)^{1/2}} = {e^{ - {t_1}/\tau }}$

Similarly, ${\left( {{1 \over 2}} \right)^3} = {e^{ - {t_2}/\tau }}$

$ \Rightarrow {{{t_1}} \over {{t_2}}} = {1 \over 6}$

$U = {1 \over 2}(k{C_0}){V^2}$

$ \Rightarrow {{U'} \over U} = 1.5$

$\Rightarrow$ Energy increases by 50%

E between two plates is ${\sigma \over {{\varepsilon _0}}}$ and due to one plate is ${\sigma \over {2{\varepsilon _0}}}$ so the force will be halved

So new force F = 5 N

${C_{eq}} = {{120} \over {26}}\mu F$

$ \Rightarrow {Q_{flown}}$ or $Q = {{13 \times 120} \over {26}}\mu C = 60\mu C$

$\Rightarrow$ Charge on 15 $\mu$F capacitor = 60 $\mu$C

As all the capacitors are in series.

Equivalent circuit is

Now, ${{{\varepsilon _0}A} \over d} = 4\,\mu F$

$ \Rightarrow {{12} \over 8}{{{\varepsilon _0}A} \over d} = 6\,\mu F$

${C_{eq}} = {{{\varepsilon _0}A} \over {d - {d \over 2} + {d \over {2k}}}} = {{{\varepsilon _0}A} \over {{d \over 2}}} = {{2{\varepsilon _0}A} \over d}$

If $C = {{{\varepsilon _0}A} \over d}$

$ \Rightarrow {C_{eq}} = 2C$ or ${{{C_{new}}} \over {{C_{old}}}} = {2 \over 1}$

Let initially the charge is q so

${1 \over 2}{{{q^2}} \over C} = {U_i}$

And ${1 \over 2}{{{{(q + 2)}^2}} \over C} = {U_f}$

Given ${{{U_f} - {U_i}} \over {{U_i}}} \times 100 = 44$

${{{{(q + 2)}^2} - {q^2}} \over q} = .44$

$ \Rightarrow q = 10C$

Field inside the dielectric $ = {\sigma \over {k{\varepsilon _0}}}$

According to the given information,

${\sigma \over {k{\varepsilon _0}}} = 3.6 \times {10^7}$

$ \Rightarrow {{{Q \over A}} \over {k{\varepsilon _0}}} = 3.6 \times {10^7}$

$ \Rightarrow k = 2.33$