Effect of Dielectric Inside Capacitor Questions in English

(B) When a dielectric material with dielectric constant $K$ is introduced between the plates of a charged capacitor,the induced charges on the dielectric surface create an internal electric field that opposes the original external electric field.
If the original electric field is $E_0$,the new electric field $E$ is given by $E = \frac{E_0}{K}$.
Since $K > 1$ for any dielectric material,the resulting electric field $E$ is less than the original electric field $E_0$.
Therefore,the electric field between the plates decreases.

(D) The dielectric constant $K$ of a medium is defined as the ratio of the capacitance of the capacitor with the dielectric medium $(C_m)$ to the capacitance of the capacitor with air or vacuum $(C_0)$.
Given:
$C_0 = 2.0 \, \mu F$
$C_m = 12 \, \mu F$
Using the formula:
$K = \frac{C_m}{C_0}$
$K = \frac{12 \, \mu F}{2.0 \, \mu F} = 6$
Therefore,the dielectric constant of the medium is $6$.

(B) The capacitance of a parallel plate capacitor is given by $C = \frac{K \varepsilon_0 A}{d}$,where $K$ is the dielectric constant,$\varepsilon_0$ is the permittivity of free space,$A$ is the area of the plates,and $d$ is the distance between them.
From the formula,we see that $C \propto \frac{K}{d}$.
Let the initial capacitance be $C_1 = \frac{K_1 \varepsilon_0 A}{d_1}$.
Given: $d_2 = \frac{d_1}{2}$ and $K_2 = 2K_1$.
The new capacitance $C_2$ is given by $C_2 = \frac{K_2 \varepsilon_0 A}{d_2} = \frac{(2K_1) \varepsilon_0 A}{d_1/2} = 4 \times \frac{K_1 \varepsilon_0 A}{d_1} = 4C_1$.
Therefore,the capacity increases by $4$ times.

(B) The given arrangement can be considered as two capacitors connected in parallel,each having an area $A/2$ and plate separation $t$.
For the first capacitor with dielectric constant $k_1$,the capacitance is $C_1 = \frac{k_1 \varepsilon_0 (A/2)}{t} = \frac{k_1 \varepsilon_0 A}{2t}$.
For the second capacitor with dielectric constant $k_2$,the capacitance is $C_2 = \frac{k_2 \varepsilon_0 (A/2)}{t} = \frac{k_2 \varepsilon_0 A}{2t}$.
Since they are in parallel,the equivalent capacitance $C = C_1 + C_2$.
$C = \frac{k_1 \varepsilon_0 A}{2t} + \frac{k_2 \varepsilon_0 A}{2t} = \frac{\varepsilon_0 A}{2t}(k_1 + k_2) = \frac{\varepsilon_0 A}{t} \cdot \frac{k_1 + k_2}{2}$.
Thus,the correct option is $B$.

(D) The capacitor can be viewed as two capacitors in series,each with plate area $A$ and plate separation $d/2$.
The capacitance of the first capacitor is $C_1 = \frac{K_1 \varepsilon_0 A}{d/2} = \frac{2 K_1 \varepsilon_0 A}{d}$.
The capacitance of the second capacitor is $C_2 = \frac{K_2 \varepsilon_0 A}{d/2} = \frac{2 K_2 \varepsilon_0 A}{d}$.
Since they are in series,the equivalent capacitance $C_{eq}$ is given by $\frac{1}{C_{eq}} = \frac{1}{C_1} + \frac{1}{C_2}$.
$\frac{1}{C_{eq}} = \frac{d}{2 K_1 \varepsilon_0 A} + \frac{d}{2 K_2 \varepsilon_0 A} = \frac{d}{2 \varepsilon_0 A} \left( \frac{1}{K_1} + \frac{1}{K_2} \right) = \frac{d}{2 \varepsilon_0 A} \left( \frac{K_1 + K_2}{K_1 K_2} \right)$.
Therefore,$C_{eq} = \frac{2 \varepsilon_0 A}{d} \left( \frac{K_1 K_2}{K_1 + K_2} \right)$.

(B) The capacitor can be viewed as a combination of three capacitors. The top half consists of two capacitors in parallel with areas $A/2$ and separation $d/2$,having capacitances $C_1 = \frac{k_1 \epsilon_0 (A/2)}{d/2} = \frac{k_1 \epsilon_0 A}{d}$ and $C_2 = \frac{k_2 \epsilon_0 (A/2)}{d/2} = \frac{k_2 \epsilon_0 A}{d}$.
These two are in parallel,so their equivalent capacitance is $C_{12} = C_1 + C_2 = \frac{\epsilon_0 A}{d} (k_1 + k_2)$.
The bottom half is a capacitor with area $A$ and separation $d/2$,having capacitance $C_3 = \frac{k_3 \epsilon_0 A}{d/2} = \frac{2k_3 \epsilon_0 A}{d}$.
Since $C_{12}$ and $C_3$ are in series,the total capacitance $C$ is given by $\frac{1}{C} = \frac{1}{C_{12}} + \frac{1}{C_3}$.
Substituting the values: $\frac{1}{C} = \frac{d}{\epsilon_0 A (k_1 + k_2)} + \frac{d}{2k_3 \epsilon_0 A} = \frac{d}{\epsilon_0 A} \left( \frac{1}{k_1 + k_2} + \frac{1}{2k_3} \right)$.
For a single dielectric $k$,$C = \frac{k \epsilon_0 A}{d}$,so $\frac{1}{C} = \frac{d}{k \epsilon_0 A}$.
Comparing the two expressions,we get $\frac{1}{k} = \frac{1}{k_1 + k_2} + \frac{1}{2k_3}$.

(D) The capacitor can be modeled as a combination of three capacitors.
Capacitors with dielectrics $K_1$ and $K_2$ are in parallel with each other,and this combination is in series with the capacitor containing dielectric $K_3$.
For the top part,the area is $A/2$ and the distance is $d/2$:
$C_1 = \frac{K_1 \varepsilon_0 (A/2)}{d/2} = \frac{K_1 \varepsilon_0 A}{d}$
$C_2 = \frac{K_2 \varepsilon_0 (A/2)}{d/2} = \frac{K_2 \varepsilon_0 A}{d}$
Since $C_1$ and $C_2$ are in parallel,their equivalent capacitance is $C_p = C_1 + C_2 = \frac{\varepsilon_0 A}{d} (K_1 + K_2)$.
For the bottom part,the area is $A$ and the distance is $d/2$:
$C_3 = \frac{K_3 \varepsilon_0 A}{d/2} = \frac{2 K_3 \varepsilon_0 A}{d}$.
Now,$C_p$ and $C_3$ are in series:
$C_{eq} = \frac{C_p C_3}{C_p + C_3} = \frac{\left[ \frac{\varepsilon_0 A}{d} (K_1 + K_2) \right] \left[ \frac{2 K_3 \varepsilon_0 A}{d} \right]}{\frac{\varepsilon_0 A}{d} (K_1 + K_2 + 2K_3)} = \frac{2 K_3 (K_1 + K_2)}{K_1 + K_2 + 2K_3} \frac{\varepsilon_0 A}{d}$.
This expression does not match any of the given options,so the correct answer is $(d)$.

(A) The initial capacitance of the air-filled capacitor is $C_0 = \frac{\varepsilon_0 A}{d} = 10\,\mu F$.
As shown in the figure,the capacitor can be considered as three capacitors in parallel: two air-filled capacitors ($C_1$ and $C_3$) each with area $A/4$ and distance $d$,and one dielectric-filled capacitor $(C_2)$ with area $A/2$,distance $d$,and dielectric constant $K=4$.
The capacitances are:
$C_1 = \frac{\varepsilon_0 (A/4)}{d} = \frac{1}{4} \left( \frac{\varepsilon_0 A}{d} \right) = \frac{1}{4} C_0 = 2.5\,\mu F$
$C_3 = \frac{\varepsilon_0 (A/4)}{d} = \frac{1}{4} C_0 = 2.5\,\mu F$
$C_2 = \frac{K \varepsilon_0 (A/2)}{d} = \frac{K}{2} \left( \frac{\varepsilon_0 A}{d} \right) = \frac{4}{2} C_0 = 2 C_0 = 20\,\mu F$
Since they are in parallel,the equivalent capacitance is:
$C_{eq} = C_1 + C_2 + C_3 = 2.5 + 20 + 2.5 = 25\,\mu F$.

(D) The initial capacitance of the parallel plate capacitor is $C = \frac{\varepsilon_0 A}{d}$.
When the space between the plates is filled with two dielectric layers of thickness $d/2$ each,they act as two capacitors in series.
The capacitance of the first layer is $C_A = \frac{K_1 \varepsilon_0 A}{d/2} = \frac{2 K_1 \varepsilon_0 A}{d} = 2 K_1 C$.
The capacitance of the second layer is $C_B = \frac{K_2 \varepsilon_0 A}{d/2} = \frac{2 K_2 \varepsilon_0 A}{d} = 2 K_2 C$.
Since they are in series,the equivalent capacitance $C_1$ is given by $\frac{1}{C_1} = \frac{1}{C_A} + \frac{1}{C_B} = \frac{1}{2 K_1 C} + \frac{1}{2 K_2 C} = \frac{K_1 + K_2}{2 K_1 K_2 C}$.
Therefore,$C_1 = \frac{2 K_1 K_2}{K_1 + K_2} C$.
The ratio $C_1/C$ is $\frac{2 K_1 K_2}{K_1 + K_2}$.

(B) The capacitance of a parallel plate capacitor with air is $C_0 = \frac{\epsilon_0 A}{d} = 100\,\mu F$.
When a dielectric slab of thickness $t$ and dielectric constant $K$ is introduced,the new capacitance $C'$ is given by the formula:
$C' = \frac{\epsilon_0 A}{d - t + \frac{t}{K}}$
Substituting $K = 5$:
$C' = \frac{\epsilon_0 A}{d - t + \frac{t}{5}} = \frac{\epsilon_0 A}{d - 0.8t}$
Since $t < d$,the denominator $(d - 0.8t)$ is less than $d$. Therefore,$C' > C_0$.
Also,if the slab were to fill the entire space $(t = d)$,the capacitance would be $K \times C_0 = 5 \times 100 = 500\,\mu F$.
Since $t < d$,the new capacitance $C'$ must be greater than $100\,\mu F$ but less than $500\,\mu F$.

(D) The negative plate is at $x = 0$ and the positive plate is at $x = 3d$. The electric field lines are always directed from the positive plate to the negative plate. Therefore,the direction of the electric field is from $x = 3d$ towards $x = 0$ (i.e.,in the negative $x$-direction) throughout the region. Thus,the direction remains the same.
The magnitude of the electric field in the air gap is $E_{air} = \frac{\sigma}{\varepsilon_0}$ and in the dielectric slab is $E_{dielectric} = \frac{\sigma}{K\varepsilon_0}$. Since $K > 1$,the magnitudes are different.
Electric potential $V$ is related to the electric field $E$ by $E = -\frac{dV}{dx}$. Since the electric field is directed towards the negative $x$-axis,$E_x < 0$. Thus,$-\frac{dV}{dx} < 0$,which implies $\frac{dV}{dx} > 0$. This means the electric potential $V$ increases continuously as we move from $x = 0$ (negative plate) to $x = 3d$ (positive plate).
Therefore,both statements $(b)$ and $(c)$ are correct.

(C) Initially,the switch $S$ is closed. Both capacitors $A$ and $B$ are connected in parallel to the battery of potential $V$. The capacitance of each capacitor is $C$.
The initial total energy stored in the system is:
$U_1 = \frac{1}{2}CV^2 + \frac{1}{2}CV^2 = CV^2$ --- $(i)$
When the switch $S$ is opened,capacitor $A$ remains connected to the battery,so its potential difference remains $V$. Capacitor $B$ is disconnected from the battery,so its charge $Q$ remains constant. The initial charge on capacitor $B$ was $Q = CV$.
After filling the space with a dielectric of constant $K = 3$,the new capacitance of each capacitor becomes $C' = KC = 3C$.
For capacitor $A$,the potential difference remains $V$. The new energy is $U_{A}' = \frac{1}{2}C'V^2 = \frac{1}{2}(3C)V^2 = \frac{3}{2}CV^2$.
For capacitor $B$,the charge $Q = CV$ remains constant. The new energy is $U_{B}' = \frac{Q^2}{2C'} = \frac{(CV)^2}{2(3C)} = \frac{1}{6}CV^2$.
The final total energy stored in the system is:
$U_2 = U_{A}' + U_{B}' = \frac{3}{2}CV^2 + \frac{1}{6}CV^2 = \left(\frac{9+1}{6}\right)CV^2 = \frac{10}{6}CV^2 = \frac{5}{3}CV^2$ --- (ii)
The ratio of total electrostatic energy before and after is:
$\frac{U_1}{U_2} = \frac{CV^2}{(5/3)CV^2} = \frac{3}{5}$

(B) When the capacitor is disconnected from the battery, the charge $Q$ on the plates remains constant.
The capacitance $C$ of a parallel plate capacitor with a dielectric is given by $C = \frac{K \epsilon_0 A}{d}$.
As the dielectric plate of length $L$ is drawn out by a distance $x$, the capacitor can be considered as two capacitors in parallel: one with dielectric (length $L-x$) and one with air (length $x$).
The equivalent capacitance $C(x)$ decreases as $x$ increases.
Since $V = \frac{Q}{C}$, and $Q$ is constant, the potential difference $V$ increases as $C$ decreases.
Specifically, $C(x) = \frac{\epsilon_0 w}{d} (L-x)K + \frac{\epsilon_0 w}{d} x = \frac{\epsilon_0 w}{d} [K(L-x) + x]$.
As $x$ increases from $0$ to $L$, $C(x)$ decreases linearly, which means $V(x) = \frac{Q}{C(x)}$ will increase non-linearly (specifically, it follows a hyperbolic curve $V \propto \frac{1}{a-bx}$).
Once the dielectric is completely removed $(x=L)$, the capacitance becomes constant $(C_{air} = \frac{\epsilon_0 A}{d})$, and thus the potential difference $V$ becomes constant.
Graph $B$ correctly shows $V$ increasing non-linearly and then becoming constant.

(A) The initial capacitance of the parallel plate capacitor is $C = \frac{\epsilon_0 A}{d}$.
In the second condition,the space between the plates is divided into two parts,each with area $A/2$ and distance $d$ between the plates.
These two parts act as two capacitors connected in parallel.
One capacitor has air as the dielectric,so its capacitance is $C_1 = \frac{\epsilon_0 (A/2)}{d} = \frac{C}{2}$.
The other capacitor has a dielectric material with constant $K$,so its capacitance is $C_2 = \frac{K \epsilon_0 (A/2)}{d} = \frac{KC}{2}$.
Since they are connected in parallel,the equivalent capacitance is $C_{eq} = C_1 + C_2$.
$C_{eq} = \frac{C}{2} + \frac{KC}{2} = \frac{C}{2}(1 + K)$.

(C) In air,the electrostatic force between two point charges $q_1$ and $q_2$ separated by distance $r$ is given by:
$F = \frac{1}{4\pi\epsilon_0} \frac{q_1 q_2}{r^2}$
When placed in a medium with dielectric constant $K$,the force $F'$ at a new distance $r'$ is given by:
$F' = \frac{1}{4\pi\epsilon_0 K} \frac{q_1 q_2}{(r')^2}$
We want the force to remain the same,so $F = F'$:
$\frac{1}{4\pi\epsilon_0} \frac{q_1 q_2}{r^2} = \frac{1}{4\pi\epsilon_0 K} \frac{q_1 q_2}{(r')^2}$
Canceling common terms:
$\frac{1}{r^2} = \frac{1}{K(r')^2}$
Rearranging for $r'$:
$(r')^2 = \frac{r^2}{K}$
Taking the square root of both sides:
$r' = \frac{r}{\sqrt{K}}$

(C) When a capacitor remains connected to a battery,the potential difference $V$ across its plates remains constant.
When a dielectric slab of constant $K$ is inserted,the capacitance increases from $C$ to $C' = KC$.
The charge on the capacitor increases from $Q = CV$ to $Q' = KCV$.
The energy stored in the capacitor changes from $U = \frac{1}{2}CV^2$ to $U' = \frac{1}{2}KCV^2$.
The change in energy is $\Delta U = U' - U = \frac{1}{2}(K-1)CV^2$,which is positive,meaning the energy stored increases.
The work done by the battery is $W_{battery} = \Delta Q \cdot V = (Q' - Q)V = (K-1)CV^2$.
Since $W_{battery} = 2\Delta U$,the battery supplies energy equal to the increase in stored energy plus an equal amount of work done by the external agent (or the force of attraction pulling the slab in).
Thus,energy from the battery is consumed to increase the stored energy and perform work.

(C) When a parallel plate capacitor is connected to a battery,the potential difference $V$ across the plates remains constant and equal to the electromotive force of the battery.
The capacitance of a parallel plate capacitor with a dielectric slab of constant $K$ is given by $C' = K C$,where $C$ is the initial capacitance.
The charge on the capacitor is given by $Q = CV$.
After introducing the dielectric slab,the new charge $Q'$ becomes:
$Q' = C' V = (K C) V = K Q$.
Since the dielectric constant $K > 1$ for all dielectric materials,it follows that $Q' > Q$.
Therefore,the new charge on the plates is greater than the initial charge.

(C) $1$. When the battery is disconnected,the charge $Q$ on the capacitor plates remains constant because there is no path for the charge to flow.
$2$. When a dielectric slab of constant $K > 1$ is inserted,the capacitance $C$ increases according to the formula $C' = KC$.
$3$. Since $Q = CV$,the new voltage $V' = Q / C'$. Since $C' > C$,the voltage $V'$ decreases $(V' = V/K)$.
$4$. The electrostatic potential energy $U$ is given by $U = Q^2 / (2C)$. Since $C$ increases and $Q$ is constant,the potential energy $U$ decreases $(U' = U/K)$.

(A) For a parallel plate capacitor,the electric field between the plates is given by $E = \frac{\sigma}{\epsilon_0}$,where $\sigma$ is the surface charge density.
When a dielectric medium with dielectric constant $K$ is introduced between the plates,the induced charges on the dielectric surface create an opposing electric field.
The net electric field inside the dielectric becomes $E' = \frac{E}{K}$.
Since $K > 1$ for any dielectric material,the new electric field $E'$ is less than the original electric field $E$.
Therefore,the intensity of the electric field decreases.

(A) The capacitance $C$ of a parallel plate capacitor with a dielectric slab of thickness $t$ and dielectric constant $K$ is given by the formula:
$C = \frac{\varepsilon_0 A}{d - t + t/K}$
This can be rearranged as:
$C = \frac{\varepsilon_0 A}{d - t(1 - 1/K)}$
Here,the effective distance between the plates is reduced due to the presence of the dielectric slab. The potential difference $V$ across the plates is the sum of the potential drop across the air gap $(d-t)$ and the dielectric slab $t$:
$V = E_{\text{air}}(d-t) + E_{\text{medium}}(t)$
Since $E_{\text{air}} = \frac{\sigma}{\varepsilon_0}$ and $E_{\text{medium}} = \frac{\sigma}{K\varepsilon_0}$,where $\sigma = \frac{Q}{A}$:
$V = \frac{Q}{A\varepsilon_0}(d-t) + \frac{Q}{A\varepsilon_0 K}(t) = \frac{Q}{A\varepsilon_0} [d - t + t/K] = \frac{Q}{A\varepsilon_0} [d - t(1 - 1/K)]$
Using $C = \frac{Q}{V}$,we get $C = \frac{\varepsilon_0 A}{d - t(1 - 1/K)}$.

(A) parallel plate capacitor with dielectric slabs of thicknesses $t_1$ and $t_2$ can be treated as two capacitors in series.
The capacitance of a parallel plate capacitor with a dielectric slab of thickness $t$ and dielectric constant $K$ is given by $C = \frac{\varepsilon_0 A}{t/K}$.
For two slabs in series,the equivalent capacitance $C_{eq}$ is given by $\frac{1}{C_{eq}} = \frac{1}{C_1} + \frac{1}{C_2}$.
Substituting the values,we get $\frac{1}{C_{eq}} = \frac{t_1}{\varepsilon_0 A K_1} + \frac{t_2}{\varepsilon_0 A K_2}$.
$\frac{1}{C_{eq}} = \frac{1}{\varepsilon_0 A} (\frac{t_1}{K_1} + \frac{t_2}{K_2})$.
Therefore,$C_{eq} = \frac{\varepsilon_0 A}{\frac{t_1}{K_1} + \frac{t_2}{K_2}}$.

(B) The initial capacitance is given by $C_0 = \frac{\epsilon_0 A}{d} = 20 \ \mu F$,where $d = 2 \ mm$.
When a dielectric slab of thickness $t = 1 \ mm$ and dielectric constant $K = 2$ is inserted,the new capacitance $C'$ is given by the formula:
$C' = \frac{\epsilon_0 A}{d - t + \frac{t}{K}}$
Substituting the values:
$C' = \frac{\epsilon_0 A}{2 \times 10^{-3} - 1 \times 10^{-3} + \frac{1 \times 10^{-3}}{2}}$
$C' = \frac{\epsilon_0 A}{1 \times 10^{-3} + 0.5 \times 10^{-3}} = \frac{\epsilon_0 A}{1.5 \times 10^{-3}}$
Since $\frac{\epsilon_0 A}{2 \times 10^{-3}} = 20 \ \mu F$,we have $\epsilon_0 A = 20 \times 10^{-6} \times 2 \times 10^{-3} = 40 \times 10^{-9} \ F \cdot m$.
$C' = \frac{40 \times 10^{-9}}{1.5 \times 10^{-3}} = \frac{40}{1.5} \times 10^{-6} \ F = 26.66 \ \mu F \approx 26.6 \ \mu F$.

(B) The capacitors $C_1$ and $C_2$ are connected in parallel to the battery. Therefore,the potential difference across both capacitors is the same,i.e.,$V_1 = V_2 = V$.
The charge on a capacitor is given by the formula $Q = CV$.
For capacitor $C_1$,the charge is $Q_1 = C_1 V$.
For capacitor $C_2$,the charge is $Q_2 = C_2 V$.
Since the space between the plates of $C_2$ is filled with a dielectric material of dielectric constant $K > 1$,its capacitance increases such that $C_2 = K C_1$,which implies $C_2 > C_1$.
Since $V$ is constant and $C_2 > C_1$,it follows that $C_2 V > C_1 V$,which means $Q_2 > Q_1$ or $Q_1 < Q_2$.

(B) The initial capacitance of a parallel plate capacitor is given by $C = \frac{\epsilon_0 A}{d}$.
When the distance between the plates is halved $(d' = d/2)$ and the space is filled with a dielectric medium of constant $K$,the new capacitance $C'$ is given by $C' = \frac{K \epsilon_0 A}{d'}$.
Substituting $d' = d/2$ into the equation,we get $C' = \frac{K \epsilon_0 A}{d/2} = \frac{2K \epsilon_0 A}{d}$.
Given that the new capacitance $C' = 3C$,we equate the expressions:
$3 \left( \frac{\epsilon_0 A}{d} \right) = \frac{2K \epsilon_0 A}{d}$.
Canceling $\frac{\epsilon_0 A}{d}$ from both sides,we get $3 = 2K$.
Therefore,$K = \frac{3}{2} = 1.5$.

(A) The capacitance of a capacitor in a medium is given by $C_{medium} = K \cdot C_{air}$,where $K$ is the dielectric constant of the medium.
Given:
$C_{air} = 50 \,\mu F$
$C_{medium} = 110 \,\mu F$
Therefore,the dielectric constant $K$ is calculated as:
$K = \frac{C_{medium}}{C_{air}}$
$K = \frac{110}{50} = 2.2$
Thus,the dielectric constant of the oil is $2.2$.

(A) The capacitance of a capacitor filled with a dielectric is given by $C = K C_0$,where $C_0$ is the capacitance of the capacitor when filled with air (or vacuum).
Given that $K = 2$ and the capacitance with the dielectric is $C$,we have $C = 2 C_0$.
Therefore,the capacitance without the dielectric (when the oil is removed) is $C_0 = \frac{C}{2}$.

(C) The equivalent capacitance $C$ of a capacitor with multiple dielectric slabs in series is given by the formula: $C = \frac{\varepsilon_0 A}{\sum \frac{d_i}{k_i}}$.
Given: $d_1 = 6 \ mm = 6 \times 10^{-3} \ m$,$k_1 = 10$,$d_2 = 4 \ mm = 4 \times 10^{-3} \ m$,$k_2 = 5$.
Substituting the values into the formula:
$C = \frac{\varepsilon_0 A}{\frac{6 \times 10^{-3}}{10} + \frac{4 \times 10^{-3}}{5}}$
$C = \frac{\varepsilon_0 A}{0.6 \times 10^{-3} + 0.8 \times 10^{-3}}$
$C = \frac{\varepsilon_0 A}{1.4 \times 10^{-3}}$
$C = \frac{1000}{1.4} \varepsilon_0 A = \frac{10000}{14} \varepsilon_0 A = \frac{5000}{7} \varepsilon_0 A$.

(B) The capacitance of a parallel plate capacitor with a dielectric medium is given by the formula: $C = \frac{\varepsilon_0 \varepsilon_r A}{d}$.
Here,$C = 2 \times 10^{-4} \, \mu F = 2 \times 10^{-4} \times 10^{-6} \, F = 2 \times 10^{-10} \, F$.
The area $A = 0.01 \, m^2$,distance $d = 2 \, mm = 2 \times 10^{-3} \, m$,and permittivity of free space $\varepsilon_0 = 8.85 \times 10^{-12} \, F/m$.
Rearranging the formula to solve for the dielectric constant $\varepsilon_r$:
$\varepsilon_r = \frac{C \cdot d}{\varepsilon_0 \cdot A}$
Substituting the values:
$\varepsilon_r = \frac{(2 \times 10^{-10}) \times (2 \times 10^{-3})}{(8.85 \times 10^{-12}) \times (0.01)}$
$\varepsilon_r = \frac{4 \times 10^{-13}}{8.85 \times 10^{-14}}$
$\varepsilon_r = \frac{40}{8.85} \approx 4.52$.

(A) The capacitor is divided into two parts in series,each with a plate separation of $d/2$ and area $A$.
For the first part,the dielectric constant is $k_1 = 1$,so the capacitance is $C_1 = \frac{k_1 \varepsilon_0 A}{d/2} = \frac{1 \cdot \varepsilon_0 A}{d/2} = \frac{2 \varepsilon_0 A}{d}$.
For the second part,the dielectric constant is $k_2 = 2$,so the capacitance is $C_2 = \frac{k_2 \varepsilon_0 A}{d/2} = \frac{2 \cdot \varepsilon_0 A}{d/2} = \frac{4 \varepsilon_0 A}{d}$.
Since they are in series,the equivalent capacitance $C_{eq}$ is given by $\frac{1}{C_{eq}} = \frac{1}{C_1} + \frac{1}{C_2}$.
Substituting the values: $\frac{1}{C_{eq}} = \frac{d}{2 \varepsilon_0 A} + \frac{d}{4 \varepsilon_0 A} = \frac{2d + d}{4 \varepsilon_0 A} = \frac{3d}{4 \varepsilon_0 A}$.
Therefore,$C_{eq} = \frac{4 \varepsilon_0 A}{3d}$.

(C) The equivalent capacitance $C$ for a parallel plate capacitor with multiple dielectric slabs of thickness $t_i$ and dielectric constant $k_i$ is given by the formula:
$C = \frac{\varepsilon_0 A}{\sum \frac{t_i}{k_i}}$
Given:
$t_1 = 6 \, mm = 6 \times 10^{-3} \, m$
$t_2 = 4 \, mm = 4 \times 10^{-3} \, m$
$k_1 = 10$
$k_2 = 5$
Substituting the values:
$C = \frac{\varepsilon_0 A}{\frac{6 \times 10^{-3}}{10} + \frac{4 \times 10^{-3}}{5}}$
$C = \frac{\varepsilon_0 A}{0.6 \times 10^{-3} + 0.8 \times 10^{-3}}$
$C = \frac{\varepsilon_0 A}{1.4 \times 10^{-3}}$
$C = \frac{1000}{1.4} \varepsilon_0 A = \frac{10000}{14} \varepsilon_0 A = \frac{5000}{7} \varepsilon_0 A$

(D) The capacitance of a parallel plate capacitor is given by $C = \frac{k \epsilon_0 A}{d}$,where $A = l \times b$.
Given: $C = 2 \ \mu F = 2 \times 10^{-6} \ F$,$k = 2.5$,$d = 0.15 \ mm = 0.15 \times 10^{-3} \ m$,$b = 400 \ mm = 0.4 \ m$,and $\epsilon_0 = 8.85 \times 10^{-12} \ F/m$.
Substituting the values into the formula:
$2 \times 10^{-6} = \frac{2.5 \times 8.85 \times 10^{-12} \times (l \times 0.4)}{0.15 \times 10^{-3}}$
$2 \times 10^{-6} = \frac{2.5 \times 8.85 \times 10^{-12} \times 0.4 \times l}{0.15 \times 10^{-3}}$
$2 \times 10^{-6} = \frac{8.85 \times 10^{-12} \times l}{0.15 \times 10^{-3}}$
$l = \frac{2 \times 10^{-6} \times 0.15 \times 10^{-3}}{8.85 \times 10^{-12}}$
$l = \frac{0.3 \times 10^{-9}}{8.85 \times 10^{-12}} = \frac{300}{8.85} \approx 33.9 \ m$.

(B) The capacitance of a parallel plate capacitor with a dielectric slab of thickness $t$ and dielectric constant $K$ is given by $C = \frac{\epsilon_0 A}{d - t + t/K}$.
For a metal plate,the dielectric constant $K = \infty$.
Given $t = d/2$,the new capacitance $C'$ is:
$C' = \frac{\epsilon_0 A}{d - d/2 + (d/2)/\infty} = \frac{\epsilon_0 A}{d/2 + 0} = \frac{2\epsilon_0 A}{d}$.
Since the original capacitance $C = \frac{\epsilon_0 A}{d}$,we have $C' = 2C$.
Therefore,the capacitance becomes double.

(A) The capacitance of a parallel plate capacitor is given by $C = \frac{K \epsilon_0 A}{d}$.
Initially,$C_1 = 2\ \mu F$,$d_1 = 0.4\ cm$,and $K_1 = 1$ (assuming air).
Finally,$d_2 = \frac{d_1}{2} = 0.2\ cm$ and $K_2 = 2.8$.
The ratio of capacitances is $\frac{C_2}{C_1} = \frac{K_2}{K_1} \times \frac{d_1}{d_2}$.
Substituting the values: $\frac{C_2}{2} = \frac{2.8}{1} \times \frac{0.4}{0.2}$.
$\frac{C_2}{2} = 2.8 \times 2 = 5.6$.
$C_2 = 5.6 \times 2 = 11.2\ \mu F$.

(C) Initially,both capacitors $A$ and $B$ are connected in parallel to a battery of potential $V$. The capacitance of each is $C$.
Total initial energy $U_i = \frac{1}{2}CV^2 + \frac{1}{2}CV^2 = CV^2$.
When the switch $S$ is opened,capacitor $A$ remains connected to the battery,so its potential remains $V$ and its capacitance becomes $K C$. Capacitor $B$ is disconnected,so its charge $Q = CV$ remains constant,and its capacitance becomes $K C$.
Final energy in capacitor $A$: $U_{Af} = \frac{1}{2}(KC)V^2 = \frac{3}{2}CV^2$.
Final energy in capacitor $B$: $U_{Bf} = \frac{Q^2}{2(KC)} = \frac{(CV)^2}{2(3C)} = \frac{CV^2}{6}$.
Total final energy $U_f = U_{Af} + U_{Bf} = \frac{3}{2}CV^2 + \frac{1}{6}CV^2 = \frac{9+1}{6}CV^2 = \frac{10}{6}CV^2 = \frac{5}{3}CV^2$.
The ratio of initial to final energy is $\frac{U_i}{U_f} = \frac{CV^2}{(5/3)CV^2} = \frac{3}{5}$.

(A) Since the capacitors are connected in parallel to the same battery,the potential difference $V$ across both capacitors is the same.
The charge on a capacitor is given by $q = CV$.
For capacitor $C_1$ (air-filled),the capacitance is $C_1 = C_0$.
For capacitor $C_2$ (dielectric-filled),the capacitance is $C_2 = KC_0$,where $K > 1$.
Thus,$q_1 = C_1 V = C_0 V$ and $q_2 = C_2 V = KC_0 V$.
Since $K > 1$,it follows that $KC_0 V > C_0 V$,which implies $q_2 > q_1$ or $q_1 < q_2$.

(A) The area of the circular plate is $A = \pi r^2 = \pi \times (8 \times 10^{-2} \ m)^2 = 64\pi \times 10^{-4} \ m^2 \approx 0.0201 \ m^2$.
The distance between the plates is $d = 1 \ mm = 1 \times 10^{-3} \ m$.
The dielectric constant is $K = 6$.
The capacitance $C$ of a parallel plate capacitor with a dielectric is given by $C = \frac{K \varepsilon_0 A}{d}$.
Substituting the values: $C = \frac{6 \times 8.854 \times 10^{-12} \times 0.0201}{1 \times 10^{-3}}$.
$C = \frac{1.068 \times 10^{-12}}{10^{-3}} = 1.068 \times 10^{-9} \ F$.

(A) The potential difference $V$ between the plates is the sum of the potential difference across the air gap and the dielectric slab.
$V = V_{air} + V_{medium} = \frac{\sigma}{\varepsilon_0}(d - t) + \frac{\sigma}{K\varepsilon_0}t$
Since $\sigma = \frac{Q}{A}$,we have:
$V = \frac{Q}{A\varepsilon_0} \left( (d - t) + \frac{t}{K} \right)$
$V = \frac{Q}{A\varepsilon_0} \left( d - t + \frac{t}{K} \right) = \frac{Q}{A\varepsilon_0} \left( d - t(1 - \frac{1}{K}) \right)$
The capacitance $C$ is given by $C = \frac{Q}{V}$:
$C = \frac{Q}{\frac{Q}{A\varepsilon_0} (d - t(1 - \frac{1}{K}))} = \frac{\varepsilon_0 A}{d - t(1 - \frac{1}{K})}$

(A) For an air-filled capacitor,the initial capacitance is $C_0 = \frac{\epsilon_0 A}{d} = 9 \ pF$.
When two dielectric slabs are inserted in series,the equivalent capacitance $C'$ is given by $\frac{1}{C'} = \frac{1}{C_1} + \frac{1}{C_2}$.
The capacitance of the first slab is $C_1 = \frac{K_1 \epsilon_0 A}{d/3} = \frac{3 K_1 \epsilon_0 A}{d} = 3 K_1 C_0 = 3 \times 3 \times 9 \ pF = 81 \ pF$.
The capacitance of the second slab is $C_2 = \frac{K_2 \epsilon_0 A}{2d/3} = \frac{3 K_2 \epsilon_0 A}{2d} = \frac{3}{2} K_2 C_0 = \frac{3}{2} \times 6 \times 9 \ pF = 81 \ pF$.
Since $C_1$ and $C_2$ are in series,the equivalent capacitance is $C' = \frac{C_1 C_2}{C_1 + C_2} = \frac{81 \times 81}{81 + 81} = \frac{81}{2} = 40.5 \ pF$.

(C) The initial capacitance is $C = \frac{\epsilon_0 A}{d} = 9 \ pF$.
The space is divided into two layers of thickness $d_1 = d/3$ and $d_2 = 2d/3$,which act as two capacitors in series.
The capacitance of the first layer is $C_1 = \frac{K_1 \epsilon_0 A}{d_1} = \frac{3 \epsilon_0 A}{d/3} = 9 \left( \frac{\epsilon_0 A}{d} \right) = 9 \times 9 = 81 \ pF$.
The capacitance of the second layer is $C_2 = \frac{K_2 \epsilon_0 A}{d_2} = \frac{6 \epsilon_0 A}{2d/3} = \frac{18}{2} \left( \frac{\epsilon_0 A}{d} \right) = 9 \times 9 = 81 \ pF$.
Since $C_1$ and $C_2$ are in series,the equivalent capacitance $C_{eq}$ is given by:
$C_{eq} = \frac{C_1 C_2}{C_1 + C_2} = \frac{81 \times 81}{81 + 81} = \frac{81 \times 81}{162} = 40.5 \ pF$.

(C) The capacitance of a parallel plate capacitor with air is $C_1 = \frac{\epsilon_0 A}{d}$.
When a dielectric slab of thickness $t$ is inserted,the new capacitance is $C_2 = \frac{\epsilon_0 A}{d - t(1 - 1/K)}$. Since $K > 1$,the denominator decreases,so $C_2 > C_1$.
When a conducting slab of thickness $t$ is inserted,the effective distance between plates becomes $(d - t)$. The new capacitance is $C_3 = \frac{\epsilon_0 A}{d - t}$. Since $(d - t) < d$,$C_3 > C_1$.
Comparing $C_2$ and $C_3$: For a conducting slab,$K = \infty$. Substituting $K = \infty$ into the dielectric formula gives $C_3 = \frac{\epsilon_0 A}{d - t}$. Since the conducting slab effectively reduces the gap more significantly than a dielectric of the same thickness,$C_3 > C_2$.
Therefore,the order is $C_3 > C_2 > C_1$.

(A) The initial capacitance of the air-filled capacitor is $C = \frac{\varepsilon_0 A}{d} = 10 \ \mu F$.
As shown in the figure,the capacitor can be considered as a parallel combination of three capacitors:
$C_1$ with area $A/4$ and air dielectric $(K=1)$,
$C_2$ with area $A/2$ and dielectric $(K=4)$,
$C_3$ with area $A/4$ and air dielectric $(K=1)$.
The capacitances are:
$C_1 = \frac{\varepsilon_0 (A/4)}{d} = \frac{1}{4} \left( \frac{\varepsilon_0 A}{d} \right) = \frac{10}{4} = 2.5 \ \mu F$
$C_2 = \frac{K \varepsilon_0 (A/2)}{d} = \frac{4 \varepsilon_0 A}{2d} = 2 \left( \frac{\varepsilon_0 A}{d} \right) = 2 \times 10 = 20 \ \mu F$
$C_3 = \frac{\varepsilon_0 (A/4)}{d} = \frac{1}{4} \left( \frac{\varepsilon_0 A}{d} \right) = \frac{10}{4} = 2.5 \ \mu F$
Since they are in parallel,the equivalent capacitance is:
$C_{eq} = C_1 + C_2 + C_3 = 2.5 + 20 + 2.5 = 25 \ \mu F$.

(D) For a parallel plate capacitor,the initial potential difference $V$ is given by $V = E \times d$.
Given $E = 3 \times 10^4\, V/m$ and $d = 0.05\, m$,we have $V = 3 \times 10^4 \times 0.05 = 1500\, V = 1.5\, kV$.
Since the capacitor is disconnected from the battery,the charge $q$ remains constant.
The potential difference $V'$ after inserting a slab of thickness $t$ and dielectric constant $K$ is given by $V' = V \times \frac{(d - t) + (t/K)}{d}$.
For a dielectric slab with $K = 2$ and $t = 0.01\, m$:
$V' = 1.5 \times \frac{(0.05 - 0.01) + (0.01/2)}{0.05}$
$V' = 1.5 \times \frac{0.04 + 0.005}{0.05}$
$V' = 1.5 \times \frac{0.045}{0.05}$
$V' = 1.5 \times 0.9 = 1.35\, kV$.

(A) The initial capacitance of the parallel plate capacitor is $C = \frac{\epsilon_0 A}{d}$.
When the space is filled with a dielectric of constant $K$ parallel to the plates,it acts as two capacitors in series,each with a plate separation of $d/2$.
The capacitance of the part with the dielectric is $C_1 = \frac{K \epsilon_0 A}{d/2} = 2KC$.
The capacitance of the part with air is $C_2 = \frac{\epsilon_0 A}{d/2} = 2C$.
Since these two capacitors are in series,the equivalent capacitance $C_{total}$ is given by:
$C_{total} = \frac{C_1 C_2}{C_1 + C_2} = \frac{(2KC)(2C)}{2KC + 2C} = \frac{4KC^2}{2C(K + 1)} = \frac{2KC}{K + 1}$.

(A) The potential difference between the plates in air is given by $V = E \cdot d = \frac{\sigma}{\varepsilon_0} d$.
When a dielectric slab of thickness $t$ and dielectric constant $K$ is inserted,the new potential difference $V'$ is given by $V' = \frac{\sigma}{\varepsilon_0} (d - t + \frac{t}{K})$.
Given that the distance between the plates is increased by $d' = 1.6 \ mm$ to maintain the same potential difference,the new potential difference becomes $V_{new} = \frac{\sigma}{\varepsilon_0} ((d + d') - t + \frac{t}{K})$.
Since the potential difference remains constant,$V = V_{new}$,which implies $d = d + d' - t + \frac{t}{K}$.
Rearranging the terms,we get $d' = t - \frac{t}{K} = t(1 - \frac{1}{K})$.
Substituting the given values $t = 2 \ mm$ and $d' = 1.6 \ mm$:
$1.6 = 2(1 - \frac{1}{K}) \implies 0.8 = 1 - \frac{1}{K} \implies \frac{1}{K} = 1 - 0.8 = 0.2$.
Therefore,$K = \frac{1}{0.2} = 5$.

(D) The electric field $E$ inside a dielectric-filled capacitor is given by $E = \frac{Q}{K A \varepsilon_0}$,where $Q$ is the charge,$K$ is the dielectric constant,$A$ is the area,and $\varepsilon_0$ is the permittivity of free space.
To avoid dielectric breakdown,the electric field must not exceed the dielectric strength $E_{max} = 1.9 \times 10^7 \ V/m$.
Thus,$Q_{max} = E_{max} \cdot K \cdot A \cdot \varepsilon_0$.
Given: $A = 100 \ cm^2 = 10^{-2} \ m^2$,$K = 5.0$,$\varepsilon_0 = 8.85 \times 10^{-12} \ F/m$,and $E_{max} = 1.9 \times 10^7 \ V/m$.
Substituting the values:
$Q_{max} = (1.9 \times 10^7) \times 5.0 \times 10^{-2} \times 8.85 \times 10^{-12}$
$Q_{max} = 8.4075 \times 10^{-6} \ C \approx 8.4 \times 10^{-6} \ C$.

(B) The initial capacitance of the air capacitor is given by $C = \frac{\epsilon_0 A}{d} = 10 \ \mu F$.
When the capacitor is divided into two equal parts as shown in the figure,the area of each part becomes $A/2$ while the distance $d$ remains the same. These two parts act as two capacitors connected in parallel.
The capacitance of the first part is $C_1 = \frac{K_1 \epsilon_0 (A/2)}{d} = K_1 \times (C/2) = 2 \times (10/2) = 10 \ \mu F$.
The capacitance of the second part is $C_2 = \frac{K_2 \epsilon_0 (A/2)}{d} = K_2 \times (C/2) = 4 \times (10/2) = 20 \ \mu F$.
Since the two capacitors are connected in parallel,the equivalent capacitance is $C_{new} = C_1 + C_2$.
Therefore,$C_{new} = 10 \ \mu F + 20 \ \mu F = 30 \ \mu F$.

(B) Let the initial capacitances be $C_1 = C$ and $C_2 = C$. When a dielectric of constant $K = 4$ is inserted into the second capacitor,its new capacitance becomes $C_2' = K \cdot C = 4C$.
In a series combination,the charge $Q$ on each capacitor is the same. Since $Q = CV$,we have $V = Q/C$,which implies $V \propto 1/C$.
The ratio of capacitances is $C_1 : C_2' = C : 4C = 1 : 4$.
Therefore,the ratio of potential differences is $V_1 : V_2 = 1/C_1 : 1/C_2' = 1/C : 1/4C = 4 : 1$.
The total potential difference is $V = 100 \ V$.
Potential difference across the first capacitor $(V_1)$: $V_1 = \frac{4}{4+1} \times 100 \ V = \frac{4}{5} \times 100 \ V = 80 \ V$.
Potential difference across the second capacitor $(V_2)$: $V_2 = \frac{1}{4+1} \times 100 \ V = \frac{1}{5} \times 100 \ V = 20 \ V$.
Thus,the potential differences are $80 \ V$ and $20 \ V$ respectively.

(C) The capacitance of a parallel plate capacitor is given by $C = Q/V$. Since the charge $Q$ remains constant when the dielectric is inserted,we have $C_0 V_0 = C V$,where $C_0$ and $V_0$ are the initial capacitance and potential,and $C$ and $V$ are the values after inserting the dielectric.
Given that $V = V_0 / 8$,we can write $C_0 V_0 = C (V_0 / 8)$.
This simplifies to $C = 8 C_0$.
The relationship between the new capacitance $C$ and the initial capacitance $C_0$ with a dielectric constant $K$ is $C = K C_0$.
Comparing the two expressions,we get $K = 8$.

(B) The capacitance of a parallel plate capacitor with air between the plates is given by $C = \frac{\epsilon_0 A}{d}$.
When a conducting slab (copper) of thickness $t$ is inserted between the plates,the new capacitance $C'$ is given by $C' = \frac{\epsilon_0 A}{d - t}$.
Given that the thickness of the copper slab is $t = d/2$,we substitute this into the formula:
$C' = \frac{\epsilon_0 A}{d - d/2} = \frac{\epsilon_0 A}{d/2} = 2 \left( \frac{\epsilon_0 A}{d} \right)$.
Since $C = \frac{\epsilon_0 A}{d}$,we have $C' = 2C$.
Therefore,the ratio $C'/C = 2$.

(A) The system consists of two capacitors in series,each with a plate separation of $d_1$ and $d_2$ such that $d_1 + d_2 = a - b$.
The capacitance of a parallel plate capacitor is given by $C = \frac{{\varepsilon _0}A}{d}$.
For two capacitors in series,the equivalent capacitance $C_{eq}$ is given by $\frac{1}{C_{eq}} = \frac{1}{C_1} + \frac{1}{C_2}$.
Substituting the values,we get $\frac{1}{C_{eq}} = \frac{d_1}{{\varepsilon _0}A} + \frac{d_2}{{\varepsilon _0}A} = \frac{d_1 + d_2}{{\varepsilon _0}A}$.
Since $d_1 + d_2 = a - b$,we have $\frac{1}{C_{eq}} = \frac{a - b}{{\varepsilon _0}A}$.
Therefore,$C_{eq} = \frac{{\varepsilon _0}A}{a - b}$.