Effect of Dielectric Inside Capacitor Questions in English

(C) The dielectric constant $K$ is defined as the ratio of the electric field in a vacuum $(E_0)$ to the electric field in the dielectric medium $(E)$.
Given:
Electric field in vacuum,$E_0 = 2 \times 10^5 \ V/m$
Electric field in dielectric,$E = 1 \times 10^5 \ V/m$
Formula:
$K = \frac{E_0}{E}$
Calculation:
$K = \frac{2 \times 10^5}{1 \times 10^5} = 2$
Therefore,the dielectric constant of the material is $2$.

(C) The capacitance of a parallel plate capacitor is given by $C = \frac{Q}{V}$.
When a dielectric slab is inserted between the plates,the charge $Q$ remains constant if the capacitor is isolated.
The new potential $V'$ is related to the original potential $V$ by the dielectric constant $K$ as $V' = \frac{V}{K}$.
Given that the new potential $V' = \frac{V}{8}$,we have $\frac{V}{K} = \frac{V}{8}$.
Comparing both sides,we get $K = 8$.

(C) When the battery is disconnected,the charge $Q$ on the plates remains constant because there is no path for the charge to flow.
When a dielectric slab of dielectric constant $K > 1$ is inserted,the capacitance $C$ increases according to the relation $C' = KC$.
Since $Q = CV$,the potential difference $V$ across the plates is given by $V = Q/C$. As $C$ increases and $Q$ remains constant,the potential difference $V$ decreases.
The energy stored in the capacitor is given by $U = Q^2 / (2C)$. Since $C$ increases and $Q$ is constant,the stored energy $U$ decreases.
Therefore,the potential difference decreases,the stored energy decreases,and the charge remains unchanged.

(D) When a dielectric slab is inserted into a parallel plate capacitor while the battery remains connected,the potential difference $V$ across the plates remains constant,i.e.,$V = {V_0}$.
Since the capacitance $C$ increases by a factor of $K$ (dielectric constant),where $C = KC_0$,the charge $Q$ on the plates increases according to $Q = CV = K C_0 V_0 = K Q_0$. Thus,$Q > {Q_0}$.
The electric field $E$ between the plates is given by $E = V/d$. Since $V$ and $d$ remain constant,$E = {E_0}$.
The energy stored in the capacitor is $U = \frac{1}{2}CV^2$. Since $C$ increases and $V$ remains constant,$U$ increases,i.e.,$U > {U_0}$.
Therefore,both statements $(a)$ and $(b)$ are correct.

(D) The capacitance of a parallel plate capacitor in a medium is given by the formula $C_{medium} = K \cdot C_{air}$,where $K$ is the dielectric constant of the medium.
Given:
$C_{air} = 50\,\mu F$
$C_{medium} = 110\,\mu F$
Rearranging the formula to solve for $K$:
$K = \frac{C_{medium}}{C_{air}}$
Substituting the values:
$K = \frac{110}{50} = 2.2$
Therefore,the dielectric constant of the oil is $2.2$.

(C) The potential difference $V$ between the plates is the sum of the potential drop across the air gap and the dielectric slab.
$V = V_{\text{air}} + V_{\text{medium}}$
$V = E_0(d - t) + E_{\text{medium}}t$
Since $E_0 = \frac{\sigma}{\varepsilon_0}$ and $E_{\text{medium}} = \frac{\sigma}{k\varepsilon_0}$,where $\sigma = \frac{Q}{A}$:
$V = \frac{\sigma}{\varepsilon_0}(d - t) + \frac{\sigma}{k\varepsilon_0}t$
$V = \frac{Q}{A\varepsilon_0} \left( d - t + \frac{t}{k} \right)$
$V = \frac{Q}{A\varepsilon_0} \left( d - t(1 - \frac{1}{k}) \right)$
The capacitance $C$ is given by $C = \frac{Q}{V}$:
$C = \frac{Q}{\frac{Q}{A\varepsilon_0} \left( d - t(1 - \frac{1}{k}) \right)}$
$C = \frac{\varepsilon_0 A}{d - t(1 - \frac{1}{k})}$

(B) The electric field between the plates of a capacitor in a vacuum or air is given by $E = \frac{\sigma}{\epsilon_0}$.
When the capacitor is immersed in a dielectric medium with dielectric constant $K$,the electric field is reduced by a factor of $K$.
The new electric field $E_{medium}$ is given by $E_{medium} = \frac{E}{K}$.
Given $K = 2$,the new electric field is $E_{medium} = \frac{E}{2}$.
Therefore,the electric field is decreased proportional to $\frac{1}{2}$.

(B) The capacitance of a parallel plate capacitor with air as the medium is given by $C_{air} = \frac{\epsilon_0 A}{d}$.
When a dielectric medium of dielectric constant $K$ is introduced between the plates,the new capacitance becomes $C_{medium} = \frac{K \epsilon_0 A}{d}$.
Comparing the two expressions,we get $C_{medium} = K \times C_{air}$.
Therefore,the capacity increases $K$ times.

(C) The capacitance of a parallel plate capacitor with air is $C_0 = \frac{\epsilon_0 A}{d}$.
When a dielectric slab of thickness $t$ and dielectric constant $K$ is introduced,the new capacitance $C$ is given by $C = \frac{\epsilon_0 A}{d - t + \frac{t}{K}}$.
The ratio of the new capacitance to the original capacitance is $\frac{C}{C_0} = \frac{d}{d - t + \frac{t}{K}}$.
Given $d = 6\,mm$,$t = 4.5\,mm$,and $K = 9$.
Substituting the values: $\frac{C}{C_0} = \frac{6}{6 - 4.5 + \frac{4.5}{9}} = \frac{6}{1.5 + 0.5} = \frac{6}{2} = 3$.
Therefore,the capacity becomes $3$ times the original capacity.

(A) The potential difference $V$ across a capacitor with multiple dielectric slabs is the sum of the potential drops across each slab.
$V = V_1 + V_2 = E_1 t_1 + E_2 t_2$
Since the electric field in a dielectric is $E = \frac{\sigma}{k\varepsilon_0} = \frac{Q}{Ak\varepsilon_0}$,we substitute this into the equation:
$V = \left( \frac{Q}{A k_1 \varepsilon_0} \right) t_1 + \left( \frac{Q}{A k_2 \varepsilon_0} \right) t_2$
$V = \frac{Q}{A\varepsilon_0} \left( \frac{t_1}{k_1} + \frac{t_2}{k_2} \right)$

(D) The initial capacitance of the parallel plate capacitor is $C = \frac{\varepsilon_0 A}{d}$,where $A = \pi r^2$.
When a dielectric slab of radius $r/2$ and thickness $d$ is inserted,the area covered by the dielectric is $A' = \pi (r/2)^2 = A/4$.
The remaining area of the capacitor is filled with air,which is $A'' = A - A/4 = 3A/4$.
This arrangement is equivalent to two capacitors connected in parallel.
The capacitance of the part filled with the dielectric is $C' = \frac{K \varepsilon_0 A'}{d} = \frac{6 \varepsilon_0 (A/4)}{d} = \frac{6}{4} C = \frac{3}{2} C$.
The capacitance of the air-filled part is $C'' = \frac{\varepsilon_0 A''}{d} = \frac{\varepsilon_0 (3A/4)}{d} = \frac{3}{4} C$.
The total capacitance is $C_{eq} = C' + C'' = \frac{3}{2} C + \frac{3}{4} C = \frac{6+3}{4} C = \frac{9}{4} C$.

(D) When a dielectric slab is introduced into a capacitor after disconnecting the battery,the charge $(Q)$ on the plates remains constant.
Initial capacitance $C_0 = \frac{\varepsilon_0 A}{d}$.
Initial potential difference $V_0 = 120\,V$.
When the dielectric slab of thickness $t = 6\,mm$ and dielectric constant $K = 6$ is inserted into the gap $d = 8\,mm$,the new capacitance $C'$ is given by $C' = \frac{\varepsilon_0 A}{d - t + \frac{t}{K}}$.
Since the charge $Q$ is constant,$Q = C_0 V_0 = C' V'$,where $V'$ is the new potential difference.
Therefore,$V' = V_0 \times \frac{C_0}{C'} = V_0 \times \frac{d - t + \frac{t}{K}}{d}$.
Substituting the values: $V' = 120 \times \frac{8 - 6 + \frac{6}{6}}{8} = 120 \times \frac{2 + 1}{8} = 120 \times \frac{3}{8} = 15 \times 3 = 45\,V$.

(C) The capacitance $C$ of a parallel plate capacitor with a dielectric slab of thickness $t$ and dielectric constant $K$ is given by the formula: $C = \frac{\varepsilon_0 A}{d - t + \frac{t}{K}}$.
Here, $A = \pi r^2 = \pi (0.12)^2\,m^2$, $d = 5 \times 10^{-3}\,m$, $t = 3 \times 10^{-3}\,m$, and $K = 6$.
Substituting the values:
$C = \frac{\varepsilon_0 \pi (0.12)^2}{5 \times 10^{-3} - 3 \times 10^{-3} + \frac{3 \times 10^{-3}}{6}}$
$C = \frac{\varepsilon_0 \pi (0.0144)}{2 \times 10^{-3} + 0.5 \times 10^{-3}} = \frac{\varepsilon_0 \pi (0.0144)}{2.5 \times 10^{-3}}$
Using $\frac{1}{4\pi\varepsilon_0} = 9 \times 10^9$, so $\varepsilon_0 = \frac{1}{36\pi \times 10^9}$.
$C = \frac{0.0144 \pi}{36\pi \times 10^9 \times 2.5 \times 10^{-3}} = \frac{0.0144}{90 \times 10^6} = 0.16 \times 10^{-9}\,F = 160\,pF$.

(B) The capacitance of an air-filled parallel plate capacitor is given by $C = \frac{\varepsilon_0 A}{d} = 1\,pF$.
When the plate separation is doubled $(d' = 2d)$ and the space is filled with a dielectric material (wax) of constant $K$,the new capacitance is $C' = \frac{K \varepsilon_0 A}{d'}$.
Substituting $d' = 2d$,we get $C' = \frac{K \varepsilon_0 A}{2d} = \frac{K}{2} \left( \frac{\varepsilon_0 A}{d} \right) = \frac{K}{2} C$.
Given $C' = 2\,pF$ and $C = 1\,pF$,we have $2 = \frac{K}{2} \times 1$.
Therefore,$K = 4$.

(B) When a dielectric of constant $K$ is introduced between the plates of a parallel plate capacitor,its capacity increases by a factor of $K$. Therefore,the new capacity $C' = K \times C_0 = 5C_0$.
Assuming the charge $q$ on the plates remains constant (isolated capacitor),the energy stored is given by $W = \frac{q^2}{2C}$.
The initial energy is $W_0 = \frac{q^2}{2C_0}$.
The new energy $W'$ is given by $W' = \frac{q^2}{2C'} = \frac{q^2}{2(5C_0)} = \frac{1}{5} \left( \frac{q^2}{2C_0} \right) = \frac{W_0}{5}$.
Thus,the new capacity is $5C_0$ and the new energy is $\frac{W_0}{5}$.

(C) The capacitance of a parallel plate capacitor with a dielectric slab of thickness $t$ is given by:
$C = \frac{A \epsilon_0}{(d - t) + \frac{t}{K}}$
where $A$ is the area of the plates,$d$ is the distance between them,and $K$ is the dielectric constant.
When a metal sheet (aluminium foil) of thickness $t$ is introduced,the dielectric constant $K$ for a conductor is $\infty$.
Substituting $K = \infty$ into the formula:
$C' = \frac{A \epsilon_0}{(d - t) + \frac{t}{\infty}} = \frac{A \epsilon_0}{d - t}$
If the thickness $t$ of the foil is negligible $(t \approx 0)$,then:
$C' = \frac{A \epsilon_0}{d - 0} = \frac{A \epsilon_0}{d}$
Since the original capacitance was $C = \frac{A \epsilon_0}{d}$,the capacitance remains unchanged.

(B) The capacitance of a spherical capacitor is given by the formula $C = 4\pi \varepsilon_0 K \left( \frac{ab}{b-a} \right)$,where $a$ and $b$ are the radii of the inner and outer spheres,respectively,and $K$ is the dielectric constant of the material between them.
From this formula,it is clear that $C \propto K$.
When a dielectric material is filled between the two spheres,the dielectric constant $K$ increases (since $K > 1$ for any dielectric material compared to vacuum where $K = 1$).
Therefore,filling the space with a dielectric material increases the capacitance of the capacitor.

(C) When a dielectric slab is placed near the plates of a charged capacitor, it experiences an attractive force directed into the region between the plates. This is because the electric field of the capacitor induces polarization charges on the dielectric, creating an attractive interaction. The potential energy of the system $U = \frac{Q^2}{2C}$ decreases as the dielectric enters the capacitor because the capacitance $C$ increases. Since every physical system tends to move towards a state of minimum potential energy, the dielectric slab is pulled into the region between the plates. Once it is fully inside, the force becomes zero (assuming a uniform field), and it remains there in equilibrium. Therefore, the correct option is $(c)$.

(D) When the battery is disconnected,the charge $q$ on the capacitor remains constant because there is no path for the charge to flow.
The capacitance of a parallel plate capacitor is given by $C = \frac{\varepsilon_0 A}{d}$.
When the distance $d$ between the plates is increased,the capacitance $C$ decreases.
Since $q = CV$,the voltage $V = \frac{q}{C}$ must increase as $C$ decreases.
The electrostatic energy stored in the capacitor is given by $U = \frac{q^2}{2C}$.
Since $q$ is constant and $C$ decreases,the energy $U$ increases.

(C) When a capacitor is charged and then disconnected from the battery,the charge $Q$ on the plates remains constant.
The capacitance of a parallel plate capacitor with air is $C_0 = \frac{\epsilon_0 A}{d}$.
When a dielectric of constant $K$ is introduced,the new capacitance becomes $C = K C_0$.
Since the charge $Q$ is constant,the relation between potential $V$ and capacitance $C$ is $V = \frac{Q}{C}$.
The new potential difference $V'$ is given by $V' = \frac{Q}{C} = \frac{Q}{K C_0} = \frac{V_0}{K}$.
Given $V_0 = 100\;V$ and $K = 10$,we get $V' = \frac{100}{10} = 10\;V$.

(C) When a dielectric slab is introduced between the plates of a capacitor,the capacitance $C$ increases by a factor of $K$ (dielectric constant),where $C' = KC$.
Since the capacitor remains connected to the battery,the potential difference $V$ across the plates remains constant.
According to the relation $Q = CV$,if $C$ increases and $V$ remains constant,the charge $Q$ on the plates must increase.
Therefore,the new charge is more than the earlier charge.

(D) The capacitance of a parallel plate capacitor with plate separation $d$ and area $A$ is given by $C_0 = \frac{\varepsilon_0 A}{d}$.
When a metal plate of thickness $t$ is introduced between the plates,the new capacitance $C$ is given by the formula $C = \frac{\varepsilon_0 A}{d - t}$.
Given $t = d/2$,we substitute this into the formula:
$C = \frac{\varepsilon_0 A}{d - d/2} = \frac{\varepsilon_0 A}{d/2} = 2 \frac{\varepsilon_0 A}{d}$.
Since $C_0 = \frac{\varepsilon_0 A}{d}$,we get $C = 2 C_0$.
Therefore,the capacitance is doubled.

(C) When a dielectric slab is inserted into a capacitor while it remains connected to a battery, the potential difference $V$ across the plates remains constant.
Since the capacitance $C$ increases $(C = K C_0)$, the charge $Q$ on the plates increases $(Q = CV)$.
The battery must supply additional charge to the capacitor to maintain the constant potential difference.
The work done by the battery is $W_{battery} = \Delta Q \cdot V = (Q_{final} - Q_{initial})V$.
The change in energy stored in the capacitor is $\Delta U = \frac{1}{2} C_{final} V^2 - \frac{1}{2} C_{initial} V^2 = \frac{1}{2} \Delta C V^2$.
Since $W_{battery} = \Delta C V^2$, the work done by the battery is twice the increase in stored energy.
The remaining half of the energy supplied by the battery is converted into the work done by the external agent inserting the slab.
Thus, the work is done at the cost of the battery.

(D) The capacitance of an air-filled parallel plate capacitor is given by $C_{air} = \frac{\varepsilon_0 A}{d} = 15\,\mu F$.
When a conducting plate (copper) of thickness $t$ is introduced between the plates,the new capacitance $C_{medium}$ is given by the formula $C_{medium} = \frac{\varepsilon_0 A}{d - t + \frac{t}{K}}$.
For a conductor,the dielectric constant $K = \infty$,so the term $\frac{t}{K} = 0$.
Thus,the new capacitance is $C_{medium} = \frac{\varepsilon_0 A}{d - t}$.
Taking the ratio,we get $\frac{C_{medium}}{C_{air}} = \frac{d}{d - t}$.
Substituting the given values: $d = 6\,mm$ and $t = 3\,mm$,we have $\frac{C_{medium}}{15} = \frac{6}{6 - 3} = \frac{6}{3} = 2$.
Therefore,$C_{medium} = 15 \times 2 = 30\,\mu F$.

(D) When a dielectric material with dielectric constant $K$ is inserted between the plates of a capacitor while it remains connected to a battery,the potential difference $V$ across the plates remains constant because it is fixed by the battery.
The capacitance of the capacitor increases according to the formula $C = \frac{K{\varepsilon _0}A}{d}$,where $K > 1$.
Since the charge $Q$ on the capacitor is given by $Q = CV$,and $V$ is constant while $C$ increases,the charge $Q$ also increases.
The electric field $E$ between the plates is given by $E = \frac{V}{d}$. Since $V$ and $d$ remain constant,the electric field $E$ remains constant.
Therefore,both the charge $Q$ and the capacitance $C$ increase.

(C) The initial capacity of the parallel plate capacitor is given by $C = \frac{\varepsilon_0 A}{d}$.
When the separation is doubled $(d' = 2d)$ and a dielectric medium of constant $K$ is introduced,the new capacity $C'$ is given by $C' = \frac{K \varepsilon_0 A}{d'}$.
Substituting $d' = 2d$ into the equation,we get $C' = \frac{K \varepsilon_0 A}{2d} = \frac{K}{2} C$.
Given that the new capacity $C' = 2C$,we equate the two expressions: $2C = \frac{K}{2} C$.
Solving for $K$,we get $K = 2 \times 2 = 4$.

(A) The given arrangement consists of three capacitors in series,each with dielectric constant $K_1, K_2, K_3$ and thickness $d_1, d_2, d_3$ respectively.
The capacitance of a parallel plate capacitor with a dielectric is given by $C = \frac{K \varepsilon_0 A}{d}$.
For the three capacitors in series,the equivalent capacitance $C_{eq}$ is given by:
$\frac{1}{C_{eq}} = \frac{1}{C_1} + \frac{1}{C_2} + \frac{1}{C_3}$
Substituting the values:
$\frac{1}{C_{eq}} = \frac{d_1}{K_1 \varepsilon_0 A} + \frac{d_2}{K_2 \varepsilon_0 A} + \frac{d_3}{K_3 \varepsilon_0 A}$
$\frac{1}{C_{eq}} = \frac{1}{\varepsilon_0 A} \left[ \frac{d_1}{K_1} + \frac{d_2}{K_2} + \frac{d_3}{K_3} \right]$
Therefore,the equivalent capacitance is:
$C_{eq} = \frac{\varepsilon_0 A}{\left( \frac{d_1}{K_1} + \frac{d_2}{K_2} + \frac{d_3}{K_3} \right)}$

(B) The capacitance of a parallel plate capacitor with a dielectric is given by $C = \frac{K \varepsilon_0 A}{d}$.
Since $C \propto K$,we can write the ratio as $\frac{C_1}{C_2} = \frac{K_1}{K_2}$.
Given $K_1 = 5$,$C_1 = C$,and $K_2 = 20$.
Substituting these values: $\frac{C}{C_2} = \frac{5}{20}$.
$\frac{C}{C_2} = \frac{1}{4}$.
Therefore,$C_2 = 4C$.

(D) When a capacitor is disconnected from a battery, the charge $Q$ on the plates remains constant because there is no path for the charge to flow.
Given initial dielectric constant $K_1 = 3$ and final dielectric constant $K_2 = 9$.
The charge remains constant: $Q' = Q_0$.
The potential difference $V$ across a capacitor is given by $V = Q/C$. Since $C = K \epsilon_0 A/d$, we have $V \propto 1/K$. Therefore, $V' = V_0 \times (K_1 / K_2) = V_0 \times (3 / 9) = V_0 / 3$.
The electric field $E$ is given by $E = V/d$. Since $V' = V_0 / 3$, the new electric field is $E' = E_0 / 3$.

(C) The capacitance $C$ of a parallel plate capacitor with multiple dielectric slabs is given by the formula:
$C = \frac{{\varepsilon _0 A}}{{\sum \frac{t_i}{k_i}}}$
Given:
$t_1 = 6\,mm = 6 \times 10^{-3}\,m$,$k_1 = 10$
$t_2 = 4\,mm = 4 \times 10^{-3}\,m$,$k_2 = 5$
Substituting the values:
$C = \frac{{\varepsilon _0 A}}{{\frac{6 \times 10^{-3}}{10} + \frac{4 \times 10^{-3}}{5}}}$
$C = \frac{{\varepsilon _0 A}}{{0.6 \times 10^{-3} + 0.8 \times 10^{-3}}}$
$C = \frac{{\varepsilon _0 A}}{{1.4 \times 10^{-3}}}$
$C = \frac{1000}{1.4} \varepsilon _0 A = \frac{10000}{14} \varepsilon _0 A = \frac{5000}{7} \varepsilon _0 A$

(C) Initially,the charge on the air capacitor is given by $Q = C \times V = 10\,\mu F \times 12\,V = 120\,\mu C$.
When the space between the plates is filled with a dielectric of constant $K = 5$,the new capacitance becomes $C' = K \times C = 5 \times 10\,\mu F = 50\,\mu F$.
The new charge on the capacitor is $Q' = C' \times V = 50\,\mu F \times 12\,V = 600\,\mu C$.
The additional charge that flows from the battery to the capacitor is $\Delta Q = Q' - Q = 600\,\mu C - 120\,\mu C = 480\,\mu C$.

(A) When a capacitor is charged and then disconnected from the battery,it becomes an isolated system.
According to the law of conservation of charge,the total charge $Q$ on the plates of an isolated capacitor cannot change.
When a dielectric slab is introduced,the capacitance $C$ increases $(C = K C_0)$,the potential difference $V$ decreases $(V = V_0 / K)$,and the stored energy $U$ decreases $(U = U_0 / K)$.
Therefore,the charge $Q$ remains constant.

(D) The capacitance of a parallel plate capacitor filled with a dielectric medium is given by $C_{medium} = \frac{K \epsilon_0 A}{d} = KC_0$,where $C_0$ is the capacitance with air/vacuum between the plates.
Given that $C_{medium} = C$ and $K = 2$,we have $C = 2C_0$.
Therefore,the capacitance with air (when oil is removed) is $C_0 = \frac{C}{2}$.
Thus,the correct option is $D$.

(D) The initial capacitance of the parallel plate capacitor is given by $C = \frac{A\varepsilon_0}{d} = 10\,\mu F$.
When a dielectric slab of thickness $t$ and dielectric constant $K$ is inserted between the plates,the new capacitance is given by $C' = \frac{A\varepsilon_0}{d - t + \frac{t}{K}}$.
Here,the dielectric fills half the distance,so $t = \frac{d}{2}$ and $K = 2$.
Substituting these values: $C' = \frac{A\varepsilon_0}{d - \frac{d}{2} + \frac{d/2}{2}} = \frac{A\varepsilon_0}{\frac{d}{2} + \frac{d}{4}} = \frac{A\varepsilon_0}{\frac{3d}{4}} = \frac{4}{3} \times \frac{A\varepsilon_0}{d}$.
Since $\frac{A\varepsilon_0}{d} = 10\,\mu F$,we have $C' = \frac{4}{3} \times 10 = 13.33\,\mu F$.

(C) The initial capacitance of the air-filled capacitor is given by $C_1 = \frac{\varepsilon_0 A}{d} = 10 \, pF = 10 \times 10^{-12} \, F$.
When the separation is doubled $(d' = 2d)$ and the space is filled with a dielectric of constant $K$,the new capacitance $C_2$ is given by $C_2 = \frac{K \varepsilon_0 A}{d'} = \frac{K \varepsilon_0 A}{2d}$.
Taking the ratio of the two capacitances: $\frac{C_2}{C_1} = \frac{K \varepsilon_0 A / 2d}{\varepsilon_0 A / d} = \frac{K}{2}$.
Given $C_2 = 40 \times 10^{-12} \, F$,we have $\frac{40 \times 10^{-12}}{10 \times 10^{-12}} = \frac{K}{2}$.
$4 = \frac{K}{2} \implies K = 8$.

(A) 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.
As a result,the net electric field $E_{medium}$ between the plates decreases and is given by the formula:
$E_{medium} = \frac{E_{air}}{K}$
Since $K > 1$ for any dielectric material,the electric field $E_{medium}$ will be less than the original electric field $E_{air}$.

(A) The maximum potential difference $(V_{max})$ that can be applied across a capacitor is given by the product of the dielectric strength $(E_{max})$ and the plate separation $(d)$.
Given:
Dielectric strength $(E_{max})$ = $19\, kV/mm = 19,000\, V/mm$.
Plate separation $(d)$ = $0.01\, mm$.
Calculation:
$V_{max} = E_{max} \times d$
$V_{max} = 19,000\, V/mm \times 0.01\, mm$
$V_{max} = 190\, V$.
Therefore,the maximum potential difference is $190\, V$.

(C) The initial capacitance of the parallel plate capacitor is given by $C = \frac{\varepsilon_0 A}{d} = 20\,\mu F$.
When a dielectric slab of thickness $t$ and dielectric constant $K$ is inserted,the new capacitance $C'$ is given by the formula:
$C' = \frac{\varepsilon_0 A}{d - t + \frac{t}{K}}$
Dividing the expression for $C'$ by $C$,we get:
$\frac{C'}{C} = \frac{d}{d - t + \frac{t}{K}}$
Given $d = 2\,mm$,$t = 1\,mm$,and $K = 2$:
$C' = C \times \frac{d}{d - t + \frac{t}{K}} = 20 \times \frac{2}{2 - 1 + \frac{1}{2}} = 20 \times \frac{2}{1.5} = 20 \times \frac{2}{3/2} = 20 \times \frac{4}{3} = \frac{80}{3} \approx 26.67\,\mu F$.
Thus,the new capacitance is approximately $26.6\,\mu F$.

(A) The capacitance of a parallel plate capacitor is given by $C = \frac{K \varepsilon_0 A}{d}$.
Initially,$K_1 = 1$ (air),$d_1 = 0.4 \, cm$,and $C_1 = 2 \, \mu F$.
Finally,$K_2 = 2.8$,$d_2 = \frac{d_1}{2} = 0.2 \, cm$,and we need to find $C_2$.
Using the ratio: $\frac{C_2}{C_1} = \frac{K_2}{K_1} \times \frac{d_1}{d_2}$.
Substituting the values: $\frac{C_2}{2} = \frac{2.8}{1} \times \frac{0.4}{0.2}$.
$\frac{C_2}{2} = 2.8 \times 2 = 5.6$.
$C_2 = 5.6 \times 2 = 11.2 \, \mu F$.

(D) The initial capacitance of the parallel plate capacitor is given by $C = \frac{\varepsilon_0 A}{d}$,where $A$ is the area of the plates and $d$ is the separation between them.
When a metal sheet of thickness $t$ is inserted between the plates,the effective separation between the plates decreases. The new capacitance $C'$ is given by the formula $C' = \frac{\varepsilon_0 A}{d - t + \frac{t}{K}}$,where $K$ is the dielectric constant.
For a metal sheet,the dielectric constant $K = \infty$.
Given that the thickness of the sheet $t = d/2$,we substitute these values into the formula:
$C' = \frac{\varepsilon_0 A}{d - d/2 + \frac{d/2}{\infty}}$
Since $\frac{d/2}{\infty} = 0$,the expression simplifies to:
$C' = \frac{\varepsilon_0 A}{d - d/2} = \frac{\varepsilon_0 A}{d/2} = 2 \left( \frac{\varepsilon_0 A}{d} \right)$
Substituting $C = \frac{\varepsilon_0 A}{d}$,we get $C' = 2C$.

(B) When a dielectric slab is inserted between the plates of a capacitor while it remains connected to a battery,the potential difference $(V)$ across the plates remains constant because it is fixed by the battery.
Since the capacitance $(C)$ increases by a factor of $K$ (where $K$ is the dielectric constant),the charge on the plates $(Q = CV)$ must increase.
Therefore,additional charge flows from the battery to the capacitor to maintain the potential difference.
Thus,option $(B)$ is correct.

(B) When a parallel plate capacitor is connected to a battery,the potential difference $V$ across its plates remains constant because it is maintained by the battery.
The electric field $E$ between the plates is given by the relation $E = \frac{V}{d}$,where $d$ is the distance between the plates.
Since the battery remains connected,$V$ is constant,and the distance $d$ between the plates does not change.
Therefore,the electric field $E$ remains equal to its initial value $E_0$,regardless of the introduction of a dielectric material.

(C) The capacitance of a parallel plate capacitor is given by $C = \frac{K \epsilon_0 A}{d}$,where $K$ is the dielectric constant,$A$ is the area of the plates,and $d$ is the distance between them.
Initially,$C_1 = \frac{K \epsilon_0 A}{d}$.
According to the problem,the new distance $d' = \frac{d}{2}$ and the new dielectric constant $K' = 2K$.
The new capacitance $C_2$ is given by $C_2 = \frac{K' \epsilon_0 A}{d'} = \frac{(2K) \epsilon_0 A}{d/2}$.
Simplifying this,$C_2 = 4 \times \frac{K \epsilon_0 A}{d} = 4C_1$.
Therefore,the capacitance increases four times.

(A) When a dielectric material with dielectric constant $K > 1$ is inserted between the plates of a capacitor (assuming the capacitor is disconnected from the battery,so charge $Q$ remains constant):
$1$. Capacitance $(C)$: The new capacitance is $C' = KC$. Since $K > 1$,the capacitance increases.
$2$. Potential Difference $(V)$: Since $V = Q/C$,and $C$ increases while $Q$ is constant,the potential difference $V$ decreases.
$3$. Potential Energy $(U)$: The potential energy is given by $U = Q^2 / (2C)$. Since $C$ increases and $Q$ is constant,the potential energy $U$ decreases.
Therefore,the correct sequence is: Increase,decrease,decrease.

(A) When a thin metal plate is inserted between the plates of a parallel plate capacitor,it acts as an equipotential surface.
This effectively divides the original capacitor into two capacitors in series,each with a plate separation of $d/2$.
The capacitance of each new capacitor is $C' = \frac{\epsilon_0 A}{d/2} = 2C$.
Since these two capacitors are in series,the equivalent capacitance $C_{eq}$ is given by $\frac{1}{C_{eq}} = \frac{1}{C'} + \frac{1}{C'} = \frac{1}{2C} + \frac{1}{2C} = \frac{1}{C}$.
Therefore,$C_{eq} = C$.
Thus,the capacitance remains unchanged.

(C) When a capacitor is charged and then isolated,the total charge $Q$ on the plates remains constant because there is no external source or path for the charge to flow.
When a dielectric material is inserted,the capacitance $C$ increases to $C' = KC$,where $K$ is the dielectric constant.
Since $Q$ is constant and $C$ increases,the potential difference $V = Q/C$ decreases.
The electric field $E = V/d$ also decreases.
The energy stored $U = Q^2/(2C)$ decreases because $C$ increases.
Therefore,the charge on the plates remains unchanged.

(D) The capacitance of a parallel plate capacitor with air is given by $C = \frac{\varepsilon_0 A}{d}$.
When a dielectric slab of thickness $t$ and dielectric constant $K$ is inserted,the new capacitance $C'$ is given by $C' = \frac{\varepsilon_0 A}{d - t + \frac{t}{K}}$.
Given that $C' = \frac{4}{3}C$ and $t = \frac{d}{2}$,we substitute these values:
$\frac{\varepsilon_0 A}{d - \frac{d}{2} + \frac{d/2}{K}} = \frac{4}{3} \times \frac{\varepsilon_0 A}{d}$.
Simplifying the equation:
$\frac{1}{\frac{d}{2} + \frac{d}{2K}} = \frac{4}{3d} \Rightarrow \frac{1}{\frac{d}{2}(1 + \frac{1}{K})} = \frac{4}{3d}$.
$\frac{2}{d(1 + 1/K)} = \frac{4}{3d} \Rightarrow \frac{2}{1 + 1/K} = \frac{4}{3}$.
$6 = 4(1 + 1/K) \Rightarrow 6 = 4 + \frac{4}{K}$.
$2 = \frac{4}{K} \Rightarrow K = 2$.

(B) The energy stored in a capacitor is given by the formula $U = \frac{Q^2}{2C}$,where $Q$ is the charge on the plates and $C$ is the capacitance.
When a dielectric substance is introduced between the plates of a charged capacitor (assuming the capacitor is isolated,so $Q$ remains constant),the capacitance $C$ increases because $C = \frac{K \epsilon_0 A}{d}$,where $K > 1$.
Since $U$ is inversely proportional to $C$ $(U \propto \frac{1}{C})$,an increase in capacitance $C$ leads to a decrease in the stored energy $U$.

(D) The capacitance of an air-filled parallel plate capacitor is given by $C = \frac{\varepsilon_0 A}{d}$ ... $(i)$.
When the distance between the plates is doubled $(d' = 2d)$ and it is immersed in a liquid of dielectric constant $K$,the new capacitance $C'$ is given by $C' = \frac{K \varepsilon_0 A}{d'} = \frac{K \varepsilon_0 A}{2d}$ ... $(ii)$.
Given that the new capacity is twice the original capacity,i.e.,$C' = 2C$.
Substituting the expressions from $(i)$ and $(ii)$ into this relation:
$\frac{K \varepsilon_0 A}{2d} = 2 \left( \frac{\varepsilon_0 A}{d} \right)$.
Simplifying the equation:
$\frac{K}{2} = 2$.
Therefore,$K = 4$.