Class 12 Physics (Part 1) Chapter 4 Capacitors and Dielectrics

Key Concept: Parallel Plate Capacitor, Partial Dielectric Filling

b) 1.475 nF
[Solution Description]
This setup creates three regions: two air gaps ($K = 1$) each of thickness $\frac{d - t}{2} = 0.2 \, \text{mm}$, and one dielectric region ($K = 3$) of thickness $t = 0.6 \, \text{mm}$. The total capacitance is calculated as series combination of these regions:
$\frac{1}{C} = \frac{0.2 \times 10^{-3}}{\varepsilon_0 A} + \frac{0.6 \times 10^{-3}}{3 \varepsilon_0 A} + \frac{0.2 \times 10^{-3}}{\varepsilon_0 A}.$
Simplifying:
$\frac{1}{C} = \frac{0.2 + 0.2 + 0.2}{\varepsilon_0 A} \times 10^{3} = \frac{0.6}{\varepsilon_0 A} \times 10^{3}.$
Thus:
$C = \frac{\varepsilon_0 A}{0.6 \times 10^{-3}} = \frac{(8.85 \times 10^{-12})(0.1)}{0.6 \times 10^{-3}} = 1.475 \times 10^{-9} \, \text{F}.$
The capacitance is $1.475 \, \text{nF}$.

Key Concept: Energy Density in Capacitor

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The assertion states that the energy density in a capacitor is given by $\frac{1}{2} \varepsilon_0 E^2$, which is correct. The reason states that energy density is defined as the energy stored per unit volume, which is also correct. Moreover, the reason correctly explains why the assertion is true, as the formula $\frac{1}{2} \varepsilon_0 E^2$ is derived from this definition.

Key Concept: Redistribution of Charges, Common Potential

c) -80\,V
[Solution Description]
Step 1: The initial charges are:
$q_1 = C_1V_1 = 4 \times 100 = 400\,\mu C$ (positive plate).
$q_2 = C_2V_2 = 6 \times 200 = 1200\,\mu C$ (negative plate).
Step 2: Since the polarities are opposite, the net charge before connection is $q_{net} = q_1 - q_2 = 400 - 1200 = -800\,\mu C$.
Step 3: The common potential $V$ after redistribution is given by $V = \frac{C_1V_1 - C_2V_2}{C_1 + C_2}$.
$V = \frac{400 - 1200}{4 + 6} = \frac{-800}{10} = -80\,V$.
Thus, the common potential is $-80\,V$.

Key Concept: Net charge on a capacitor

c) Zero
[Solution Description]
The net charge on a capacitor is the sum of the charges on both plates. Since the plates have equal and opposite charges ($+Q$ and $-Q$), the net charge is $Q + (-Q) = 0$.

Key Concept: Assertion-Reason

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The assertion states that when a dielectric is placed in an electric field, it gets polarized. This is true because the electric field causes the charges within the dielectric to redistribute, leading to polarization. The reason explains that this polarization occurs due to the alignment of either permanent or induced dipole moments in the direction of the applied field, which is also correct and directly explains why the assertion is true. Hence, both the assertion and reason are true, and the reason correctly explains the assertion.

Key Concept: Energy Stored in a Charged Capacitor

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The energy stored in a capacitor is derived from the work done to move charges against the electric field, which results in the formula $\frac{1}{2} C V^2$. This energy is indeed stored as electric potential energy between the plates. Therefore, both Assertion and Reason are true, and Reason correctly explains the Assertion.

Key Concept: Series and Parallel Combinations, Equivalent Capacitance

c) 8 V
[Solution Description]
First, find the equivalent capacitance of the series combination:
$\frac{1}{C_{eq}} = \frac{1}{4} + \frac{1}{6} + \frac{1}{12} = \frac{3}{12} + \frac{2}{12} + \frac{1}{12} = \frac{6}{12} = \frac{1}{2}$
So, $C_{eq} = 2 \mu F$.
The total charge on each capacitor (since they are in series):
$Q = C_{eq} V = 2 \times 24 = 48 \mu C$.
Now, the potential difference across the 6 \mu F capacitor:
$V_2 = \frac{Q}{C_2} = \frac{48}{6} = 8 V$.

Key Concept: Energy Stored in a Charged Capacitor, Energy Loss in Capacitor

d) Assertion is false, but Reason is true.
[Solution Description]
Let the capacitance of each capacitor be $C$ and the potential be $V$. Initial energy stored in each capacitor is $\frac{1}{2}CV^2$, so total initial energy is $2 \times \frac{1}{2}CV^2 = CV^2$. When connected in parallel, the equivalent capacitance becomes $2C$ and the common potential becomes $V$ (since they are identical and at the same potential initially). Thus, the total energy stored becomes $\frac{1}{2}(2C)V^2 = CV^2$, which is equal to the initial total energy. Hence, no energy is lost. However, if the capacitors were charged to different potentials, energy would be lost during redistribution. Here, Assertion (A) is false because no energy is lost when identical capacitors at the same potential are connected in parallel. Reason (R) is true as energy loss does occur during charge redistribution in general cases, but it is not applicable here.

Key Concept: Capacitance with Dielectric, Energy Storage

a) 5:8
[Solution Description]
Let the initial capacitance be $C_0 = 100$ pF and the final capacitance be $C = 160$ pF.
Since the capacitor remains connected to the voltage source, the potential difference $V$ remains constant.
Initial energy:
$U_0 = \frac{1}{2} C_0 V^2$
Final energy:
$U = \frac{1}{2} C V^2$
The ratio is:
$\frac{U_0}{U} = \frac{C_0}{C} = \frac{100}{160} = \frac{5}{8}$

Key Concept: Drift velocity of free electrons

c) $1.47 \times 10^{-4}$ m/s
[Solution Description]
The formula for drift velocity $(v_d)$ is given by:
$v_d = \frac{I}{nAe}$
where $I$ is the current, $n$ is the number density of free electrons, $A$ is the cross-sectional area, and $e$ is the charge of an electron ($1.6 \times 10^{-19}$ C). Substituting the given values:
$v_d = \frac{2}{(8.5 \times 10^{28}) \times (1 \times 10^{-6}) \times (1.6 \times 10^{-19})}$
$v_d = \frac{2}{13.6 \times 10^3} \approx 0.147 \times 10^{-3} \text{ m/s}$
$v_d \approx 1.47 \times 10^{-4} \text{ m/s}$

Key Concept: Vacuum or Air Between Plates, Dielectric Slab Filling Entire Space Between Plates

c) 6
[Solution Description]
The initial energy stored in the capacitor is $U_0 = \frac{Q^2}{2C_0}$, where $Q$ is the charge on the plates. After inserting the dielectric, the capacitance becomes $C = K C_0 = 6 C_0$. The charge remains the same (assuming an isolated system), so the new energy is:
$U = \frac{Q^2}{2C} = \frac{Q^2}{2 \times 6 C_0} = \frac{U_0}{6}$
The energy decreases by a factor of $6$.

Key Concept: Dielectric Strength

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The assertion states that the maximum voltage in a capacitor depends on the dielectric strength. This is true because exceeding the dielectric strength causes breakdown. The reason correctly defines dielectric strength as the threshold for breakdown, which directly explains why it limits the maximum voltage. Hence, both are true, and the reason explains the assertion.

Key Concept: Parallel Plate Capacitor with Dielectric Slab

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
When a dielectric material is placed between the plates of a capacitor, it reduces the effective electric field because the dielectric polarizes and creates an opposing electric field. This allows the capacitor to store more charge for the same potential difference, hence increasing the capacitance. Therefore, both Assertion (A) and Reason (R) are true, and Reason (R) correctly explains Assertion (A).

Key Concept: Effect of dielectric material on capacitance

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The assertion is true because when a dielectric slab is introduced between the plates of a parallel-plate capacitor, the capacitance increases as it is proportional to the dielectric constant ($C \propto K$). The reason is also true because the dielectric reduces the effective electric field by creating an opposing field due to polarization, which allows the capacitor to store more charge for the same potential difference. Moreover, the reason correctly explains the assertion.

Key Concept: Effect of Dielectric on Capacitance

c) It increases by a factor of 4
[Solution Description]
The capacitance of a parallel-plate capacitor with a dielectric is given by:
$C_{\text{new}} = K \times C_{\text{air}}$
Since $K = 4$, the new capacitance will be four times the original capacitance when air was the medium.

Key Concept: Conductors and Insulators

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The assertion states that free electrons in a conductor cannot escape from its surface without external energy. This is true because, although free electrons move randomly within the conductor, they are still bound to the conductor as a whole due to electrostatic forces.
The reason explains why free electrons do not escape: they remain bound to the conductor as a whole, even though they are not attached to any specific atom. This is correct and directly explains the assertion.
Therefore, both the assertion and reason are true, and the reason correctly explains the assertion.

Key Concept: Effect of Dielectric on Stored Energy

b) It decreases by a factor of $K$
[Solution Description]
When the battery is disconnected, the charge remains the same, but the capacitance increases to $C = K C_0$. The energy stored decreases because $U = \frac{U_0}{K}$, where $U_0$ is the initial energy.

Key Concept: Dependence of Capacitance on Dielectric Constant

b) $\frac{8}{5}C_0$
[Solution Description]
The original capacitance without the dielectric is $C_0 = \frac{\varepsilon_0 A}{d}$. When a dielectric slab of thickness $t = \frac{d}{2}$ and dielectric constant $K = 4$ is inserted, the capacitance becomes:
$C = \frac{\varepsilon_0 A}{d - t + \frac{t}{K}} = \frac{\varepsilon_0 A}{d - \frac{d}{2} + \frac{d}{8}} = \frac{\varepsilon_0 A}{\frac{5d}{8}} = \frac{8}{5}C_0$
Therefore, the new capacitance is $\frac{8}{5}C_0$.

Key Concept: Dielectric Constant and Capacitance

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The assertion states that the capacitance of a capacitor increases when a dielectric material is placed between its plates, which is true because according to the syllabus, $C = K C_0$, where $K$ is the dielectric constant ($K > 1$ for most dielectrics). The reason defines the dielectric constant as $K = \frac{C}{C_0}$, which is also correct and directly explains why the capacitance increases (since $K$ is greater than 1). Thus, both the assertion and reason are true, and the reason correctly explains the assertion.

Key Concept: Electric Potential Energy Formula

c) $\frac{1}{2} \frac{Q^2}{C}$
[Solution Description]
From the syllabus, the electric potential energy stored in a capacitor is given by:
$U = \frac{1}{2} \frac{Q^2}{C}$

Key Concept: Conductors and Insulators

c) Copper
[Solution Description]
Conductors are materials that allow electric charge to flow easily through them. Metals, human body, earth, mercury, and electrolytes are examples of conductors. Among the given options, copper is a metal and hence a conductor.

Key Concept: Polar Dielectrics, Thermal Agitation

b) $P$ decreases
[Solution Description]
In polar dielectrics, the alignment of dipoles is opposed by thermal agitation. When temperature increases, the thermal agitation increases, leading to less perfect alignment of dipoles in the direction of the applied field. Hence, the net dipole moment per unit volume decreases with increasing temperature.

Key Concept: Energy Density in Capacitor

b) $\frac{1}{2} \varepsilon_0 E^2$
[Solution Description]
The problem requires deriving the energy density formula for a capacitor.
Given:
- Electric field = $E$
- Permittivity of free space = $\varepsilon_0$
Energy density ($u$) is defined as energy per unit volume.
For a parallel-plate capacitor:
1. Total energy stored:
$U = \frac{1}{2} C V^2$
Since $C = \frac{\varepsilon_0 A}{d}$ and $V = E d$, we substitute:
$U = \frac{1}{2} \left( \frac{\varepsilon_0 A}{d} \right)(E d)^2 = \frac{1}{2} \varepsilon_0 E^2 A d$
2. Volume between plates = $A d$
Therefore, energy density:
$u = \frac{U}{A d} = \frac{1}{2} \varepsilon_0 E^2$
The correct option must match this final expression.

Key Concept: Behavior of free electrons, Capacitance

b) Both Assertion and Reason are true, but Reason is NOT the correct explanation of Assertion.
[Solution Description]
The assertion states that the capacitance of a conductor increases when it is placed in a dielectric medium compared to vacuum. This is true according to the relation $C = K C_0$, where $K > 1$ for dielectric materials. The reason states that the presence of a dielectric medium reduces the effective electric field between the plates of a capacitor. While this is also true, the reduction in the electric field does not directly explain why capacitance increases. The increase in capacitance is due to the polarization of the dielectric material, which allows more charge to be stored for the same potential difference. Hence, both Assertion and Reason are true, but Reason is NOT the correct explanation of Assertion.

Key Concept: Dielectric Strength

c) Mica
[Solution Description]
Dielectric strength refers to the maximum electric field a material can withstand without breaking down. Materials like mica and certain plastics have very high dielectric strengths compared to air or water.
Among typical dielectrics, mica is known for its high dielectric strength (~118 kV/mm), much higher than air (3 kV/mm) or water (varies but generally lower than mica).

Key Concept: Assertion-Reason

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The assertion states that the dielectric constant of a material is always greater than or equal to 1. This is true because for vacuum K = 1 (minimum value), and for any other dielectric material, K > 1. The reason given is also correct as it correctly states that for vacuum, K = 1 by definition. Further, this reason explains why the assertion is true (since K cannot be less than 1).

Key Concept: Electric field in dielectric

c) $\frac{Q}{4 \epsilon_0 A}$
[Solution Description]
The electric field in the absence of dielectric is:
$E_0 = \frac{Q}{\epsilon_0 A}$
With the dielectric, the electric field becomes:
$E = \frac{E_0}{K} = \frac{Q}{K \epsilon_0 A}$
Substituting $K = 4$, we get:
$E = \frac{Q}{4 \epsilon_0 A}$

Key Concept: Polar and Non-polar Molecules, Dipole Moment

c) CO$_2$ and CH$_4$ develop induced dipole moments, whereas H$_2$O aligns its permanent dipole moment
[Solution Description]
CO$_2$ is a non-polar molecule due to its linear symmetry ($\vec{p} = 0$). H$_2$O is polar with a permanent dipole moment. CH$_4$ is non-polar due to its tetrahedral symmetry. When placed in an external electric field, CO$_2$ and CH$_4$ develop an induced dipole moment, while H$_2$O aligns its existing dipole moment.

Key Concept: Parallel Plate Capacitor, Metal Slab Insertion

b) 5.90 nF
[Solution Description]
When a metal slab of thickness $t$ is inserted, the effective distance between plates reduces to $d - t$. The formula becomes:
$C = \frac{\varepsilon_0 A}{d - t}.$
Substituting the given values:
$C = \frac{(8.85 \times 10^{-12})(0.2)}{(5 \times 10^{-4}) - (2 \times 10^{-4})} = \frac{1.77 \times 10^{-12}}{3 \times 10^{-4}} = 5.9 \times 10^{-9} \, \text{F}.$
The new capacitance is $5.9 \, \text{nF}$.

Key Concept: Capacitance of a spherical conductor in a medium, Proportionality

b) 1.5
[Solution Description]
The capacitance of a spherical conductor in vacuum is given by:
$C_0 = 4 \pi \varepsilon_0 a$
In a medium with dielectric constant $K$, the capacitance becomes:
$C_K = 4 \pi \varepsilon_0 K a$
According to the problem, $C_K = 1.5 C_0$. Substituting the values:
$4 \pi \varepsilon_0 K a = 1.5 \times 4 \pi \varepsilon_0 a$
Simplifying:
$K = 1.5$

Key Concept: Capacitors in Series and Parallel

b) Both Assertion and Reason are true, but Reason is NOT the correct explanation of Assertion.
[Solution Description]
The assertion states that the charge on each capacitor in a series combination is the same. This is true because in a series connection, the charge is conserved and redistributes equally across all capacitors due to electrostatic induction.
The reason states that the equivalent capacitance in a series combination is less than the smallest individual capacitance, which is also true as per the formula for series combination:
$\frac{1}{C_{eq}} = \frac{1}{C_1} + \frac{1}{C_2} + \frac{1}{C_3}.$
However, the reason does not correctly explain the assertion. The equality of charge in a series combination arises from charge conservation and redistribution, not directly from the relationship between equivalent and individual capacitances.
Therefore, both the assertion and reason are true, but the reason does not correctly explain the assertion.

Key Concept: Voltage Division in Series

b) 6 V
[Solution Description]
Since the capacitors are in series, they share the same charge $q$. First, find the equivalent capacitance:
$\frac{1}{C} = \frac{1}{6} + \frac{1}{12} = \frac{2}{12} + \frac{1}{12} = \frac{3}{12} = \frac{1}{4}$
So,
$C = 4 \text{ pF}$
Now, calculate the total charge:
$q = C \times V = 4 \times 10^{-12} \times 9 = 36 \times 10^{-12} \text{ C} = 36 \text{ pC}$
The potential difference across $C_1$ is:
$V_1 = \frac{q}{C_1} = \frac{36 \text{ pC}}{6 \text{ pF}} = 6 \text{ V}$
The potential difference across the $6 \text{ pF}$ capacitor is $6 \text{ V}$.

Key Concept: Energy stored, Distance between plates

c) Energy doubles
[Solution Description]
The energy stored in a capacitor is given by $U = \frac{Q^2}{2C}$. Since the capacitor is disconnected, $Q$ remains constant. As $C \propto 1/d$, doubling $d$ halves $C$. Therefore, $U$ becomes $U' = \frac{Q^2}{2(C/2)} = \frac{Q^2}{C} = 2U$. Hence, the energy stored doubles.

Key Concept: Capacitance formula

b) $4 \pi \epsilon_0 a$
[Solution Description]
The capacitance $C$ of an isolated spherical conductor in vacuum is given by the formula:
$C = 4 \pi \epsilon_0 a$
where $\epsilon_0$ is the permittivity of free space and $a$ is the radius of the sphere.

Key Concept: Dependence of Capacitance on Separation and Dielectric Thickness

c) The capacitance increases
[Solution Description]
The effective capacitance of the capacitor with a dielectric slab is given by:
$C = \frac{\varepsilon_0 A}{\left(d - t + \frac{t}{K}\right)}$
If the slab thickness increases to $2t$, the new capacitance becomes:
$C' = \frac{\varepsilon_0 A}{\left(d - 2t + \frac{2t}{K}\right)}$
Since the denominator decreases (as $\frac{2t}{K} 1$), $C'$ increases compared to $C$.

Key Concept: Energy storage, capacitance, energy density

b) $4.26 \, \text{m}^3$
[Solution Description]
First, find the capacitance $C$ using $U = \frac{1}{2} C V^2$:
$C = \frac{2 U}{V^2} = \frac{2 \times 2}{(1000)^2} = 4 \times 10^{-6} \, \text{F}$.
Next, find the electric field $E$ using $u = \frac{1}{2} \varepsilon_0 E^2$:
$E = \sqrt{\frac{2 u}{\varepsilon_0}} = \sqrt{\frac{2 \times 0.5}{8.85 \times 10^{-12}}} \approx 1.06 \times 10^5 \, \text{V/m}$.
Now, calculate the plate separation $d$ using $E = \frac{V}{d}$:
$d = \frac{V}{E} = \frac{1000}{1.06 \times 10^5} \approx 9.43 \times 10^{-3} \, \text{m}$.
Finally, the volume $V_{\text{space}} = A d = \frac{C d}{\varepsilon_0} = \frac{4 \times 10^{-6} \times 9.43 \times 10^{-3}}{8.85 \times 10^{-12}} \approx 4.26 \, \text{m}^3$.

Key Concept: Polar and Non-polar Molecules

c) Water
[Solution Description]
Polar dielectrics have molecules with a permanent dipole moment due to the separation of positive and negative charges. Among the given options, water is a polar molecule because it has a bent structure leading to an uneven distribution of charge.

Key Concept: Capacitance with Dielectric Slab Inserted

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]

Key Concept: Free vs. Bound Charges

c) Free charges move freely, while bound charges remain fixed.
[Solution Description]
In conductors, free charges refer to the electrons that are not bound to any specific atom and can move freely, facilitating electrical conduction. Bound charges refer to the positive ions that are fixed within the atomic lattice and cannot move.

Key Concept: Capacitance with dielectric slab

b) $141.6 \, \text{pF}$
[Solution Description]
First, calculate the initial capacitance without the dielectric ($C_0$):
$C_0 = \frac{\varepsilon_0 A}{d} = \frac{(8.85 \times 10^{-12})(100 \times 10^{-4})}{1 \times 10^{-3}} = 8.85 \times 10^{-11} \, \text{F}$.
Now, use the formula for capacitance with a dielectric slab:
$C = \frac{\varepsilon_0 A}{(d - t) + \frac{t}{K}} = \frac{(8.85 \times 10^{-12})(100 \times 10^{-4})}{(1 - 0.5) \times 10^{-3} + \frac{0.5 \times 10^{-3}}{4}}$
Simplifying denominator: $(0.5 \times 10^{-3}) + (0.125 \times 10^{-3}) = 0.625 \times 10^{-3} \, \text{m}$
Thus, $C = \frac{8.85 \times 10^{-14}}{0.625 \times 10^{-3}} = 1.416 \times 10^{-10} \, \text{F}$ or $141.6 \, \text{pF}$.

Key Concept: Work done in charging a capacitor

c) It becomes four times
[Solution Description]
The energy stored in a capacitor is given by $U = \frac{1}{2}\frac{Q^2}{C}$. If the charge $Q$ is doubled, the new energy becomes:
$U' = \frac{1}{2}\frac{(2Q)^2}{C} = \frac{1}{2}\frac{4Q^2}{C} = 4 \left(\frac{1}{2}\frac{Q^2}{C}\right) = 4U$
Thus, the energy stored becomes four times the original energy.

Key Concept: Potential Difference and Distance

b) The divergence increases
[Solution Description]
The divergence of the electroscope leaves indicates the potential difference between the plates. As the distance ($d$) between the plates increases, the capacitance ($C$) decreases because $C \propto \frac{1}{d}$. Since the charge ($Q$) remains constant (assuming the plates are isolated), the potential difference ($V$) increases as $V = \frac{Q}{C}$. Therefore, the divergence of the electroscope leaves increases.

Key Concept: Force between the Plates of a Charged Capacitor

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
To verify this assertion and reason:
\begin{itemize}

Key Concept: Bound vs Free Charge

c) Free charges are electrons, bound charges are positive ions
[Solution Description]
In metals, free charges are the electrons that can move freely throughout the volume of the metal, while bound charges are the positively charged ions that remain fixed in their positions within the lattice.

Key Concept: Capacitance and Distance

b) Capacitance doubles
[Solution Description]
The capacitance $C$ of a parallel-plate capacitor is inversely proportional to the distance $d$ between the plates, given by $C = \frac{\varepsilon_0 A}{d}$. If the distance $d$ is halved, the capacitance doubles because $C \propto \frac{1}{d}$.

Key Concept: Series Combination of Capacitors

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
For a series combination of capacitors:
1. The equivalent capacitance $C$ is given by $\frac{1}{C} = \frac{1}{C_1} + \frac{1}{C_2} + \dots + \frac{1}{C_n}$.
2. Since we are adding reciprocals, the resultant $C$ will always be less than the smallest capacitor in the combination. Hence, the Assertion is true.
3. In a series combination, the charge on each capacitor is the same ($Q$). From $V = \frac{Q}{C}$, the potential difference $V$ across each capacitor is inversely proportional to its capacitance $C$. Thus, the Reason is also true and correctly explains the Assertion.

Key Concept: Effect of Introducing a Dielectric Slab

d) Decreases by a factor of $K$
[Solution Description]
The question examines how inserting a dielectric affects the force between capacitor plates when the battery is disconnected.
Initial conditions:
- Capacitance without dielectric = $C_0$
- Dielectric constant = $K$
When the dielectric is inserted while the battery is disconnected:
1. Charge on plates ($Q$) remains unchanged.
2. New capacitance becomes $C = K C_0$.
3. Potential difference decreases to $V = \frac{V_0}{K}$.
4. Electric field inside the dielectric decreases to $E = \frac{E_0}{K}$.
5. The force between the plates is given by:
$F = \frac{Q^2}{2 A \varepsilon_0 K}$
Thus, the force decreases by a factor of $K$.

Key Concept: Work done in charging a capacitor

b) $\frac{1}{2} \frac{Q^2}{C}$
[Solution Description]
The work done in charging a capacitor is equal to the energy stored in it. The energy stored is given by:
$W = \frac{1}{2} \frac{Q^2}{C}$
Here, $Q$ is the final charge on the capacitor and $C$ is its capacitance. This expression is derived from integrating the small work done at each step during charging.

Key Concept: Definition of Dielectric Constant

a) Ratio of capacitance in vacuum to capacitance with dielectric
[Solution Description]
The dielectric constant (K) is defined as the ratio of the capacitance of a capacitor when filled with the dielectric material ($C$) to the capacitance of the same capacitor in vacuum ($C_0$). Mathematically, it is expressed as:
$K = \frac{C}{C_0}$

Key Concept: Energy stored in capacitor

c) 4U
[Solution Description]
The energy stored in a capacitor is given by $U = \frac{1}{2} C V^2$. When voltage doubles, energy becomes $U' = \frac{1}{2} C (2V)^2 = 4 \times \frac{1}{2} C V^2 = 4U$.

Key Concept: Capacitance of an Isolated Spherical Conductor

c) $4 \pi \epsilon_0 a$
[Solution Description]
The capacitance $C$ of an isolated spherical conductor of radius $a$ in vacuum is given by:
$C = 4 \pi \epsilon_0 a$
where $\epsilon_0$ is the permittivity of free space.

Key Concept: Dielectric Strength Definition

b) The maximum electric field a material can tolerate without breakdown
[Solution Description]
Dielectric strength is defined as the maximum electric field a material can tolerate without breakdown. It is expressed in kV/mm or V/m.

Key Concept: Metallic slab insertion, Capacitance change, Charge conservation

c) $100 \, \text{V}$
[Solution Description]
Initial capacitance $C_0 = \frac{\varepsilon_0 A}{d} = \frac{(8.85 \times 10^{-12})(2)}{6 \times 10^{-3}} = 2.95 \times 10^{-9} \, \text{F}$. Charge stored $Q = C_0 V_0 = 8.85 \times 10^{-7} \, \text{C}$. With the metallic slab inserted, the new capacitance is $C = \frac{\varepsilon_0 A}{d - t} = \frac{(8.85 \times 10^{-12})(2)}{2 \times 10^{-3}} = 8.85 \times 10^{-7} \, \text{F}$. The new potential difference is $V = \frac{Q}{C} = \frac{8.85 \times 10^{-7}}{8.85 \times 10^{-9}} = 100 \, \text{V}$.

Key Concept: Dielectric Strength

c) The maximum electric field a dielectric can withstand without breakdown.
[Solution Description]
Dielectric strength is the maximum electric field a dielectric can tolerate without breaking down. It represents the threshold beyond which the dielectric material loses its insulating properties and starts conducting. It is measured in kV/mm.

Key Concept: Energy stored in electric field, Energy density

d) Assertion is false, but Reason is true.
[Solution Description]
**Step 1:** The energy stored in a capacitor when charge $Q$ is kept constant is given by $U = \frac{1}{2} \frac{Q^2}{C}$.
**Step 2:** The capacitance $C$ of a parallel-plate capacitor is $C = \frac{\epsilon_0 A}{d}$, where $d$ is the separation between the plates. If $d$ decreases, $C$ increases.
**Step 3:** Substituting $C$ into the energy formula, $U = \frac{1}{2} \frac{Q^2 d}{\epsilon_0 A}$. Thus, as $d$ decreases, $U$ decreases, which contradicts the Assertion (A). Therefore, Assertion (A) is false.
**Step 4:** The energy density $u$ in the electric field is given by $u = \frac{1}{2} \epsilon_0 E^2$, which is indeed proportional to the square of the electric field strength. Hence, Reason (R) is true.

Key Concept: Capacitance, Plate Area Calculation

b) 904 \, \text{m}^2
[Solution Description]
The formula to calculate the area of the plates of a parallel-plate capacitor is:
$A = \frac{C d}{\varepsilon_0}$
Given:
$C = 4 \times 10^{-6} \, \text{F}, \quad d = 2 \times 10^{-3} \, \text{m}, \quad \varepsilon_0 = 8.85 \times 10^{-12} \, \text{F m}^{-1}$
Substituting the values into the formula:
$A = \frac{4 \times 10^{-6} \times 2 \times 10^{-3}}{8.85 \times 10^{-12}} = \frac{8 \times 10^{-9}}{8.85 \times 10^{-12}} \approx 904 \, \text{m}^2$

Key Concept: Capacitance of a Parallel-Plate Capacitor with Dielectric Slab

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The assertion states that the capacitance of a parallel-plate capacitor increases when a dielectric slab is inserted between the plates. This is true because the effective electric field decreases due to polarization in the dielectric, which allows more charge to be stored for the same potential difference, thereby increasing capacitance. The reason correctly explains why the capacitance increases, as the reduction in the electric field ($E_i = \frac{E_0}{K}$) leads to a higher capacitance value ($C = \frac{\varepsilon_0 A}{d - t + \frac{t}{K}}$). Therefore, both Assertion and Reason are true, and the Reason is the correct explanation of the Assertion.

Key Concept: Capacitance of a Spherical Conductor

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
From the syllabus, it is clear that for an isolated spherical conductor, the capacitance is given by $C = 4 \pi \epsilon_0 K a$. Here, $C \propto K$ when other parameters are fixed. Hence, increasing the dielectric constant ($K$) will increase the capacitance. Therefore, both Assertion (A) and Reason (R) are true, and Reason (R) correctly explains Assertion (A).

Key Concept: Free electron drift, electric field effect

c) $1.42 \times 10^{-15}$ s
[Solution Description]
The drift velocity $v_d$ is given by:
$v_d = \frac{eE\tau}{m}$
Here, $e$ is the electronic charge, $E$ is the electric field, $\tau$ is the relaxation time (time between collisions), and $m$ is the mass of the electron. Rearranging for $\tau$:
$\tau = \frac{v_d \cdot m}{eE}$
Substituting the given values:
$\tau = \frac{(1.25 \times 10^{-4}) \times (9.1 \times 10^{-31})}{(1.6 \times 10^{-19}) \times 0.5}$
$\tau = \frac{1.1375 \times 10^{-34}}{0.8 \times 10^{-19}}$
$\tau = 1.421875 \times 10^{-15} \text{ s}$

Key Concept: Capacitance with Dielectric Slab Inserted

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
When a dielectric slab is inserted between the plates of a parallel-plate capacitor, the capacitance increases because the dielectric reduces the effective electric field between the plates. This happens due to the polarization of the dielectric material, which aligns dipoles opposite to the applied electric field. The formula for the new capacitance $C$ is given by: $C = \frac{\varepsilon_0 A}{d - t + \frac{t}{K}}$, where $K$ is the dielectric constant. Since $K > 1$, the term $\frac{t}{K}$ is less than $t$, leading to an increase in capacitance compared to the case with no dielectric. Thus, both Assertion and Reason are true, and Reason correctly explains the Assertion.

Key Concept: Dielectric Strength Definition

c) The maximum electric field a dielectric can tolerate without breakdown
[Solution Description]
Dielectric strength is defined as the maximum value of the electric field that a dielectric material can tolerate without breaking down. It is the minimum potential gradient required to puncture a dielectric and is typically expressed in $kV/mm$.

Key Concept: Parallel Combination

d) 140 $\mu$C
[Solution Description]
For capacitors in parallel, the equivalent capacitance $C$ is given by:
$C = C_1 + C_2 + C_3.$
Substituting the given values: $C_1 = 2\, \mu\text{F}$, $C_2 = 4\, \mu\text{F}$, and $C_3 = 8\, \mu\text{F}$,
$C = 2 + 4 + 8 = 14\, \mu\text{F}.$
The total charge $q$ stored in the combination when connected to a 10 V battery is calculated using $q = CV$:
$q = 14\, \mu\text{F} \times 10\, \text{V} = 140\, \mu\text{C}.$

Key Concept: Bound Charges, Electric Field

a) $7.08 \times 10^{-8} \, \text{C/m}^2$
[Solution Description]
Bound charge density $\sigma_b$ is related to polarisation $P$ as $\sigma_b = P$.
Polarisation is:
$P = \epsilon_0 \chi_e E$
where $\chi_e = K - 1$ and $E = \frac{E_0}{K}$.
Substituting:
$P = \epsilon_0 (5 - 1) \left(\frac{10^4}{5}\right) = \epsilon_0 \times 4 \times 2000 = 8000 \epsilon_0$
Calculate $\sigma_b$:
$\sigma_b = 8000 \times 8.85 \times 10^{-12} = 7.08 \times 10^{-8} \, \text{C/m}^2$

Key Concept: Concept of Polarization in Dielectrics

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The Assertion states that a non-polar dielectric acquires an induced dipole moment when placed in an electric field, which is true because the electric field causes a displacement of charges within the molecules. The Reason explains this phenomenon by stating that the positive and negative charges experience opposing forces, leading to a separation of their centers of gravity. This separation creates a dipole moment aligned with the field. Thus, both the Assertion and Reason are true, and the Reason correctly explains the Assertion.

Key Concept: Energy Stored in a Charged Capacitor

b) 0.05 J
[Solution Description]
The energy stored in a charged capacitor is given by $U = \frac{1}{2} C V^2$.
Step 1: Convert the capacitance from microfarads (\$\mu F\$) to farads (F).
$C = 10 \mu F = 10 \times 10^{-6} F = 1 \times 10^{-5} F$
Step 2: Plug in the values into the formula.
$U = \frac{1}{2} \times 1 \times 10^{-5} F \times (100 V)^2$
$U = \frac{1}{2} \times 10^{-5} \times 10000$
$U = 0.05 J$

Key Concept: Capacitance with metallic slab

b) $118 \, \text{pF}$
[Solution Description]
For a metallic slab ($K \to \infty$), the capacitance formula simplifies to:
$C = \frac{\varepsilon_0 A}{d - t} = \frac{(8.85 \times 10^{-12})(200 \times 10^{-4})}{(2 - 0.5) \times 10^{-3}}$
$C = \frac{1.77 \times 10^{-13}}{1.5 \times 10^{-3}} = 1.18 \times 10^{-10} \, \text{F}$ or $118 \, \text{pF}$.

Key Concept: Energy density formula

c) $u = \frac{1}{2} \varepsilon_0 E^2$
[Solution Description]
The energy density (u) in an electric field is given by $u = \frac{1}{2} \varepsilon_0 E^2$ where $\varepsilon_0$ is the permittivity of free space and E is the electric field strength.

Key Concept: Energy Density, Electric Field

c) It becomes four times.
[Solution Description]
Step 1: Let initial energy density be $u_1 = \frac{1}{2} \varepsilon_0 E^2$.
Step 2: When the electric field is doubled, new energy density becomes $u_2 = \frac{1}{2} \varepsilon_0 (2E)^2 = 4 \times \frac{1}{2} \varepsilon_0 E^2 = 4u_1$.
Step 3: Since the volume is constant, the total stored energy is proportional to the energy density. Thus, the energy increases by a factor of 4.

Key Concept: Capacitance with Dielectric Slab

b) $\frac{K \varepsilon_0 A}{d}$
[Solution Description]
The capacitance $C$ of a parallel plate capacitor with a dielectric slab of thickness $t = d$ and dielectric constant $K$ is given by:
$C = \frac{K \varepsilon_0 A}{d}$
This is derived from the general formula for capacitance with a dielectric slab, where substituting $t = d$ simplifies the expression to $\frac{K \varepsilon_0 A}{d}$.

Key Concept: Effect of Dielectric

c) Capacitance increases.
[Solution Description]
If the battery remains connected while inserting the dielectric, the potential difference remains unchanged. The capacitance increases by a factor of $K$, where $K$ is the dielectric constant. Thus, the new capacitance is $C = K C_0$.

Key Concept: Polarization Charge

b) Polarization charges opposing $E_0$
[Solution Description]
The external field $E_0$ polarizes the dielectric, inducing polarization charges on its surfaces. These charges produce an opposing field $E'$, reducing the net field inside to $E = E_0 - E' = E_0 / K$.

Key Concept: Energy stored in a capacitor

b) $\frac{1}{2} Q V$
[Solution Description]
The electrostatic potential energy stored in a capacitor is given by:
$U = \frac{1}{2} Q V$
Here, $Q$ is the charge on the plates and $V$ is the potential difference between them. This energy represents the work done to move charge $Q$ across the potential difference $V$.
Therefore, the correct expression for the stored energy is:
$U = \frac{1}{2} Q V$

Key Concept: Energy Density, Electric Field, Parallel-Plate Capacitor

c) It remains the same
[Solution Description]
The electric field $E$ is given by $E = \frac{V}{d}$ initially. After separation, $V'$ changes but $Q$ remains constant. The new capacitance is $C' = \frac{\epsilon_0 A}{2d}$. The electric field $E'$ remains $\frac{Q}{\epsilon_0 A}$ because $E = \frac{\sigma}{\epsilon_0} = \frac{Q}{\epsilon_0 A}$, which is independent of $d$. Energy density $u = \frac{1}{2} \epsilon_0 E^2$ thus remains the same.

Key Concept: Real-world application, Unit conversion

c) 3555 \$\mu F\$
[Solution Description]
The capacitance of Earth in vacuum is:
$C_0 = 4 \pi \varepsilon_0 a = \frac{a}{k}, \text{ where } k = 9 \times 10^9$
Given $a = 6400 \times 10^3 \, m$, substituting:
$C_0 = \frac{6400 \times 10^3}{9 \times 10^9} = 711 \times 10^{-6} \, F = 711 \mu F$
In a medium with $K = 5$, the capacitance becomes:
$C_K = K C_0 = 5 \times 711 \mu F = 3555 \mu F$

Key Concept: Dependence of Capacitance on Area, Distance, Dielectric Medium

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
1. **Assertion Analysis**:
- Initial capacitance: $C_1 = \frac{\varepsilon_0 K A}{d}$.
- New area: $A' = 2A$.
- New distance: $d' = \frac{d}{2}$.
- Final capacitance: $C_2 = \frac{\varepsilon_0 K (2A)}{d/2} = \frac{2\varepsilon_0 K A}{d/2} = 4 \left(\frac{\varepsilon_0 K A}{d}\right) = 4C_1$.
Thus, the assertion is true as capacitance becomes four times the original.
2. **Reason Analysis**:
- The formula $C = \frac{\varepsilon_0 K A}{d}$ correctly shows that $C$ is directly proportional to $A$ and inversely proportional to $d$. When $A$ doubles and $d$ halves, the product indicates a four-fold increase. Therefore, the reason is correct and fully explains the assertion.
Conclusion: Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.

Key Concept: Free and Bound Charges Inside a Conductor, Capacitance

d) Assertion is false, but Reason is true.
[Solution Description]
Capacitance ($C$) of a conductor is defined as $C = \frac{Q}{V}$, where $Q$ is the charge and $V$ is the potential. The capacitance is independent of the charge $Q$ or potential $V$; it depends only on the geometry of the conductor and the surrounding medium. Free charges move under an external electric field, but they do not determine the capacitance. Thus, Assertion (A) is false. However, Reason (R) is true because when an external electric field is applied, free electrons drift against the field, leading to a redistribution of charges and a change in potential.

Key Concept: Units of capacitance

b) Both Assertion and Reason are true, but Reason is NOT the correct explanation of Assertion.
[Solution Description]
The Assertion states the dimensional formula of capacitance, which is indeed $\text{M}^{-1} \text{L}^{-2} \text{T}^4 \text{A}^2$. This is derived from the relationship between capacitance (farad), charge (coulomb), and voltage (volt). The Reason correctly defines the farad in terms of its electrical properties: 1 farad equals 1 coulomb per volt. However, while both statements are true, the Reason does not directly explain why the dimensional formula of capacitance is $\text{M}^{-1} \text{L}^{-2} \text{T}^4 \text{A}^2$. Therefore, the Reason is not the correct explanation of the Assertion.

Key Concept: Dependence of Capacitance on Plate Area

c) the area of the plates
[Solution Description]
The capacitance $C$ of a parallel-plate capacitor is given by the formula $C = \frac{A K \varepsilon_0}{d}$. From the formula, it is clear that capacitance is directly proportional to the area $A$ of the plates.

Key Concept: Dielectric constant, Capacitance, Multiple dielectrics

b) $\frac{K_1 + K_2}{2}$
[Solution Description]
The system can be considered as two capacitors in parallel. The capacitance of each part is $C_1 = \frac{K_1\epsilon_0 (A/2)}{d}$ and $C_2 = \frac{K_2\epsilon_0 (A/2)}{d}$. Total capacitance is $C = C_1 + C_2 = \frac{\epsilon_0 A}{2d}(K_1 + K_2)$. Comparing with $C = K_{eff}C_0 = K_{eff}\frac{\epsilon_0 A}{d}$, we get $K_{eff} = \frac{K_1 + K_2}{2}$.

Key Concept: Capacitance in Different Media

a) $1.67 \times 10^{-11} \, \text{F}$
[Solution Description]
The capacitance of a spherical conductor in a medium is:
$C_K = 4 \pi \epsilon_0 K a$
Here, $a = 5 \, \text{cm} = 0.05 \, \text{m}$ and $K = 3$. Substituting the values:
$C_K = 4 \pi (8.85 \times 10^{-12}) (3) (0.05) = 1.67 \times 10^{-11} \, \text{F}$

Key Concept: Capacitance, Charge Distribution

b) $16 \mu C$ on both $C_1$ and $C_2$
[Solution Description]
For capacitors in series, the equivalent capacitance $C_{eq}$ is given by $\frac{1}{C_{eq}} = \frac{1}{C_1} + \frac{1}{C_2} = \frac{1}{2} + \frac{1}{4} = \frac{3}{4}$. Thus, $C_{eq} = \frac{4}{3} \mu F$. The total charge $Q = C_{eq} V = \frac{4}{3} \times 12 = 16 \mu C$. Since the charge is the same for capacitors in series, both $C_1$ and $C_2$ have a charge of $16 \mu C$.

Key Concept: Units of Dielectric Strength

d) Kilovolts per millimeter (kV/mm)
[Solution Description]
Dielectric strength is expressed in kilovolts per millimeter (kV/mm).

Key Concept: Electric Field and Potential Difference with Dielectric

b) $\frac{E_0}{4}$
[Solution Description]
The electric field within a dielectric is reduced by the factor $K$. The formula is $E = \frac{E_0}{K}$. Given $K = 4$, we substitute to get $E = \frac{E_0}{4}$.

Key Concept: Drift velocity, current density

d) $1.25 \times 10^{-4}$ m/s
[Solution Description]
The current density $J$ is given by:
$J = n \cdot e \cdot v_d$
Where $n$ is the electron density, $e$ is the charge of an electron, and $v_d$ is the drift velocity. The current $I$ is related to current density by:
$I = J \cdot A$
Combining these equations:
$v_d = \frac{I}{n \cdot e \cdot A}$
Substituting the given values:
$v_d = \frac{1.6}{8 \times 10^{28} \times 1.6 \times 10^{-19} \times 10^{-6}}$
$v_d = \frac{1.6}{12.8 \times 10^{3}}$
$v_d = 1.25 \times 10^{-4} \text{ m/s}$

Key Concept: Energy density calculation

c) 0.177 J/m$^3$
[Solution Description]
First find capacitance: $C = \frac{\varepsilon_0 A}{d} = \frac{8.85 \times 10^{-12} \times 50 \times 10^{-4}}{1 \times 10^{-3}} = 4.425 \times 10^{-11}$ F
Then calculate energy stored: $U = \frac{1}{2}CV^2 = \frac{1}{2} \times 4.425 \times 10^{-11} \times (200)^2 = 8.85 \times 10^{-7}$ J
Volume between plates: $Ad = 50 \times 10^{-4} \times 1 \times 10^{-3} = 5 \times 10^{-6}$ m$^3$
Energy density: $u = \frac{U}{Ad} = \frac{8.85 \times 10^{-7}}{5 \times 10^{-6}} = 0.177$ J/m$^3$

Key Concept: Capacitance of Spherical Conductor

d) $3C$
[Solution Description]
The capacitance of an isolated spherical conductor in vacuum is given by $C = 4\pi\varepsilon_0 R$. When placed in a medium with dielectric constant $K$, the capacitance becomes:
$C_K = K \cdot C = 3 \cdot 4\pi\varepsilon_0 R = 3C$
Hence, the new capacitance is $3C$.

Key Concept: Electric polarization

d) Electrons are displaced slightly, creating induced dipoles
[Solution Description]
When an external electric field is applied to a dielectric, the electrons are not free but are displaced slightly from their equilibrium positions. This displacement creates induced dipoles, leading to electric polarization. The material remains non-conductive as the electrons are still bound to their atoms.

Key Concept: Dielectric effect

b) Decreases by factor K
[Solution Description]
From the syllabus, introduction of dielectric reduces the electric field by factor $K$:
$E = \frac{E_0}{K}$
This occurs because the capacitance increases while potential difference remains constant when battery stays connected

Key Concept: Bound charges, Electric field in dielectric

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
From the syllabus, the electric field within the dielectric slab is given by $E = \frac{Q}{K \epsilon_0 A}$, whereas the field in the absence of the dielectric is $E_0 = \frac{Q}{\epsilon_0 A}$. This shows that the field is reduced by a factor $K$. The reason is also true because the induced bound charges $-Q'$ and $+Q'$ on the dielectric surfaces generate an opposing electric field $E' = \frac{Q'}{\epsilon_0 A}$. Using $Q' = Q \left(1 - \frac{1}{K}\right)$, this opposing field reduces the net field in the dielectric. Thus, the Assertion is correct, and the Reason correctly explains it.

Key Concept: Dielectric Strength

c) 3 kV/mm
[Solution Description]
Dielectric strength is defined as the maximum electric field a material can endure without breaking down. For air at normal temperature and pressure (NTP), the dielectric strength is given as 3 kV/mm.

Key Concept: Energy Density

c) Electric field strength
[Solution Description]
From the syllabus, the energy density formula is:
$u = \frac{1}{2} \varepsilon_0 E^2$
So it depends only on the permittivity of free space ($\varepsilon_0$) and electric field strength ($E$).

Key Concept: Concept of conduction and polarization

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
Polar dielectrics are characterized by molecules that possess a permanent dipole moment due to the separation of centers of positive and negative charges by a finite distance. This is what makes them polar. Therefore, both the Assertion (A) and Reason (R) are true, and Reason correctly explains the Assertion.

Key Concept: Effect of dielectric on electric field, Potential energy

a) $\frac{1}{2}$
[Solution Description]
The original electric field is:
$E_0 = \frac{V_0}{d}$
After inserting the dielectric, the capacitance becomes:
$C = K C_0$
The charge remains the same since the battery is disconnected ($Q = C_0 V_0$). The new potential difference is:
$V = \frac{Q}{C} = \frac{C_0 V_0}{K C_0} = \frac{V_0}{K}$
The new electric field inside the dielectric is:
$E = \frac{V}{d} = \frac{V_0 / K}{d} = \frac{E_0}{K}$
Thus, the ratio is:
$\frac{E}{E_0} = \frac{1}{K} = \frac{1}{2}$

Key Concept: Bound charges and dielectric constant

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
According to the formula for bound charge, $Q' = Q \left(1 - \frac{1}{K}\right)$. Given $K = 2$, substituting the value gives:
$Q' = Q \left(1 - \frac{1}{2}\right) = Q \times \frac{1}{2}$
This means $Q'$ is indeed half of the original charge $Q$. Thus, the Assertion is true. The Reason is also correct and provides the exact formula that justifies the Assertion. Both statements are true, and the Reason correctly explains the Assertion.

Key Concept: Capacitance, Dielectric Slab

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
For a metallic slab ($K = \infty$), the capacitance formula simplifies to $C = \frac{\varepsilon_0 A}{d - t}$, indicating a direct reduction in effective plate separation. In contrast, for a dielectric slab ($K > 1$), the capacitance is given by $C = \frac{\varepsilon_0 A}{d - t + \frac{t}{K}}$, where $\frac{t}{K}$ is non-zero and finite. Thus, the Assertion is true because the capacitance increase is indeed more significant with a metallic slab. The Reason correctly explains this because the field inside the conductor vanishes, which is equivalent to eliminating the thickness $t$ from the separation, while a dielectric only reduces the field by a factor of $K$.

Key Concept: Area Calculation

a) 1130 km$^2$
[Solution Description]
Using the formula $A = \frac{Cd}{\varepsilon_0}$, where $C = 2$ F, $d = 0.5 \times 10^{-2}$ m, and $\varepsilon_0 = 8.85 \times 10^{-12}$ F/m:
$A = \frac{2 \times 0.5 \times 10^{-2}}{8.85 \times 10^{-12}} = 0.113 \times 10^{10} \text{ m}^2 \approx 1130 \text{ km}^2$

Key Concept: Polarization of Dielectrics, Non-polar vs Polar Dielectrics, Electric Field Reduction

c) Assertion is true, but Reason is false.
[Solution Description]
The assertion (A) is true because when a non-polar dielectric is placed in an external electric field, the positive and negative charges experience forces in opposite directions, leading to a small separation of charge centers and thus creating an induced dipole moment in the direction of the field.
However, the reason (R) is false because, even though the dipole moment aligns with the field, perfect alignment does not occur due to thermal agitation which causes random motion of molecules. Thus, while the assertion is correct, the reason incorrectly states that the alignment is perfect.

Key Concept: Charge Distribution in Series

b) $\frac{600}{11} \times 10^{-6} \, C$
[Solution Description]
First, find the equivalent capacitance for the series combination:
$\frac{1}{C} = \frac{1}{1} + \frac{1}{2} + \frac{1}{3} = \frac{6}{6} + \frac{3}{6} + \frac{2}{6} = \frac{11}{6}$
Thus,
$C = \frac{6}{11} \, \mu F$
The total charge $q$ supplied by the battery is:
$q = C \times V = \left( \frac{6}{11} \times 10^{-6} \right) \times 100 = \frac{600}{11} \times 10^{-6} \, C$
In a series combination, the charge on each capacitor is equal to the total charge $q$. So, each capacitor has:
$q = \frac{600}{11} \times 10^{-6} \, C$

Key Concept: Electric field, energy density, non-parallel plates

c) $0.177 \, \text{J/m}^3$
[Solution Description]
The average electric field $E_{\text{avg}}$ is the arithmetic mean of the minimum and maximum fields:
$E_{\text{avg}} = \frac{1 \times 10^5 + 3 \times 10^5}{2} = 2 \times 10^5 \, \text{V/m}$.
The average energy density $u_{\text{avg}}$ is then:
$u_{\text{avg}} = \frac{1}{2} \varepsilon_0 E_{\text{avg}}^2 = \frac{1}{2} \times 8.85 \times 10^{-12} \times (2 \times 10^5)^2$.
Calculate step-by-step:
$(2 \times 10^5)^2 = 4 \times 10^{10}$,
$8.85 \times 10^{-12} \times 4 \times 10^{10} = 0.354$,
$\frac{1}{2} \times 0.354 = 0.177 \, \text{J/m}^3$.

Key Concept: Electric Potential Energy

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The assertion (A) states that the electric potential energy stored in a capacitor is given by $U = \frac{1}{2}CV^2$, which is correct as derived from the work done in charging the capacitor.
The reason (R) states that the work done in charging the capacitor from $0$ to $Q$ is equal to the energy stored in the capacitor, which is also correct and directly explains why the assertion is true.
Therefore, both Assertion and Reason are true, and Reason is the correct explanation of Assertion.

Key Concept: Series Combination of Capacitors

a) $\frac{144}{11} \mu C$
[Solution Description]
For capacitors in series, the equivalent capacitance $C_{eq}$ is given by:
$\frac{1}{C_{eq}} = \frac{1}{2 \mu F} + \frac{1}{4 \mu F} + \frac{1}{6 \mu F}$
Simplify the equation:
$\frac{1}{C_{eq}} = \frac{6}{12 \mu F} + \frac{3}{12 \mu F} + \frac{2}{12 \mu F} = \frac{11}{12 \mu F}$
Therefore,
$C_{eq} = \frac{12 \mu F}{11}$
The charge $q$ stored on each capacitor is the same and is calculated as:
$q = C_{eq} V = \frac{12 \mu F}{11} \times 12 V = \frac{144}{11} \mu C$

Key Concept: Polarization in Dielectrics

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The assertion states that the electric field inside the dielectric reduces when a dielectric slab is inserted between the plates of a parallel-plate capacitor. This is true because the dielectric becomes polarized due to the external electric field, and the induced polarization charges create an opposing field ($E'$) that weakens the resultant field ($E$) inside the dielectric. The reason correctly explains that these polarization charges are responsible for reducing the net field inside the dielectric. Therefore, both the assertion and reason are true, and the reason is the correct explanation of the assertion.

Key Concept: Definition of Capacitance

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The assertion (A) states the definition of capacitance correctly: $C = Q/V$. The reason (R) explains why capacitance does not depend on charge or potential but on physical properties like size, shape, and the surrounding medium. Thus, both statements are true, and the reason correctly explains the assertion.

Key Concept: Dielectric constant definition

b) 2
[Solution Description]
The dielectric constant $K$ is defined as the ratio of the capacitance with the dielectric $C$ to the capacitance without the dielectric $C_0$. Here, $C = 2C_0$, so $K = \frac{C}{C_0} = \frac{2C_0}{C_0} = 2$.

Key Concept: Capacitance, Charge, Units

b) $1 nC$
[Solution Description]
Using the formula for capacitance:
$Q = C \times V$
Here, $C = 2 pF = 2 \times 10^{-12} F$ and $V = 500 V$. Substituting these values:
$Q = 2 \times 10^{-12} \times 500 = 10^{-9} C = 1 nC$

Key Concept: Effective capacitance, Metal slab

b) $88.5 \, \text{pF}$
[Solution Description]
For a metal slab ($K = \infty$), the capacitance formula reduces to:
$C = \frac{\varepsilon_0 A}{d - t}$
Given: $d = 8 \times 10^{-3} \, \text{m}$, $t = 3 \times 10^{-3} \, \text{m}$, $A = 0.05 \, \text{m}^2$.
First, verify $C_0$:
$C_0 = \frac{\varepsilon_0 A}{d} \implies \varepsilon_0 = \frac{C_0 d}{A} = \frac{55.3 \times 10^{-12} \times 8 \times 10^{-3}}{0.05} = 8.85 \times 10^{-12} \, \text{F/m}.$
Now compute $C$:
$C = \frac{8.85 \times 10^{-12} \times 0.05}{(8 - 3) \times 10^{-3}} = \frac{4.425 \times 10^{-13}}{5 \times 10^{-3}} = 88.5 \times 10^{-12} \, \text{F} = 88.5 \, \text{pF}.$

Key Concept: Electric Field in Dielectrics

c) $\frac{E_0}{4}$
[Solution Description]
The electric field inside a dielectric is reduced by a factor equal to the dielectric constant. The formula is $E = \frac{E_0}{K}$. Substituting $K = 4$, we get:
$E = \frac{E_0}{4}$

Key Concept: Non-polar Dielectrics, Polarisation

c) The molecules develop an induced dipole moment aligned with the field.
[Solution Description]
In a non-polar dielectric, the centres of positive and negative charges coincide initially. Under an external electric field, these charge centres separate, creating induced dipole moments aligned with the field. This results in polarisation.

Key Concept: Conductivity and free electrons

d) Free electrons drift against the direction of the applied electric field.
[Solution Description]
Free electrons in conductors are responsible for electrical conductivity as they can move freely under an external electric field. They are not bound to any particular atom but remain confined within the conductor unless external energy is supplied. The presence of these free electrons increases the conductivity of the material.

Key Concept: Polar and Non-polar Molecules

c) They have a permanent dipole moment due to charge separation
[Solution Description]
Polar molecules are those where the center of gravity of positive charges (protons) is separated from the center of gravity of negative charges (electrons) by a finite distance. This separation results in a permanent dipole moment.

Key Concept: Capacitance of a Parallel-Plate Capacitor with Dielectric Slab

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
When a dielectric slab is introduced between the plates of a parallel-plate capacitor, the molecules in the dielectric get polarized. This polarization creates an intrinsic electric field $\vec{E}_i$ that opposes the external electric field $\vec{E}_0$, reducing the net electric field inside the dielectric to $\vec{E} = \frac{\vec{E}_0}{K}$. The potential difference between the plates decreases because $V = E_0(d - t) + E_i t$, where $E_i = \frac{E_0}{K}$. As a result, the capacitance $C = \frac{Q}{V}$ increases because $V$ decreases for the same charge $Q$. Hence, both the Assertion and Reason are true, and the Reason correctly explains the Assertion.

Key Concept: Capacitance Definition

d) $C = \frac{Q}{V}$
[Solution Description]
The capacitance of a conductor is defined as the ratio of the charge $Q$ given to the conductor to the resulting rise in its potential $V$. Mathematically, it is expressed as $C = \frac{Q}{V}$.

Key Concept: Polarization Charges

c) On both faces of the slab, opposite to the adjacent plates.
[Solution Description]
When a dielectric slab is placed in an electric field between capacitor plates, polarization charges appear on the surfaces of the slab facing the plates. The charges are negative on the face near the positive plate and positive on the face near the negative plate. The net charge within the interior of the slab remains zero.

Key Concept: Capacitors in parallel, Equivalent capacitance

b) 13.17 $\mu F$
[Solution Description]
First calculate the equivalent capacitance for each branch:
1. For two $5 \mu F$ in series: $\frac{1}{C} = \frac{1}{5} + \frac{1}{5} = \frac{2}{5}$ $\Rightarrow C = 2.5 \mu F$
2. Single $10 \mu F$ remains as $10 \mu F$
3. For three $2 \mu F$ in series: $\frac{1}{C} = \frac{1}{2} + \frac{1}{2} + \frac{1}{2} = \frac{3}{2}$ $\Rightarrow C = \frac{2}{3} \mu F$
Now combine all branches in parallel:
$C_{eq} = 2.5 + 10 + \frac{2}{3} = 12.5 + 0.666... = 13.166... \mu F \approx 13.17 \mu F$

Key Concept: Electric Polarisation in Dielectrics

c) Assertion is true, but Reason is false.
[Solution Description]
The assertion is true because when a dielectric (polar or non-polar) is placed in an electric field, it becomes polarized, resulting in a net dipole moment in the direction of the field. However, the reason is false because even in polar dielectrics, perfect alignment does not occur due to thermal agitation of the molecules. The alignment improves with stronger fields or lower temperatures but is never perfect.

Key Concept: Capacitance

b) Farad
[Solution Description]
The SI unit of capacitance is the farad (F), defined as 1 coulomb per volt ($1 \, \text{F} = 1 \, \text{C/V}$).

Key Concept: Charge Carriers in Conductors

c) Free electrons
[Solution Description]
In metallic conductors, the charge carriers are free electrons. These electrons are loosely bound and can move freely within the metal structure, allowing the flow of electric current. The correct answer is 'Free electrons.'

Key Concept: Insulators

b) Rubber
[Solution Description]
Insulators do not allow electric charge to flow through them. Examples include glass, rubber, plastics, ebonite, mica, wax, paper, and dry wood. Among the given options, rubber is an insulator.

Key Concept: Dielectric constant

b) 1
[Solution Description]
The dielectric constant ($K$) of a vacuum is defined as 1 by definition. It serves as a reference for comparing other materials.

Key Concept: Force between plates

c) $\frac{1}{2} \frac{Q^2}{\varepsilon_0 A}$
[Solution Description]
From the syllabus, the force between plates is derived as:
$F = \frac{1}{2} \frac{Q^2}{\varepsilon_0 A}$
This is obtained by substituting $E = \frac{\sigma}{\varepsilon_0}$ and $\sigma = \frac{Q}{A}$ into $F = \frac{1}{2} Q E$

Key Concept: Electric Polarisation, Dielectric Strength

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The Assertion states that dielectric strength decreases with increasing temperature for polar dielectrics, which is true because higher thermal agitation reduces the effective alignment of dipoles, making it harder for the material to sustain high electric fields without breakdown. The Reason correctly explains this by pointing out that thermal agitation disrupts dipole alignment, necessitating a stronger external field for the same polarization. Thus, both Assertion and Reason are true, and Reason correctly explains the Assertion.

Key Concept: Polar vs. Non-Polar Molecules

d) H$_2$O
[Solution Description]
Polar molecules have a permanent dipole moment due to asymmetric charge distribution. H$_2$O is polar, while CO$_2$, CH$_4$, and N$_2$ are non-polar as their dipoles cancel out.

Key Concept: Multiple Dielectrics

c) $\frac{2 \varepsilon_0 A}{d}$
[Solution Description]
The total thickness of the dielectrics is $t_1 + t_2 = \frac{d}{4} + \frac{d}{2} = \frac{3d}{4}$. The remaining air gap is $d - \frac{3d}{4} = \frac{d}{4}$. The capacitance is given by:
$C = \frac{\varepsilon_0 A}{d - (t_1 + t_2) + \left( \frac{t_1}{K_1} + \frac{t_2}{K_2} \right)} = \frac{\varepsilon_0 A}{\frac{d}{4} + \left( \frac{d/4}{2} + \frac{d/2}{4} \right)} = \frac{\varepsilon_0 A}{\frac{d}{4} + \frac{d}{8} + \frac{d}{8}} = \frac{\varepsilon_0 A}{\frac{d}{2}} = \frac{2 \varepsilon_0 A}{d}$

Key Concept: Force between plates, Energy stored in capacitor

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The assertion states the correct formula for the force between the plates: $F = \frac{1}{2} \frac{Q^2}{\varepsilon_0 A}$. This is derived from the work-energy principle. The reason claims that the $\frac{1}{2}$ factor appears because the electric field acting on one plate is due solely to the other plate. This is true because although the total field between the plates is $E = \frac{\sigma}{\varepsilon_0}$, the field experienced by one plate is $\frac{E}{2} = \frac{\sigma}{2\varepsilon_0}$ due to the superposition principle. Thus, the reason correctly explains the assertion.

Key Concept: Bound Charges in Insulators

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
When an external electric field is applied to a dielectric, the charges within the dielectric experience a force, causing a slight displacement between the positive and negative charges. This displacement leads to the formation of induced dipole moments, resulting in polarization. The bound charges appear on the surfaces of the dielectric as a result of this polarization. Hence, both Assertion (A) and Reason (R) are true, and Reason (R) correctly explains Assertion (A).

Key Concept: Dielectric strength

b) Maximum electric field before breakdown
[Solution Description]
Dielectric strength is the maximum electric field a material can tolerate without breaking down, measured in kV/mm. For example, air at NTP has a dielectric strength of 3 kV/mm.

Key Concept: Series Combination of Capacitors

d) 1 $\mu F$
[Solution Description]
For a series combination, the equivalent capacitance $C$ is given by:
$\frac{1}{C} = \frac{1}{C_1} + \frac{1}{C_2} + \frac{1}{C_3}$
Substituting the values:
$\frac{1}{C} = \frac{1}{2} + \frac{1}{3} + \frac{1}{6} = \frac{3}{6} + \frac{2}{6} + \frac{1}{6} = \frac{6}{6} = 1$
Therefore, $C = 1 \mu F$.

Key Concept: Charge Distribution in Series

c) 32 pC
[Solution Description]
First, find the equivalent capacitance:
$\frac{1}{C} = \frac{1}{4} + \frac{1}{8} = \frac{2}{8} + \frac{1}{8} = \frac{3}{8}$
So,
$C = \frac{8}{3} \text{ pF}$
Next, calculate the total charge using $q = C \times V$:
$q = \left(\frac{8}{3} \times 10^{-12}\right) \times 12 = 32 \times 10^{-12} \text{ C} = 32 \text{ pC}$
Since the capacitors are in series, the charge on each capacitor is equal to the total charge:
$q = 32 \text{ pC}$
Each capacitor stores $32 \text{ pC}$ of charge.

Key Concept: Unit Conversion

b) 0.005 $\mu$F
[Solution Description]
We know that:
$1 \, pF = 10^{-12} \, F$
$1 \, \mu F = 10^{-6} \, F$
To convert 5000 pF to $\mu$F:
$5000 \, pF = 5000 \times 10^{-12} \, F$
Now, convert farads to microfarads:
$5000 \times 10^{-12} \, F = 5000 \times 10^{-12} \div 10^{-6} \, \mu F$
$= 5000 \times 10^{-6} \, \mu F$
$= 5 \times 10^{-3} \, \mu F$
$= 0.005 \, \mu F$
So, 5000 pF is equal to 0.005 $\mu$F.

Key Concept: Dimensional Analysis of Capacitance

b) $[M^{-1} L^{-2} T^4 A^2]$
[Solution Description]
The dimensional formula for capacitance is derived from:
$C = \frac{Q}{V}$
Where $Q$ is charge ($[A T]$) and $V$ is potential ($[M L^2 T^{-3} A^{-1}]$). Thus:
$[C] = \frac{[A T]}{[M L^2 T^{-3} A^{-1}]} = [M^{-1} L^{-2} T^4 A^2]$

Key Concept: Dielectric Constant, Capacitance

d) 40 $\mu F$
[Solution Description]
When a dielectric material with dielectric constant $K$ is inserted between the plates of a capacitor while it remains connected to the battery, the capacitance increases by a factor of $K$. The original capacitance in air is $C_0 = 10 \mu F$. The new capacitance $C$ when the dielectric is inserted is given by:
$C = K \times C_0 = 4 \times 10 \mu F = 40 \mu F.$

Key Concept: Capacitance and Distance

b) The capacitance increases to $20 \mu F$
[Solution Description]
The capacitance ($C$) of a parallel plate capacitor is inversely proportional to the distance ($d$) between the plates, given by $C \propto \frac{1}{d}$. Initially, the capacitance is $10 \mu F$ at $d = 1 mm$. When the distance is halved to $0.5 mm$, the capacitance doubles because $\frac{1}{d}$ becomes $\frac{1}{0.5d} = 2 \cdot \frac{1}{d}$. Therefore, the new capacitance is $2 \times 10 \mu F = 20 \mu F$.

Key Concept: Special case: metallic slab

c) $C = \infty$
[Solution Description]
For a metallic slab, the dielectric constant $K$ is infinity. Substituting $K = \infty$ and $t = d$ in the general formula $C = \frac{\epsilon_0 A}{d - t + \frac{t}{K}}$, the term $\frac{t}{K}$ becomes zero, leaving $C = \frac{\epsilon_0 A}{0}$. However, since $t = d$, the denominator becomes zero, resulting in infinite capacitance.

Key Concept: Conductors, Free Charges

b) Drift of free electrons against the field
[Solution Description]
In metals, the free electrons (conduction electrons) are the primary charge carriers. When an external electric field is applied, these free electrons drift against the direction of the field, resulting in a net current. The positive ions remain stationary in the lattice and do not contribute to the current.

Key Concept: Free and Bound Charges in a Conductor

c) Free electrons can move freely while bound charges remain fixed in their positions.
[Solution Description]
Free electrons are the valence electrons that are loosely bound to their parent atoms and move freely within the conductor when an electric field is applied. Bound charges, on the other hand, refer to the positive ions fixed in their lattice positions and cannot move. This makes option c the correct answer.

Key Concept: Units of Capacitance

c) $10^{-9} F$
[Solution Description]
From the syllabus, the smaller units of capacitance are defined as follows:
$1 \text{ microfarad} (1 \mu F) = 10^{-6} F$
$1 \text{ nanofarad} (1 nF) = 10^{-9} F$
$1 \text{ picofarad} (1 pF) = 10^{-12} F$.
Therefore, $1$ nanofarad equals $10^{-9} F$.

Key Concept: Force between the Plates of a Charged Parallel-Plate Capacitor

c) $\frac{Q^2}{2 A \varepsilon_0}$
[Solution Description]
The solution involves calculating the force using the formula derived from energy considerations.
Given:
- Charge on each plate = $Q$
- Area of each plate = $A$
- Permittivity of free space = $\varepsilon_0$
The force between the plates is given by:
$F = \frac{Q^2}{2 A \varepsilon_0}$
Step-by-step derivation:
1. From electrostatic potential energy, work done = $W = F d$.
2. This work is stored as energy in the capacitor: $U = \frac{1}{2} Q E d$.
3. Equating work and energy: $F d = \frac{1}{2} Q E d$ → $F = \frac{1}{2} Q E$.
4. Substituting $E = \frac{\sigma}{\varepsilon_0}$, where $\sigma = \frac{Q}{A}$:
$F = \frac{1}{2} Q \left( \frac{Q}{A \varepsilon_0} \right) = \frac{Q^2}{2 A \varepsilon_0}$.
Thus, the correct answer includes this expression.

Key Concept: Dielectric Constant

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The assertion states that the dielectric constant ($K$) of a material is always $\geq 1$. This is true because the dielectric constant is defined as $K = \frac{C}{C_0}$, where $C$ is the capacitance with dielectric and $C_0$ is the capacitance in vacuum. Introducing a dielectric between the plates of a capacitor increases its capacitance, so $C \geq C_0$, implying $K \geq 1$. The reason correctly explains why the dielectric constant is always $\geq 1$.

Key Concept: Definition of Capacitance

c) The ratio of charge given to a conductor to its potential change
[Solution Description]
The capacitance of a conductor is defined as the ratio of the charge given to the conductor to the rise in its potential. Mathematically, it is given by $C = Q/V$, where $Q$ is the charge and $V$ is the potential.

Key Concept: Series Combination of Capacitors

a) $\frac{12}{11} \, \mu F$
[Solution Description]
For capacitors in series, the equivalent capacitance $C$ is given by:
$\frac{1}{C} = \frac{1}{C_1} + \frac{1}{C_2} + \frac{1}{C_3}$
Substituting the given values:
$\frac{1}{C} = \frac{1}{2} + \frac{1}{4} + \frac{1}{6}$
Find a common denominator (12):
$\frac{1}{C} = \frac{6}{12} + \frac{3}{12} + \frac{2}{12} = \frac{11}{12}$
Therefore,
$C = \frac{12}{11} \, \mu F \approx 1.09 \, \mu F$

Key Concept: Dielectric strength, Breakdown

d) $10 \, \text{kV}$
[Solution Description]
The dielectric strength is the maximum electric field before breakdown occurs. Given the dielectric strength is $5 \, \text{kV/mm}$ and the plate separation is $2 \, \text{mm}$, the maximum potential difference $V$ is calculated as:
$V = \text{Electric field} \times \text{distance} = 5 \, \text{kV/mm} \times 2 \, \text{mm} = 10 \, \text{kV}$
Thus, the correct answer is $10 \, \text{kV}$.

Key Concept: Conduction, Polar Dielectric

c) Polarization increases but saturates due to competition between field alignment and thermal agitation.
[Solution Description]
In a polar dielectric, molecules have permanent dipole moments. In the absence of a field, these are randomly oriented due to thermal agitation. When an external field is applied, the dipoles tend to align with the field, but thermal motion opposes perfect alignment. As the field strength increases, the alignment improves, enhancing the net polarization. However, it never reaches perfect alignment due to residual thermal effects.