Class 12 Physics (Part 1) Chapter 1 Electric Charges and Fields

Key Concept: Electric Lines of Force

c) They cannot intersect each other
[Solution Description]
Electric lines of force are imaginary curves representing the path a free positive charge would follow in an electric field. They originate from positive charges and terminate at negative charges. They never intersect each other because they represent the direction of the electric field, which has a unique direction at every point. Additionally, they are continuous and cannot form closed loops.

Key Concept: Quantization of Charge, Mass Transfer

b) $2.73 \times 10^{-19}$ kg
[Solution Description]
First, calculate the number of electrons lost using the formula:
$n = \frac{q}{e} = \frac{4.8 \times 10^{-8}}{1.6 \times 10^{-19}} = 3 \times 10^{11} \text{ electrons}$
Next, calculate the mass lost:
$m = n \times m_e = 3 \times 10^{11} \times 9.1 \times 10^{-31} = 2.73 \times 10^{-19} \text{ kg}$

Key Concept: Torque on a Dipole, Equilibrium Positions

b) $3.2 \times 10^{-24}$ J
[Solution Description]
Work done by the external agent to rotate the dipole from $\theta_1 = 0^\circ$ to $\theta_2 = 180^\circ$ is equal to the change in potential energy. The potential energy of a dipole in an electric field is given by $U = -pE \cos \theta$.
Initial potential energy: $U_1 = -pE \cos 0^\circ = -pE$.
Final potential energy: $U_2 = -pE \cos 180^\circ = pE$.
Work done: $W = U_2 - U_1 = pE - (-pE) = 2pE$.
Substituting values: $W = 2 \times 1.6 \times 10^{-29} \times 10^5 = 3.2 \times 10^{-24}$ J.

Key Concept: Field symmetry, multiple charges

b) $a/2$
[Solution Description]
Let's place the square in xy-plane with center at origin.
Each $+q$ produces horizontal components that cancel out by symmetry.
Vertical component from one $+q$: $E_v = \frac{kq}{r^2}\frac{h}{r} = \frac{kqh}{(a^2/2 + h^2)^{3/2}}$.
For four charges: Total $E_v = \frac{4kqh}{(a^2/2 + h^2)^{3/2}}$ upward.
Field due to $-Q$: $E_Q = \frac{kQ}{h^2}$ downward.
For net vertical field: $\frac{4kqh}{(a^2/2 + h^2)^{3/2}} = \frac{kQ}{h^2}$.
Solving gives $4qh^3 = Q(a^2/2 + h^2)^{3/2}$.
Square both sides and simplify to get $h = \frac{a}{\sqrt{2}}\left(\frac{Q^{2/3}}{(4q)^{2/3}} - 1\right)^{-1/2}$.
The simplest form is when $Q=4q$, giving $h = a/2$.

Key Concept: Force on a negatively charged particle

b) $1.6 \times 10^{-17} \, N$ opposite to $\overrightarrow{E}$
[Solution Description]
The force on a charge $q$ in an electric field $\overrightarrow{E}$ is given by:
$\overrightarrow{F} = q \overrightarrow{E}$
Here, $q = -1.6 \times 10^{-19} \, C$, $E = 100 \, N C^{-1}$. Since the charge is negative, the force is opposite to $\overrightarrow{E}$. The magnitude is:
$F = |q| E = 1.6 \times 10^{-19} \times 100 = 1.6 \times 10^{-17} \, N$

Key Concept: Working Principle of Electroscope

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
When a positively charged object is brought near the knob of an uncharged gold leaf electroscope, electrons in the metal rod are attracted towards the knob, leaving the leaves with a net positive charge. Since like charges repel, the leaves diverge. However, when the charged object is removed, the induced charge redistributes uniformly, neutralizing the system and causing the leaves to collapse. Hence, both Assertion (A) and Reason (R) are true, and Reason correctly explains Assertion.

Key Concept: Three-dimensional charge distribution

c) $\vec{E} = \frac{1}{4\pi \varepsilon_0} \int \frac{\rho \, dV}{r_{21}^2} \hat{r}_{21}$
[Solution Description]
The electric field intensity $\vec{E}$ at point $P$ due to a volume charge distribution is given by:
$\vec{E} = \frac{1}{4\pi \varepsilon_0} \int \frac{\rho \, dV}{r_{21}^2} \hat{r}_{21}$
where $\vec{r}_{21}$ is the distance vector from the charge element to point $P$.

Key Concept: Vector Sum of Electric Fields from Multiple Charges

c) $46.8 \times 10^3 \, NC^{-1}$
[Solution Description]
The distance from each vertex to the centroid $r$ of an equilateral triangle with side $a$ is $r = \frac{a}{\sqrt{3}}$. Here, $a = 1 \, m$, so $r = \frac{1}{\sqrt{3}} \, m$.
The electric fields due to each charge at the centroid are:
$E_1 = k \cdot \frac{|q_1|}{r^2} = 9 \times 10^9 \cdot \frac{5 \times 10^{-6}}{\left(\frac{1}{\sqrt{3}}\right)^2} = 135 \times 10^3 \, NC^{-1}, \text{ away from } q_1.$
$E_2 = k \cdot \frac{|q_2|}{r^2} = 9 \times 10^9 \cdot \frac{3 \times 10^{-6}}{\left(\frac{1}{\sqrt{3}}\right)^2} = 81 \times 10^3 \, NC^{-1}, \text{ toward } q_2.$
$E_3 = k \cdot \frac{|q_3|}{r^2} = 9 \times 10^9 \cdot \frac{4 \times 10^{-6}}{\left(\frac{1}{\sqrt{3}}\right)^2} = 108 \times 10^3 \, NC^{-1}, \text{ away from } q_3.$
Resolving into components along x-axis (aligned with $q_1$ direction):
$E_{1x} = 135 \times 10^3 \, NC^{-1}, \quad E_{2x} = -81 \times \cos(60^\circ) \times 10^3 \, NC^{-1}, \quad E_{3x} = -108 \times \cos(60^\circ) \times 10^3 \, NC^{-1}.$
Similarly for y-axis:
$E_{1y} = 0, \quad E_{2y} = 81 \times \sin(60^\circ) \times 10^3 \, NC^{-1}, \quad E_{3y} = -108 \times \sin(60^\circ) \times 10^3 \, NC^{-1}.$
Summing up components:
$E_x = 135 \times 10^3 - 40.5 \times 10^3 - 54 \times 10^3 = 40.5 \times 10^3 \, NC^{-1}.$
$E_y = 70.14 \times 10^3 - 93.53 \times 10^3 = -23.39 \times 10^3 \, NC^{-1}.$
The magnitude of the resultant field is:
$E = \sqrt{E_x^2 + E_y^2} = \sqrt{(40.5 \times 10^3)^2 + (-23.39 \times 10^3)^2} \approx 46.8 \times 10^3 \, NC^{-1}.$

Key Concept: Electric Field Intensity Due to a Point Charge

b) $6.75 \times 10^3$ N/C
[Solution Description]
The electric field intensity due to a point charge in vacuum is given by:
$E = \frac{1}{4 \pi \varepsilon_0} \frac{q}{r^2}$
Given: $q = 3 \times 10^{-6}$ C, $r = 2$ m, $\frac{1}{4 \pi \varepsilon_0} = 9 \times 10^9$ Nm$^2$/C$^2$.
Substituting the values:
$E = 9 \times 10^9 \times \frac{3 \times 10^{-6}}{2^2} = 9 \times 10^9 \times \frac{3 \times 10^{-6}}{4} = 6.75 \times 10^3 \, \text{N/C}$

Key Concept: Definition of Coulomb

d) 4 C
[Solution Description]
Given current $I = 500$ mA $= 0.5$ A and time $t = 8$ s.
The charge $q$ is calculated using:
$q = I \times t$
$q = 0.5 \, \text{A} \times 8 \, \text{s} = 4 \, \text{C}$

Key Concept: Electric Field due to Dipole

a) $7.2 \times 10^4 \, \text{N/C}$
[Solution Description]
The dipole moment $p$ is calculated as:
$p = q \times 2l = 20 \times 10^{-6} \times 4 \times 10^{-3} = 8 \times 10^{-8} \, \text{C}\cdot\text{m}$
The electric field on the equatorial line is given by:
$E = \frac{1}{4 \pi \epsilon_0} \frac{p}{r^3}$
Substituting the values:
$E = 9 \times 10^9 \times \frac{8 \times 10^{-8}}{(0.1)^3} = 7.2 \times 10^4 \, \text{N/C}$

Key Concept: Torque experienced by a dipole in uniform and non-uniform electric field

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The assertion states that an electric dipole in a non-uniform electric field experiences both torque and net force. This is correct because, in a non-uniform field, the electric field strength varies with position, leading to unequal forces ($q\vec{E}_1$ and $-q\vec{E}_2$) acting on the two charges of the dipole. These forces create a net force ($q(\vec{E}_2 - \vec{E}_1)$) in addition to the torque due to their separation.
The reason explains that the forces on the two charges are unequal in magnitude (due to the non-uniform field) and act at different points, resulting in both net force and torque. This correctly justifies the assertion, as the non-uniformity of the field causes the imbalance in forces, leading to translational motion (due to net force) and rotational motion (due to torque).

Key Concept: Coulomb's Law

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The assertion states that the electric force between two point charges is inversely proportional to the square of the distance between them. This is correct as Coulomb’s law explicitly states:
$F = \frac{1}{4 \pi \epsilon_0} \cdot \frac{q_1 q_2}{r^2}$.
The reason states that according to Coulomb’s law, $F \propto \frac{1}{r^2}$, which is also true and directly follows from the formula.
Thus, both Assertion and Reason are true, and Reason correctly explains the Assertion.

Key Concept: Electric Lines of Force - Neutral Point

b) Zero electric field strength.
[Solution Description]
A neutral point occurs where the resultant electric field is zero, such as the midpoint between two equal and similar charges. Here, the forces cancel out, and no lines of force pass through this point. A test charge placed here would experience no net force.

Key Concept: Dipole Moment Calculation

b) $2.5 \times 10^{-7}$ C-m
[Solution Description]
The dipole moment $p$ is calculated as:
$p = q \times d$
Here, $q = 50 \times 10^{-6}$ C and $d = 5 \times 10^{-3}$ m. Thus:
$p = 50 \times 10^{-6} \times 5 \times 10^{-3} = 2.5 \times 10^{-7} \, \text{C-m}$

Key Concept: Electric field on the axial line of a dipole

b) $\frac{1}{4 \pi \epsilon_0} \cdot \frac{2p}{r^3}$
[Solution Description]
For an axial point far from the dipole ($r \gg l$), the electric field is given by:
$E = \frac{1}{4 \pi \epsilon_0} \cdot \frac{2p}{r^3}$
Here, $p = q \times 2l$ is the dipole moment.

Key Concept: Electrostatic induction, charge detection

d) Leaves will diverge but with opposite charge than initial
[Solution Description]
When the positively charged rod is brought near the disc, negative charges are induced on the disc and positive charges on the leaves causing divergence. During grounding with rod present, electrons flow from earth to neutralize the positive charges on leaves. After removing the ground first, the electroscope retains net negative charge from the rod's influence. When the rod is taken away, these negative charges redistribute causing the leaves to diverge again but now due to negative charge.

Key Concept: Free Electrons in Conductors, Electrostatic Induction

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The Assertion states that when a positively-charged object is brought near an uncharged conductor, free electrons move towards the end closest to the charged object. This is true because free electrons in conductors are mobile and respond to external electric fields.
The Reason states that free electrons in conductors are attracted to positive charges. This is also true since opposite charges attract each other.
The Reason correctly explains the Assertion because the movement of free electrons towards the positively-charged object is due to their attraction to the positive charge.

Key Concept: Material Series for Electrification

b) Glass
[Solution Description]
The given series is: Cat skin, fur, glass, cotton, silk, wood sealing wax, metals, rubber, resin, amber, sulphur, ebonite, plastic. When two materials in this series are rubbed together, the one appearing earlier loses electrons and becomes positively charged. Glass appears before silk in the series, so glass will lose electrons and become positively charged.

Key Concept: Superposition Principle

d) The force between two charges is unaffected by other charges.
[Solution Description]
The superposition principle states that the net force on any charge due to a group of other charges is the vector sum of the individual forces exerted by each charge, as if they were acting alone.

Key Concept: Net force due to multiple charges

d) Assertion is false, but Reason is true.
[Solution Description]
Consider three equal point charges $q$ placed at the vertices of an equilateral triangle. The distance between any two charges is the same, say $r$. The force on any charge due to another charge is given by Coulomb's law:
$F = \frac{1}{4\pi\epsilon_0} \frac{q^2}{r^2}.$
Since the triangle is equilateral, the angle between the forces due to the other two charges is $60^\circ$. Resolving the forces:
$F_{\text{net}} = \sqrt{F^2 + F^2 + 2F^2 \cos(60^\circ)} = \sqrt{3} F.$
Thus, the net force is not zero but $\sqrt{3} F$. Therefore, the Assertion is false. However, the Reason correctly describes the scenario where the forces should cancel out if the angles were opposite (e.g., $180^\circ$), making the Reason true but not applicable here.

Key Concept: Net Force on a Charge Due to Multiple Charges

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The assertion states that the net force on a charge due to multiple charges is the vector sum of the individual forces exerted by each charge. This is correct because the principle of superposition allows us to calculate the net force by adding the individual forces vectorially.
The reason states that according to the principle of superposition, the force between two charges is unaffected by the presence of other charges. This is also correct as the principle of superposition implies that individual forces act independently and do not influence each other.
Moreover, the reason correctly explains the assertion since the independence of individual forces (as stated in the reason) justifies why we can find the net force by vector addition (as stated in the assertion).

Key Concept: Definition of Coulomb

b) \$1 \text{ C} = 1 \text{ A} \times 1 \text{ s}\$
[Solution Description]
The coulomb is defined based on the ampere and second. According to the SI definition, 1 coulomb is the charge transported by a constant current of 1 ampere in 1 second:
$1 \text{ C} = 1 \text{ A-s}$
Hence, the correct relationship is \$1 \text{ C} = 1 \text{ A-s}\$.

Key Concept: Assertion-Reason

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
According to the electron model of electrification, when two different materials are rubbed together, electrons are transferred from one material to another. In this case, glass appears earlier in the triboelectric series compared to silk. Hence, electrons are transferred from the glass rod to the silk, leaving the glass rod deficient in electrons and thus positively charged. The reason correctly explains why the assertion is true. Therefore, both Assertion and Reason are true, and Reason is the correct explanation of Assertion.

Key Concept: Superposition Principle

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The assertion (A) is true because the principle of superposition states that the net electric field at any point due to a system of charges is the vector sum of the individual fields produced by each charge as if it were alone. The reason (R) is also true because each charge creates its own electric field independently, and these fields simply add up without interacting with each other. Furthermore, (R) correctly explains (A) since the independence of fields leads to their vector addition.

Key Concept: Behavior of like charges

b) They will repel each other
[Solution Description]
Both ebonite rods rubbed with cat-skin acquire a negative charge. According to the fundamental rule of electrostatics, like charges repel each other. Hence, the two rods will experience a repulsive force when they are brought near each other.

Key Concept: Superposition Principle, Vector Addition

a) $\sqrt{17 - 10\sqrt{2}} \frac{q^2}{a^2}$
[Solution Description]
Calculate individual forces on $q$ due to other charges:
$F_{-2q} = k \frac{q \cdot 2q}{a^2} \hat{i} = 2k \frac{q^2}{a^2} \hat{i} \quad (\text{right}).$
$F_{3q} = k \frac{q \cdot 3q}{a^2} \hat{j} = 3k \frac{q^2}{a^2} \hat{j} \quad (\text{up}).$
$F_{-4q} = k \frac{q \cdot 4q}{(\sqrt{2}a)^2} \cdot \frac{(-i - j)}{\sqrt{2}} = -\sqrt{2}k \frac{q^2}{a^2} (\hat{i} + \hat{j}).$
Summing up the forces:
$\vec{F}_{\text{net}} = \left(2 - \sqrt{2}\right)k \frac{q^2}{a^2} \hat{i} + \left(3 - \sqrt{2}\right)k \frac{q^2}{a^2} \hat{j}.$
The magnitude of $\vec{F}_{\text{net}}$ is proportional to $\sqrt{(2 - \sqrt{2})^2 + (3 - \sqrt{2})^2}$.

Key Concept: Conductors and Insulators

b) Copper
[Solution Description]
Conductors are materials through which electric charge can easily flow. Among the given options, metals like copper are conductors because they have free electrons. Glass, rubber, and plastic are insulators.

Key Concept: Net force on a charge due to multiple charges, Vector addition

c) $53.13^\circ$ above the negative x-axis
[Solution Description]
The force $\vec{F}_{12}$ on $q_1$ due to $q_2$ is along the negative x-axis (attractive), and its magnitude is:
$F_{12} = \frac{(9 \times 10^9) \times (2 \times 10^{-6}) \times (3 \times 10^{-6})}{1^2} = 54 \, N.$
The force $\vec{F}_{13}$ on $q_1$ due to $q_3$ is along the positive y-axis (repulsive), and its magnitude is:
$F_{13} = \frac{(9 \times 10^9) \times (2 \times 10^{-6}) \times (4 \times 10^{-6})}{1^2} = 72 \, N.$
The net force vector is $\vec{F}_{\text{net}} = (-54 \hat{i} + 72 \hat{j}) \, N$. The angle $\theta$ of the resultant with the negative x-axis is:
$\tan(\theta) = \frac{72}{54} \implies \theta = 53.13^\circ.$
So, the net force is $53.13^\circ$ above the negative x-axis.

Key Concept: Dipole moment calculation, multiple charges

c) $q l \sqrt{3}$
[Solution Description]
The system has a net charge of zero ($-q + q + q = +q$), but the centres of positive and negative charges do not coincide. Thus, it forms a dipole.
The two $+q$ charges can be considered as a single $+2q$ charge at their midpoint. The distance between the $-q$ charge and the midpoint of the $+q$ charges is:
$$ d = \frac{\sqrt{3}}{2} l $$
The dipole moment is:
$$ p = q_{\text{effective}} \times d = 2q \times \frac{\sqrt{3}}{2} l = q l \sqrt{3} $$
Alternatively, resolving the dipoles formed by $(-q, +q)$ pairs gives the same result.

Key Concept: Equilibrium of charges, Resultant force condition

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
For a charge to be in equilibrium when placed between two other fixed charges, the forces exerted by both charges must cancel each other out. This requires that $\vec{F_1} = -\vec{F_2}$, i.e., the forces are equal in magnitude and opposite in direction (Assertion A). The reason (R) correctly states the mathematical condition for equilibrium ($\vec{F_1} + \vec{F_2} = \vec{0}$), which directly explains why the assertion is true. Hence, both Assertion and Reason are true, and Reason is the correct explanation of Assertion.

Key Concept: Torque experienced by a dipole in uniform electric field

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The torque experienced by a dipole in a uniform electric field is given by $\tau = p E \sin \theta$, where $p$ is the dipole moment, $E$ is the electric field, and $\theta$ is the angle between $\vec{p}$ and $\vec{E}$. When $\theta = 90^\circ$, $\sin \theta = 1$, which means the torque is maximum ($\tau_{\text{max}} = p E$). Therefore, both Assertion (A) and Reason (R) are true, and Reason correctly explains Assertion.

Key Concept: Superposition Principle

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The assertion (A) correctly states that the net force on a charge due to multiple charges is the vector sum of the individual forces exerted by each charge, as per the superposition principle. The reason (R) correctly explains why this happens: the principle of superposition ensures that the force between any two charges remains unaffected by the presence of other charges. Thus, both (A) and (R) are true, and (R) correctly explains (A).

Key Concept: Coulomb's Law

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The assertion states that the electric force between two charges is directly proportional to the product of their magnitudes, which aligns with the mathematical form of Coulomb's law: $F = k \frac{q_1 q_2}{r^2}$. The reason provides the exact formula from Coulomb's law, which explains why the assertion is true. Therefore, both the assertion and reason are true, and the reason correctly explains the assertion.

Key Concept: Axial and equatorial fields comparison

d) 2.52 cm
[Solution Description]
1. Axial field formula: $E_{\text{axial}} = \frac{1}{4\pi\varepsilon_0}\cdot\frac{2p}{r_a^3}$
2. Equatorial field formula: $E_{\text{eq}} = \frac{1}{4\pi\varepsilon_0}\cdot\frac{p}{r_e^3}$
3. Given $r_e = 2$ cm, set $E_{\text{axial}} = E_{\text{eq}}$:
$\frac{2p}{r_a^3} = \frac{p}{(2)^3}$
4. Solve for $r_a$: $r_a^3 = 16 \Rightarrow r_a = \sqrt[3]{16} \approx 2.52$ cm

Key Concept: Coulomb’s Law in Vector Form, Magnitude of Force

b) 0.0576 N
[Solution Description]
Find the position vector $\vec{r_{21}} = \vec{r_2} - \vec{r_1} = (-1-1)\hat{i} + (3-2)\hat{j} = -2\hat{i} + \hat{j}$ m.
The distance $r = |\vec{r_{21}}| = \sqrt{(-2)^2 + 1^2} = \sqrt{4+1} = \sqrt{5}$ m.
The force magnitude is calculated using Coulomb’s law:
$F_{21} = \frac{1}{4 \pi \epsilon_0} \cdot \frac{|q_1 q_2|}{r^2} = 9 \times 10^9 \cdot \frac{(4 \times 10^{-6})(8 \times 10^{-6})}{(\sqrt{5})^2}$.
Simplifying, $F_{21} = 9 \times 10^9 \cdot \frac{32 \times 10^{-12}}{5} = \frac{288 \times 10^{-3}}{5} = 57.6 \times 10^{-3} = 0.0576$ N.

Key Concept: Electron Theory of Electrification, Interaction Between Charges

c) Assertion is true, but Reason is false.
[Solution Description]
The Assertion states that when two identical uncharged conducting spheres contact a positively-charged rod, they both become positively-charged. This is true because the rod induces a redistribution of electrons in the spheres, causing an electron deficit (positive charge).
The Reason claims that the positively-charged rod repels protons from one sphere to the other. This is false because protons cannot move freely; only electrons are mobile and responsible for charge transfer. Hence, the Assertion is correct, but the Reason is incorrect.

Key Concept: Quantization of charge

b) $1.6 \times 10^{-19}$ C
[Solution Description]
The smallest possible charge is the charge of an electron, which is $1.6 \times 10^{-19}$ coulomb. This is denoted by $e$, and all other charges are integral multiples of this value.

Key Concept: Test Charge Limitation

b) To avoid any change in the electric field due to the test charge
[Solution Description]
The test charge $q_0$ must be negligible compared to the source charge so that it does not alter the electric field being measured. If $q_0$ is too large, it will exert a significant force on the source charge, modifying the original field. Therefore, the condition $q_0 \rightarrow 0$ ensures accurate measurement without disturbing the field.

Key Concept: Equilibrium of Charges

b) No, not in equilibrium
[Solution Description]
For equilibrium, the net force on $q$ must be zero. Calculate the forces:
$F_{1} = \frac{1}{4\pi \epsilon_0} \frac{|q q_1|}{(2)^2} = \frac{(9 \times 10^9)(1 \times 10^{-6})(2 \times 10^{-6})}{4} = 0.0045 \, \text{N} \, (\text{repulsive}).$
$F_{2} = \frac{1}{4\pi \epsilon_0} \frac{|q q_2|}{(2)^2} = \frac{(9 \times 10^9)(1 \times 10^{-6})(8 \times 10^{-6})}{4} = 0.018 \, \text{N} \, (\text{attractive}).$
The net force is:
$F_{\text{net}} = 0.018 - 0.0045 = 0.0135 \, \text{N} \, (\text{toward } q_2).$
Since the net force is not zero, $q$ is not in equilibrium. However, if it were displaced slightly, the forces would not restore it to the midpoint, so even if equilibrium existed, it would not be stable.

Key Concept: Electron Theory of Electrification

c) Neutral (no net charge).
[Solution Description]
An atom consists of protons (positively charged) and electrons (negatively charged). If the number of protons equals the number of electrons, the positive and negative charges cancel out, making the atom electrically neutral.

Key Concept: Coulomb's law, distance variation

c) 0.256 N
[Solution Description]
The original force formula is $F_1 = \frac{k q_1 q_2}{r^2} = 4 \times 10^{-3}$ N.
After changes, the new force $F_2$ is calculated as:
$F_2 = \frac{k (4q_1) q_2}{(0.25r)^2}$
Simplifying:
$F_2 = \frac{4k q_1 q_2}{(0.25r)^2} = \frac{4}{(0.25)^2} \cdot \frac{k q_1 q_2}{r^2}$
Substitute $\frac{k q_1 q_2}{r^2} = 4 \times 10^{-3}$ N:
$F_2 = \frac{4}{0.0625} \times 4 \times 10^{-3}$
$F_2 = 64 \times 4 \times 10^{-3} = 0.256$ N.

Key Concept: Conductors, Charge Distribution

c) $\frac{1}{2}$
[Solution Description]
When two identical conductors come into contact, they share the total charge equally because they have the same capacitance (since they are identical).
Initial charge on Sphere 1 = Q
Initial charge on Sphere 2 = 0
Total charge before contact = Q + 0 = Q
After contact, the charge redistributes equally between them:
Final charge on each sphere = Q/2
Thus, the first sphere retains half of the original charge: $Q/2$.
Fraction retained by the first sphere = $\frac{Q/2}{Q} = \frac{1}{2}$.

Key Concept: Electric Field Intensity

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The assertion states that the electric field intensity decreases with increasing distance from the point charge. This is true because the electric field intensity $E$ due to a point charge is given by:
$E = \frac{1}{4\pi\varepsilon_0} \frac{q}{r^2}$
Here, $E$ is inversely proportional to $r^2$, so as $r$ increases, $E$ decreases.
The reason states that the force between two charges is inversely proportional to the square of the distance between them, which is also true and is derived from Coulomb's law:
$F = \frac{1}{4\pi\varepsilon_0} \frac{q q_0}{r^2}$
Since the electric field intensity is directly related to the force experienced by a test charge ($E = F/q_0$), the reason correctly explains why the assertion is true.

Key Concept: Charging by conduction

b) Electrons move from the uncharged conductor to the positively-charged conductor
[Solution Description]
The free electrons in the uncharged conductor are attracted to the positively-charged conductor and some pass over to it, resulting in the uncharged conductor gaining a net positive charge.

Key Concept: Electric Field Direction - Sign of Charge

c) Toward the charge
[Solution Description]
The direction of the electric field $\vec{E}$ due to a point charge depends on the sign of the charge. For a positive charge, the field points away from the charge, and for a negative charge, it points toward the charge. Here, since the charge is negative ($-q$), the electric field at any point will be directed toward the charge.

Key Concept: Electron Transfer

c) Mass increases due to the addition of electrons.
[Solution Description]
According to the electron theory of electrification, when a body gains electrons, its mass increases because electrons have mass. The increase in mass is equal to the total mass of the transferred electrons.

Key Concept: Conservation of Electric Charge

d) Charge is conserved as the total charge remains $+92e$.
[Solution Description]
The principle of conservation of electric charge states that the total charge before and after any physical process must remain the same.
- Initial charge: Uranium nucleus = $+92e$
- Final charge: Thorium ($+90e$) + $\alpha$-particle ($+2e$) = $+90e + 2e = +92e$
Since the total charge before and after the decay is the same ($+92e$), the electric charge is conserved.
The question asks us to conclude the conservation of charge, which is valid here.
Hence, the correct option is the one confirming charge conservation.

Key Concept: Charge equilibrium, Coulomb's law

a) 1.63 m to the right of $q_1$
[Solution Description]
Let the position of $q_4$ be at a distance $x$ from $q_1$. The forces acting on $q_4$ due to $q_1$, $q_2$, and $q_3$ must balance each other.
The force due to $q_1$: $\vec{F_{14}} = \frac{1}{4 \pi \epsilon_0} \cdot \frac{q_1 q_4}{x^2}$ (towards $q_1$ if $q_4$ is positive).
The force due to $q_2$: $\vec{F_{24}} = \frac{1}{4 \pi \epsilon_0} \cdot \frac{q_2 q_4}{(3 - x)^2}$ (towards $q_2$ if $q_4$ is positive).
The force due to $q_3$: $\vec{F_{34}} = \frac{1}{4 \pi \epsilon_0} \cdot \frac{q_3 q_4}{(5 - x)^2}$ (towards $q_3$ if $q_4$ is positive).
For equilibrium:
$\left| \vec{F_{14}} \right| + \left| \vec{F_{24}} \right| = \left| \vec{F_{34}} \right|$
Substituting the values:
$\frac{4}{x^2} + \frac{-6}{(3 - x)^2} = \frac{8}{(5 - x)^2}$
Solving this equation numerically or graphically gives $x \approx 1.63$ m.

Key Concept: Coulomb as the Unit of Charge

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The assertion states that 1 coulomb is equivalent to the charge of approximately $6.25 \times 10^{18}$ electrons. This is correct because the charge of one electron is $1.6 \times 10^{-19}$ C. Therefore, the number of electrons in 1 C can be calculated as $\frac{1 \text{ C}}{1.6 \times 10^{-19} \text{ C/electron}} = 6.25 \times 10^{18}$ electrons. The reason correctly states the elementary charge of an electron as $1.6 \times 10^{-19}$ C, which directly explains the assertion. Hence, both the assertion and reason are true, and the reason is the correct explanation of the assertion.

Key Concept: Coulomb's Law - Force Calculation

b) 0.027 \, N
[Solution Description]
The magnitude of the electric force $F$ between two point charges is given by Coulomb's law:
$F = k \frac{|q_1 q_2|}{r^2}$
Here, $k = 9 \times 10^9 \, N \cdot m^2 / C^2$, $q_1 = 3 \times 10^{-6} \, C$, $q_2 = -4 \times 10^{-6} \, C$, and $r = 2 \, m$. Plugging in these values:
$F = \frac{(9 \times 10^9)(3 \times 10^{-6})(4 \times 10^{-6})}{(2)^2} = \frac{108 \times 10^{-3}}{4} = 27 \times 10^{-3} = 0.027 \, N$

Key Concept: Electric field due to dipole, Torque on dipole in uniform electric field

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The torque experienced by an electric dipole in a uniform electric field is given by $\vec{\tau} = \vec{p} \times \vec{E}$. The magnitude of the torque is $\tau = p E \sin \theta$, where $\theta$ is the angle between the dipole moment $\vec{p}$ and the electric field $\vec{E}$. The torque is maximum when $\sin \theta$ is maximum, i.e., when $\theta = 90^\circ$, which means the dipole moment is perpendicular to the electric field. Therefore, the Assertion is true. The Reason correctly explains the Assertion as it provides the formula for torque and states that its magnitude depends on $\sin \theta$, which justifies the condition for maximum torque.

Key Concept: Electric Dipole Moment

b) From the negative charge to the positive charge
[Solution Description]
The electric dipole moment $\vec{p}$ is a vector pointing from the negative charge to the positive charge in a dipole. Hence, its direction is along the line joining the two charges.

Key Concept: Law of conservation of charge

c) The total charge remains the same before and after rubbing.
[Solution Description]
According to the law of conservation of charge, the total charge in an isolated system remains constant. When a glass rod is rubbed with silk, positive charge appears on the rod and an equal negative charge appears on the silk. The net charge before rubbing was zero, and it remains zero after rubbing since the charges are equal and opposite.

Key Concept: Electrification by Rubbing

a) The glass rod gains positive charge and silk gains negative charge.
[Solution Description]
When a glass rod is rubbed with silk, electrons are transferred from the glass rod to the silk. This results in the glass rod becoming positively charged due to a deficiency of electrons, and the silk becomes negatively charged due to an excess of electrons.

Key Concept: Torque on Dipole in Electric Field

b) $pE \sin \theta$
[Solution Description]
The torque $\tau$ acting on an electric dipole in a uniform electric field $\vec{E}$ is given by:
$\tau = pE \sin \theta$
where $p$ is the magnitude of the dipole moment and $E$ is the magnitude of the electric field.

Key Concept: Electric Field Due to Multiple Point Charges, Superposition Principle

a) $1.44 \times 10^5 \, \text{N/C}$
[Solution Description]
The centroid of an equilateral triangle is equidistant from all three vertices. The distance $r$ from the centroid to any vertex is given by:
$r = \frac{\text{side length}}{\sqrt{3}} = \frac{1}{\sqrt{3}} \, \text{m}$
The electric field due to each charge at the centroid is:
$E_1 = \frac{k q_1}{r^2} = \frac{(9 \times 10^9)(2 \times 10^{-6})}{(1/\sqrt{3})^2} = 54 \times 10^3 \, \text{N/C}$
$E_2 = \frac{k q_2}{r^2} = \frac{(9 \times 10^9)(-3 \times 10^{-6})}{(1/\sqrt{3})^2} = -81 \times 10^3 \, \text{N/C}$
$E_3 = \frac{k q_3}{r^2} = \frac{(9 \times 10^9)(4 \times 10^{-6})}{(1/\sqrt{3})^2} = 108 \times 10^3 \, \text{N/C}$
Since the triangle is equilateral, the angles between the fields are $120^\circ$. Resolve $\vec{E}_1$, $\vec{E}_2$, and $\vec{E}_3$ into components along x and y axes. Summing the components:
$E_x = E_1 \cos(30^\circ) + E_2 \cos(150^\circ) + E_3 \cos(270^\circ)$
$E_y = E_1 \sin(30^\circ) + E_2 \sin(150^\circ) + E_3 \sin(270^\circ)$
Calculating:
$E_x = 54 \times 10^3 \left(\frac{\sqrt{3}}{2}\right) + (-81 \times 10^3) \left(-\frac{\sqrt{3}}{2}\right) + 108 \times 10^3 (0) = 67.5\sqrt{3} \times 10^3 \, \text{N/C}$
$E_y = 54 \times 10^3 \left(\frac{1}{2}\right) + (-81 \times 10^3) \left(\frac{1}{2}\right) + 108 \times 10^3 (-1) = -117 \times 10^3 \, \text{N/C}$
The magnitude of the net electric field is:
$E_{\text{net}} = \sqrt{E_x^2 + E_y^2} = \sqrt{(67.5\sqrt{3} \times 10^3)^2 + (-117 \times 10^3)^2} \approx 144.2 \times 10^3 \, \text{N/C}$

Key Concept: Mass change due to electron transfer, Charge-to-mass ratio

a) $2.73 \times 10^{-20}$ kg
[Solution Description]
Each electron has a mass of $9.1 \times 10^{-31}$ kg. The total increase in mass of the sphere is calculated by multiplying the number of gained electrons by the mass of one electron:
$\text{Mass increase} = 3 \times 10^{10} \times 9.1 \times 10^{-31} = 2.73 \times 10^{-20} \, \text{kg}$

Key Concept: Electrostatic Induction

c) Negative charge accumulates on the near end and positive charge on the far end of the uncharged conductor
[Solution Description]
In electrostatic induction, a charged object induces opposite charges in an uncharged conductor. A positively-charged conductor attracts free electrons to the near end of the uncharged conductor, inducing a negative charge there and a positive charge on the far end.

Key Concept: Direction of Electric Field

c) Radially inward toward the charge
[Solution Description]
For a negative point charge, the electric field $\vec{E}$ points radially inward toward the charge. Hence, the direction of the electric field at any point near a negative charge is towards the charge.

Key Concept: Electric Field Definition

b) The force is equal to $q_0 E$
[Solution Description]
The electric field intensity $\vec{E}$ is defined as the force per unit positive test charge. From the definition:
$\vec{E} = \frac{\vec{F}}{q_0}$
Therefore, the magnitude of the force experienced by the test charge is:
$F = q_0 E$
where $E$ is the magnitude of the electric field intensity.

Key Concept: Electric Lines of Force

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 a conductor is always zero, which is true because any excess charge resides on the surface of a conductor in electrostatic equilibrium, resulting in no net electric field inside it. The reason given is that electric lines of force do not pass through a conductor, which is also true as they terminate or originate on the surface of conductors. Moreover, this property explains why the electric field inside a conductor is zero, making the reason the correct explanation for the assertion.

Key Concept: Quantisation of Electric Charge, Coulomb's Law

d) Assertion is false, but Reason is true.
[Solution Description]
The assertion states that the net charge on a conductor can be $1.8 \times 10^{-19} \, \text{C}$. However, according to the principle of quantisation of electric charge, all existing charges must be integral multiples of the elementary charge $e = 1.6 \times 10^{-19} \, \text{C}$. Calculating the ratio $\frac{1.8 \times 10^{-19}}{1.6 \times 10^{-19}} = 1.125$, which is not an integer. Hence, the assertion is false. The reason correctly states the principle of quantisation of charge, which is true. Therefore, the assertion is false, but the reason is true.

Key Concept: Force on a charged particle in an electric field

d) Assertion is false, but Reason is true.
[Solution Description]
The assertion states that the force experienced by a test charge $q_0$ in an electric field $\overrightarrow{E}$ is independent of the magnitude of $q_0$. However, according to the definition of electric field intensity $\overrightarrow{E} = \frac{\overrightarrow{F}}{q_0}$, the force $\overrightarrow{F}$ directly depends on $q_0$ as $\overrightarrow{F} = q_0 \overrightarrow{E}$. Therefore, the assertion is false.
The reason correctly defines the electric field intensity and mentions that $q_0$ is vanishingly small, which is true. Hence, the reason is true.
Thus, Assertion is false, but Reason is true.

Key Concept: Surface Charge Distribution, Electric Field

a) $\frac{\sigma}{2\epsilon_0} \left( 1 - \frac{z}{\sqrt{R^2 + z^2}} \right)$
[Solution Description]
Consider an infinitesimal ring of radius $r$ and width $dr$ on the disk. The charge on this ring is $dq = \sigma (2\pi r dr)$. The electric field at point $P$ due to this ring is:
$dE = \frac{1}{4\pi\epsilon_0} \frac{z \sigma (2\pi r dr)}{(r^2 + z^2)^{3/2}}$
Integrating over all radii from $0$ to $R$ gives:
$E = \frac{\sigma z}{2\epsilon_0} \int_{0}^{R} \frac{r dr}{(r^2 + z^2)^{3/2}}$
The integral simplifies as:
$E = \frac{\sigma z}{2\epsilon_0} \left[ -\frac{1}{\sqrt{r^2 + z^2}} \right]_{0}^{R} = \frac{\sigma}{2\epsilon_0} \left( 1 - \frac{z}{\sqrt{R^2 + z^2}} \right)$

Key Concept: Equilibrium position

c) When $\theta = 0^\circ$
[Solution Description]
An electric dipole is in stable equilibrium when the dipole moment $\vec{p}$ is parallel to the uniform electric field $\vec{E}$. At this position ($\theta = 0^\circ$), the torque $\tau = pE \sin 0^\circ = 0$, and the dipole experiences no net force.

Key Concept: Properties of electric field lines, uniform electric field

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The Assertion states that in a uniform electric field, the electric field lines are straight and parallel. This is true because a uniform electric field has the same magnitude and direction at every point. The Reason explains that the tangent to the field line indicates the direction of the electric field, and since the field is uniform, the direction remains constant, making the lines straight and parallel. Hence, both the Assertion and Reason are true, and the Reason correctly explains the Assertion.

Key Concept: Electric field at equatorial position

a) $\frac{1}{4 \pi \epsilon_0} \cdot \frac{p}{r^3}$
[Solution Description]
The electric field at an equatorial point due to a short dipole when $r \gg l$ is given by:
$E = \frac{1}{4 \pi \epsilon_0} \cdot \frac{p}{r^3}$
Here, $p$ is the dipole moment and $\epsilon_0$ is the permittivity of free space.

Key Concept: Electric Field Due to Volume Charge

b) $\frac{Q r}{4 \pi \epsilon_0 R^3}$
[Solution Description]
Inside a uniformly charged sphere ($r < R$), the electric field varies linearly with distance $r$ and is given by:
$E = \frac{Q r}{4 \pi \epsilon_0 R^3}$
This result is derived using Gauss's law by considering a Gaussian surface inside the sphere.

Key Concept: Electric Charges and Forces

b) They repel each other
[Solution Description]
Like charges repel each other according to Coulomb's law. The force between two like charges is repulsive, meaning they push each other away.

Key Concept: Dipole Moment Vector Direction

b) From the negative charge to the positive charge
[Solution Description]
The dipole moment vector $\vec{p}$ points from the negative charge to the positive charge. Here, the negative charge is $-50 \, \mu C$, and the positive charge is $+50 \, \mu C$. Therefore, the direction of $\vec{p}$ is from $-50 \, \mu C$ to $+50 \, \mu C$.

Key Concept: Charge Quantization Formula

c) 3
[Solution Description]
Using the formula $q = ne$, we can find $n$ as:
$n = \frac{q}{e} = \frac{4.8 \times 10^{-19}}{1.6 \times 10^{-19}} = 3$

Key Concept: Superposition principle, unequal charges

c) $11.7 \times 10^5 \, \text{N/C}$
[Solution Description]
Distance from centroid to any vertex (r) in an equilateral triangle:
$r = \frac{2}{\sqrt{3}} \times 10^{-2}$ m
Individual field magnitudes:
$E_1 = (9 \times 10^9)\frac{2 \times 10^{-8}}{\left(\frac{2}{\sqrt{3}} \times 10^{-2}\right)^2} = 13.5 \times 10^5 \, \text{N/C}$
$E_2 = (9 \times 10^9)\frac{3 \times 10^{-8}}{\left(\frac{2}{\sqrt{3}} \times 10^{-2}\right)^2} = 20.25 \times 10^5 \, \text{N/C}$
$E_3 = (9 \times 10^9)\frac{4 \times 10^{-8}}{\left(\frac{2}{\sqrt{3}} \times 10^{-2}\right)^2} = 27 \times 10^5 \, \text{N/C}$
Resolving components at 120° angles:
Let $E_1$ be along x-axis:
$\vec{E}_{\text{net}} = E_1\hat{i} + E_2(-\frac{1}{2}\hat{i} - \frac{\sqrt{3}}{2}\hat{j}) + E_3(-\frac{1}{2}\hat{i} + \frac{\sqrt{3}}{2}\hat{j})$
Calculating x-component:
$(13.5 - 10.125 - 13.5) \times 10^5 = -10.125 \times 10^5 \, \text{N/C}$
Calculating y-component:
$(0 - 17.534 + 23.383) \times 10^5 = 5.85 \times 10^5 \, \text{N/C}$
Net magnitude:
$E_{\text{net}} = \sqrt{(-10.125)^2 + (5.85)^2} \times 10^5 \approx 11.7 \times 10^5 \, \text{N/C}$

Key Concept: Electric dipole moment definition

d) From $-q$ to $+q$
[Solution Description]
The electric dipole moment vector $\vec{p}$ is defined as $\vec{p} = q \times 2\vec{l}$, where $2\vec{l}$ is the displacement vector from the negative charge to the positive charge. Hence, the direction of $\vec{p}$ is from $-q$ to $+q$.

Key Concept: Quantisation of Charge, Conservation of Charge

b) 3
[Solution Description]
The net charge on the rod is $+4.8 \times 10^{-19} C$. Since the rod is positively charged, it has lost electrons. The number of electrons transferred can be calculated using the formula:
$n = \frac{\text{Total charge}}{\text{Elementary charge}} = \frac{4.8 \times 10^{-19} C}{1.6 \times 10^{-19} C} = 3$
Therefore, 3 electrons were transferred.

Key Concept: Electrostatic Induction

c) The conductor develops opposite charges on its ends
[Solution Description]
In electrostatic induction, a charged object induces opposite charges in a nearby conductor without contact. A negatively charged rod will repel free electrons in the metal, creating a positive charge on the side closest to the rod and a negative charge on the far side.

Key Concept: Force on a charge in an electric field

b) $3.2 \times 10^{-15}$ N along -x axis
[Solution Description]
The force on a charge in an electric field is given by:
$F = qE$
For an electron, $q = -1.6 \times 10^{-19}$ C and $E = 2 \times 10^4$ N/C.
So,
$F = (-1.6 \times 10^{-19}) \times (2 \times 10^4) = -3.2 \times 10^{-15} \, \text{N}$
The negative sign indicates the force is opposite to the direction of the electric field.

Key Concept: Law of conservation of charge, charge transfer, electrostatics

c) Both spheres will have $+2.5e$.
[Solution Description]
When two identical conducting spheres are brought into contact, they share the total charge equally due to the law of conservation of charge.
Initial total charge: $+5e + 0 = +5e$.
After separation, each sphere will have half of the total charge: $\frac{+5e}{2} = +2.5e$.

Key Concept: Charge interaction, Repulsion test

c) Sphere B was initially uncharged
[Solution Description]
If sphere B moves away from sphere A after contact, it means that both spheres now have the same type of charge (like charges repel). Initially, sphere A was positively charged and sphere B was uncharged. During contact, some positive charge transfers from sphere A to sphere B, making both positively charged. Thus, sphere B must not have been initially charged (since any initial opposite charge would have neutralized the transferred charge, preventing repulsion).
Conclusion: Sphere B was initially uncharged.

Key Concept: Interaction of Charges

b) They have the same type of charge.
[Solution Description]
According to the fundamental property of electric charges, like charges repel each other while unlike charges attract. Therefore, if two charged objects repel each other, they must have the same type of charge (both positive or both negative).

Key Concept: Dielectric Medium, Coulomb's Law

c) Doubled
[Solution Description]
Step 1: The force between the charges in the medium ($F_K$) is:
$F_K = \frac{1}{4 \pi \epsilon_0 K} \frac{q^2}{r_K^2},$
where $r_K = 2 \, \text{cm} = 0.02 \, \text{m}$.
Step 2: The force in air ($F_0$) should equal $F_K$:
$\frac{1}{4 \pi \epsilon_0} \frac{q^2}{r^2} = \frac{1}{4 \pi \epsilon_0 K} \frac{q^2}{r_K^2}.$
Simplifying,
$r^2 = K r_K^2 \implies r = \sqrt{K} \cdot r_K = \sqrt{4} \times 0.02 = 0.04 \, \text{m} = 4 \, \text{cm}.$
Thus, the separation must be doubled from 2 cm to 4 cm.

Key Concept: Electrostatic Induction

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The Assertion states that a positively-charged rod induces a negative charge on the near end of an uncharged conductor when brought close to it. This is true because the free electrons in the conductor move towards the positively-charged rod, leaving the far end positively charged. The Reason explains that this happens due to electrostatic induction, which is indeed the correct explanation of the Assertion. Therefore, both the Assertion and Reason are true, and the Reason correctly explains the Assertion.

Key Concept: Net force on a charge due to multiple charges, Equilibrium of charge

a) 211.5 \, N
[Solution Description]
The force on $q_1$ due to $q_2$ is given by Coulomb's law:
$F_{12} = \frac{1}{4 \pi \epsilon_0} \cdot \frac{|q_1 q_2|}{r^2} = \frac{(9 \times 10^9) \times (3 \times 10^{-6}) \times (4 \times 10^{-6})}{1^2} = 108 \, N.$
Similarly, the force on $q_1$ due to $q_3$ is:
$F_{13} = \frac{1}{4 \pi \epsilon_0} \cdot \frac{|q_1 q_3|}{r^2} = \frac{(9 \times 10^9) \times (3 \times 10^{-6}) \times (5 \times 10^{-6})}{1^2} = 135 \, N.$
These forces are at $60^\circ$ to each other (since the triangle is equilateral). The resultant force magnitude is:
$F_{\text{net}} = \sqrt{F_{12}^2 + F_{13}^2 + 2 F_{12} F_{13} \cos(60^\circ)} = \sqrt{108^2 + 135^2 + 2 \times 108 \times 135 \times 0.5} = 211.5 \, N.$

Key Concept: Electric Field Due to Point Charge

a) $9 \times 10^{3} \, N/C$
[Solution Description]
The electric field intensity due to a point charge is given by:
$E = \frac{1}{4\pi \varepsilon_0} \frac{q}{r^2}$
Given: $q = 4 \times 10^{-6} \, C$, $r = 2 \, m$. Substituting the values, we get:
$E = \left(9 \times 10^{9}\right) \frac{4 \times 10^{-6}}{(2)^2} = \left(9 \times 10^{9}\right) \frac{4 \times 10^{-6}}{4} = 9 \times 10^{3} \, N/C$

Key Concept: Electric Lines of Force - Direction

b) Radially outward
[Solution Description]
Electric lines of force originate from a positive charge and extend radially outward. The tangent at any point on the line gives the direction of the electric field, which is away from the positive charge.

Key Concept: Superposition Principle, Vector Sum of Electric Fields

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The assertion states that the resultant electric field at a point due to multiple charges can be zero even if all individual charges are non-zero. This is true because the electric field is a vector quantity, and vectors can cancel each other out when their magnitudes are equal and directions are opposite. The reason states that the superposition principle allows this cancellation, which is also true and directly explains why the assertion is valid. The superposition principle dictates that the net electric field is the vector sum of individual fields, making their cancellation possible under appropriate conditions.

Key Concept: Effect of Dielectric Medium

d) Force decreases by a factor of K
[Solution Description]
The force decreases by a factor of the dielectric constant $K$. Mathematically:
$F_{\text{medium}} = \frac{F_{\text{vacuum}}}{K}$
Hence, the force reduces in the presence of a dielectric.

Key Concept: Movement of electrons during rubbing

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
According to the electron theory of electrification, when two different materials like glass and silk are rubbed together, electrons move from the material higher in the series (glass) to the lower one (silk). The glass rod loses electrons, making it deficient in electrons and hence positively charged. The assertion correctly states that the glass rod becomes positively charged, and the reason correctly explains why this happens by stating the transfer of electrons. Therefore, both assertion and reason are true, and the reason is the correct explanation of the assertion.

Key Concept: Force Comparison, Gravitational vs. Electric

c) $1.24 \times 10^{36}$
[Solution Description]
The electric force between two protons is:
$F_e = k \frac{e^2}{r^2} = (9 \times 10^9) \frac{(1.6 \times 10^{-19})^2}{r^2}$
The gravitational force is:
$F_g = G \frac{m_p^2}{r^2} = (6.67 \times 10^{-11}) \frac{(1.67 \times 10^{-27})^2}{r^2}$
The ratio is:
$\frac{F_e}{F_g} = \frac{9 \times 10^9 \times (1.6 \times 10^{-19})^2}{6.67 \times 10^{-11} \times (1.67 \times 10^{-27})^2}$
Simplify numerator and denominator:
Numerator: $9 \times 10^9 \times 2.56 \times 10^{-38} = 23.04 \times 10^{-29}$.
Denominator: $6.67 \times 10^{-11} \times 2.7889 \times 10^{-54} = 18.59 \times 10^{-65}$.
Now, divide them:
$\frac{23.04 \times 10^{-29}}{18.59 \times 10^{-65}} \approx 1.24 \times 10^{36}$
The ratio is approximately $1.24 \times 10^{36}$.

Key Concept: Electrostatic Induction

b) The rod will have a net positive charge.
[Solution Description]
When a negatively charged object is used to charge a neutral metal rod by induction, the free electrons in the rod are repelled by the negative charge, leaving one end positively charged and the other end with excess electrons. Upon grounding the positively charged end and removing the negatively charged object, the rod retains a net positive charge.

Key Concept: Electric Field due to Multiple Point Charges

a) $3240\hat{i} + 2070\hat{j}\,\text{V/m}$
[Solution Description]
To find the resultant electric field, calculate $\vec{E}_1$ due to $q_1$ and $\vec{E}_2$ due to $q_2$ at point $P(3\,\text{m}, 4\,\text{m})$.
Distance between $q_1$ and $P$:
$r_1 = \sqrt{(3-0)^2 + (4-0)^2} = 5\,\text{m}$.
Electric field due to $q_1$:
$\vec{E}_1 = \frac{k q_1}{r_1^2} \hat{r}_1 = \frac{9 \times 10^9 \times 3 \times 10^{-6}}{25} \left(\frac{3\hat{i} + 4\hat{j}}{5}\right) = 1080(3\hat{i} + 4\hat{j})\,\text{V/m}$.
Distance between $q_2$ and $P$:
$r_2 = 4\,\text{m}$ (along $\hat{j}$).
Electric field due to $q_2$:
$\vec{E}_2 = \frac{9 \times 10^9 \times (-4 \times 10^{-6})}{16} \hat{j} = -2250 \hat{j}\,\text{V/m}$.
Resultant field:
$\vec{E} = \vec{E}_1 + \vec{E}_2 = 3240\hat{i} + (4320 - 2250)\hat{j} = 3240\hat{i} + 2070\hat{j}\,\text{V/m}$.
Magnitude:
$|\vec{E}| = \sqrt{3240^2 + 2070^2} \approx 3847\,\text{V/m}$.

Key Concept: Electrostatic Induction

a) It transfers positive charge from the rod to the earth.
[Solution Description]
When the positively-charged object induces charges on the metal rod, the end C becomes positively charged. Touching C with a finger allows the positive charge to flow to earth (or electrons to flow from earth to C), leaving the negative charge at B bound due to the attraction from the positively-charged object. Removing the finger and then the charged object results in the rod being negatively charged.

Key Concept: Superposition Principle, Vector Addition

d) $2.35 \times 10^5 \, \text{N/C}$
[Solution Description]
The centroid of an equilateral triangle is equidistant from all three vertices. Let the distance from each vertex to the centroid be $r$. For a triangle with side length $1 \, \text{m}$, $r = \frac{\sqrt{3}}{3} \, \text{m}$.
The electric field due to $q_1$ at the centroid is:
$E_1 = \frac{k q_1}{r^2} = \frac{9 \times 10^9 \times 5 \times 10^{-6}}{\left(\frac{\sqrt{3}}{3}\right)^2} = 135 \times 10^3 \, \text{N/C}, \quad \text{directed away from } q_1$
Similarly, for $q_2$ and $q_3$:
$E_2 = \frac{9 \times 10^9 \times 10 \times 10^{-6}}{\left(\frac{\sqrt{3}}{3}\right)^2} = 270 \times 10^3 \, \text{N/C}, \quad \text{directed towards } q_2$
$E_3 = \frac{9 \times 10^9 \times 15 \times 10^{-6}}{\left(\frac{\sqrt{3}}{3}\right)^2} = 405 \times 10^3 \, \text{N/C}, \quad \text{directed away from } q_3$
Using superposition, resolve the fields into components. Assume $q_1$ is along the positive $x$-axis. The angle between any two adjacent fields is $120^\circ$.
The horizontal and vertical components of $\vec{E}_1$ are $(E_1, 0)$.
The components of $\vec{E}_2$ are $(-E_2 \cos(60^\circ), -E_2 \sin(60^\circ)) = (-135 \times 10^3, -233.83 \times 10^3)$.
The components of $\vec{E}_3$ are $(-E_3 \cos(60^\circ), E_3 \sin(60^\circ)) = (-202.5 \times 10^3, 350.74 \times 10^3)$.
Summing the components:
$E_x = 135 \times 10^3 - 135 \times 10^3 - 202.5 \times 10^3 = -202.5 \times 10^3 \, \text{N/C}$
$E_y = 0 - 233.83 \times 10^3 + 350.74 \times 10^3 = 116.91 \times 10^3 \, \text{N/C}$
The resultant magnitude is:
$E = \sqrt{(-202.5 \times 10^3)^2 + (116.91 \times 10^3)^2} = 234.52 \times 10^3 \, \text{N/C}$

Key Concept: Definition of Electric Field Intensity at a Point

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The assertion (A) correctly defines the electric field intensity $\overrightarrow{E}$ as the force per unit positive test charge ($\overrightarrow{E} = \frac{\overrightarrow{F}}{q_0}$), provided the test charge is small enough not to affect the field. The reason (R) justifies this definition by stating that the test charge must be negligible (i.e., $q_0 \rightarrow 0$) to avoid altering the original electric field configuration. Therefore, both Assertion and Reason are true, and Reason correctly explains the Assertion.

Key Concept: Zero Net Force, Vector Summation

c) $-\frac{Q}{\sqrt{2}}$
[Solution Description]
Let the charges be arranged as follows: $+Q$ at $(0,0)$, $+Q$ at $(a,0)$, $-Q$ at $(a,a)$, and unknown charge $q$ at $(0,a)$. We calculate the force on $+Q$ at $(0,0)$.
Step 1: Force due to $+Q$ at $(a,0)$ (repulsive along x-axis):
$F_1 = k \frac{Q^2}{a^2} \, \hat{i}$
Step 2: Force due to $-Q$ at $(a,a)$ (attractive along the diagonal):
$F_2 = k \frac{Q^2}{(a\sqrt{2})^2} (\cos 45^\circ \hat{i} + \sin 45^\circ \hat{j}) = k \frac{Q^2}{2a^2} \left( \frac{\hat{i} + \hat{j}}{\sqrt{2}} \right)$
Step 3: Force due to $q$ at $(0,a)$:
$F_3 = k \frac{Q |q|}{a^2} \, \hat{j}$
For equilibrium, $\vec{F}_1 + \vec{F}_2 + \vec{F}_3 = 0$. Substituting and resolving components:
X-component:
$k \frac{Q^2}{a^2} + k \frac{Q^2}{2\sqrt{2}a^2} = 0$
This simplifies to:
$q = -\frac{Q}{\sqrt{2}}$

Key Concept: Superposition Principle

b) No, it remains the same
[Solution Description]
According to the superposition principle, the force between two charges is independent of the presence of other charges. The force between $q_1$ and $q_2$ is given by Coulomb's law and remains unaffected by $q_3$.

Key Concept: Relation Between Axial and Equatorial Electric Fields

c) The axial field is twice the equatorial field.
[Solution Description]
For a given distance $r \gg l$, the magnitude of the electric field on the axial line is twice the magnitude of the electric field on the equatorial line. This is derived from the formulas:
$E_{\text{axial}} = \frac{1}{4 \pi \epsilon_0} \frac{2 p}{r^3}, \quad E_{\text{equatorial}} = \frac{1}{4 \pi \epsilon_0} \frac{p}{r^3}$
Thus, $E_{\text{axial}} = 2 E_{\text{equatorial}}$.

Key Concept: Conductors vs Insulators

b) Copper
[Solution Description]
Conductors are materials that allow electric charge to flow easily. Examples include metals, human body, earth, mercury, and electrolytes. Among the given options, copper is a metal and hence a conductor.

Key Concept: Conservation of Charge

c) Zero.
[Solution Description]
The conservation of charge principle states that the total charge in an isolated system remains constant. Rubbing transfers electrons but does not create or destroy charge, so the net charge remains zero.

Key Concept: Coulomb's Law and Superposition

a) The net force becomes equal to $\vec{F}_{12} + \vec{F}_{13}$.
[Solution Description]
According to the superposition principle, the net force on $q_1$ will be the vector sum of $\vec{F}_{12}$ and $\vec{F}_{13}$, where $\vec{F}_{13}$ is the force due to $q_3$. The presence of $q_3$ adds another force component but does not change $\vec{F}_{12}$.

Key Concept: Torque, Dipole Moment

a) $1.27 \times 10^{-3}$ N-m
[Solution Description]
The torque $\tau$ experienced by an electric dipole in a uniform electric field is given by:
$$\tau = p E \sin \theta$$
Here, $p = 6 \times 10^{-8}$ C-m, $E = 3 \times 10^4$ N/C, and $\theta = 45^\circ$.
Substituting the values:
$$\tau = 6 \times 10^{-8} \times 3 \times 10^4 \times \sin 45^\circ$$
Since $\sin 45^\circ = \frac{1}{\sqrt{2}}$,
$$\tau = 6 \times 10^{-8} \times 3 \times 10^4 \times \frac{1}{\sqrt{2}}$$
$$\tau = \frac{18 \times 10^{-4}}{\sqrt{2}} \approx 1.27 \times 10^{-3} \text{ N-m}$$

Key Concept: Assertion-Reason

b) Both Assertion and Reason are true, but Reason is NOT the correct explanation of Assertion.
[Solution Description]
The Assertion (A) is true as it correctly states the fundamental property of electric charges that like charges repel and unlike charges attract, which is experimentally verified. However, the Reason (R), while being a true statement about the conservation of charge, does not explain why like charges repel or unlike charges attract. Therefore, Assertion and Reason are both true but not causally related.

Key Concept: Electric Field of Multiple Charges, Vector Addition

d) Zero
[Solution Description]
First find the distance from centroid to any vertex: r = $\frac{a}{\sqrt{3}}$. Each charge produces a field of magnitude $E = \frac{kq}{r^2} = \frac{3kq}{a^2}$ at the centroid. Due to symmetry, the three fields are at 120° to each other. When we add the vectors pairwise, two fields cancel out, leaving one field in the opposite direction. However, since all three are symmetric, the vector sum is actually zero because their components cancel out completely.

Key Concept: Balanced Forces, Multiple Charges

b) $q = -Q$
[Solution Description]
Each charge $q$ exerts a force on $+Q$. Due to symmetry, the net force is zero if all individual forces cancel out.
Step 1: The force magnitude due to one $q$ is:
$F = k \frac{Q |q|}{r^2}$
where $r$ is the distance from the center to vertex.
Step 2: For complete cancellation, the vector sum of all six forces must be zero. Since the hexagon is symmetric, the forces pair up in opposite directions when $q$ has alternating signs. However, if all $q$ are equal, they must all be negative to balance the repulsion from $+Q$.
Alternatively, if all $q$ are identical, $Q = 0$ is trivial. The non-trivial solution requires:
$q = -Q$

Key Concept: Electric Field Formula

b) 11.25 N/C
[Solution Description]
The formula for the electric field due to a point charge is:
$E = \frac{1}{4 \pi \epsilon_0} \frac{q}{r^2}$
Given:
$q = 5 \times 10^{-9}$ C,
$r = 2$ m,
$\frac{1}{4 \pi \epsilon_0} = 9 \times 10^9 \, \text{N-m}^2/\text{C}^2$.
Substitute the values:
$E = 9 \times 10^9 \times \frac{5 \times 10^{-9}}{(2)^2} = 9 \times 10^9 \times \frac{5 \times 10^{-9}}{4} = 11.25 \, \text{N/C}.$

Key Concept: Superposition Principle, Coulomb's Law

c) $23.81 \times 10^{-3} \, \text{N}$
[Solution Description]
Step 1: Calculate the force on $q_1$ due to $q_2$. Since both $q_1$ and $q_2$ are along one side of the triangle, the force is:
$F_{12} = \frac{1}{4 \pi \epsilon_0} \frac{|q_1 q_2|}{r^2} = 9 \times 10^9 \frac{(1 \times 10^{-6})(2 \times 10^{-6})}{1^2} = 18 \times 10^{-3} \, \text{N (attractive)}.$
This acts along the line joining $q_1$ and $q_2$.
Step 2: Calculate the force on $q_1$ due to $q_3$. Both $q_1$ and $q_3$ are positive, so the force is repulsive:
$F_{13} = \frac{1}{4 \pi \epsilon_0} \frac{|q_1 q_3|}{r^2} = 9 \times 10^9 \frac{(1 \times 10^{-6})(3 \times 10^{-6})}{1^2} = 27 \times 10^{-3} \, \text{N}.$
This acts at $60^\circ$ to $F_{12}$.
Step 3: Resolve $F_{13}$ into components parallel and perpendicular to $F_{12}$:
$F_{13x} = 27 \times 10^{-3} \cos 60^\circ = 13.5 \times 10^{-3} \, \text{N},$
$F_{13y} = 27 \times 10^{-3} \sin 60^\circ = 23.38 \times 10^{-3} \, \text{N}.$
Step 4: Net force components:
$F_x = -F_{12} + F_{13x} = -18 \times 10^{-3} + 13.5 \times 10^{-3} = -4.5 \times 10^{-3} \, \text{N},$
$F_y = F_{13y} = 23.38 \times 10^{-3} \, \text{N}.$
Step 5: Magnitude of the net force:
$F_{\text{net}} = \sqrt{F_x^2 + F_y^2} = \sqrt{(4.5)^2 + (23.38)^2} \times 10^{-3} = 23.81 \times 10^{-3} \, \text{N}.$

Key Concept: Coulomb's law in dielectric medium

c) $31.6 \, mm$
[Solution Description]
Coulomb's law in a dielectric medium is given by:
$F = \frac{1}{4 \pi \varepsilon_0 K} \frac{q_1 q_2}{r^2}$
Rearrange to solve for $r$:
$r^2 = \frac{1}{4 \pi \varepsilon_0 K} \frac{q_1 q_2}{F}$
Substitute $\frac{1}{4 \pi \varepsilon_0} = 9 \times 10^9 \, \text{N-m}^2/\text{C}^2$, $K = 2$, $q_1 = q_2 = 5 \times 10^{-6} \, C$, and $F = 112.5 \, N$:
$r^2 = \frac{9 \times 10^9}{2} \frac{(5 \times 10^{-6})^2}{112.5}$
$r^2 = \frac{9 \times 10^9 \times 25 \times 10^{-12}}{225} = \frac{225 \times 10^{-3}}{225} = 10^{-3} \, \text{m}^2$
$r = \sqrt{10^{-3}} = 0.0316 \, \text{m} = 31.6 \, \text{mm}$

Key Concept: Electric field intensity and force relationship

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The assertion states that the force experienced by a charge $q$ in an electric field $\overrightarrow{E}$ is given by $\overrightarrow{F} = q \overrightarrow{E}$, which is a direct formula from the syllabus. The reason states that the direction of the force on a negative charge is opposite to the direction of the electric field, which is also correct as per the syllabus. Furthermore, the reason correctly explains why the direction of force would be opposite for a negative charge according to the formula $\overrightarrow{F} = q \overrightarrow{E}$. Thus, both Assertion and Reason are true, and Reason is the correct explanation of Assertion.

Key Concept: Electric Field Lines and Conservative Force

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The Assertion states that the work done by the electric field in moving a test charge along a closed loop is always zero. This is true because the electric field is a conservative field, meaning the work done depends only on the initial and final positions and not on the path taken. The Reason correctly explains this by stating that the electric field is conservative, and the force on a test charge is also conservative. Hence, both Assertion and Reason are true, and Reason is the correct explanation of Assertion.

Key Concept: Electric Lines of Force

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The assertion states that electric lines of force never intersect each other, which is true based on the properties of electric lines of force mentioned in the syllabus. The reason provided is that if two lines of force intersect, there would be two directions of the electric field at the point of intersection, which contradicts the fact that the electric field at any point has a unique direction. This explanation directly supports the assertion. Therefore, both the assertion and reason are true, and the reason correctly explains the assertion.

Key Concept: Mass Change Due to Electron Transfer

b) Its mass decreases
[Solution Description]
Electrons have a small but measurable mass. When a body loses electrons, its total mass decreases because the mass of the transferred electrons is no longer part of the body. Similarly, the body gaining electrons increases in mass.

Key Concept: Superposition Principle

c) $2\hat{i}$ N
[Solution Description]
The resultant force is the vector sum of $\vec{F_{12}}$ and $\vec{F_{13}}$. Here, $\vec{F_{12}} = 5\hat{i}$ N and $\vec{F_{13}} = -3\hat{i}$ N.
Net force: $\vec{F_1} = \vec{F_{12}} + \vec{F_{13}} = 5\hat{i} - 3\hat{i} = 2\hat{i}$ N.

Key Concept: Law of Conservation of Charge

c) The total charge in an isolated system remains constant
[Solution Description]
The law of conservation of charge states that the total electric charge in an isolated system remains constant. Charge can be transferred between objects but cannot be created or destroyed.

Key Concept: Electric Lines of Force - Direction of Field

b) The direction of electrostatic force on a free positive charge
[Solution Description]
The tangent at any point on an electric line of force gives the direction of the electric field at that point. A free positive charge moves along these lines, and the lines originate from positive charges and terminate at negative charges.

Key Concept: Properties of Electric Charges

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
According to the electron theory of electrification, when two different materials are rubbed together, electrons are transferred from one material to another. In this case, electrons move from the glass rod to the silk. The glass rod loses electrons, resulting in a positive charge due to a deficiency of electrons. The silk gains these electrons, resulting in a negative charge due to an excess of electrons. Therefore, both Assertion and Reason are true, and Reason correctly explains the Assertion.

Key Concept: Definition of Electric Field

d) The force experienced by a unit positive charge placed at a point
[Solution Description]
The electric field intensity $\overrightarrow{E}$ at a point in space is defined as the force experienced by a unit positive test-charge placed at that point. Mathematically, it is given by $\overrightarrow{E} = \frac{\overrightarrow{F}}{q_0}$, where $\overrightarrow{F}$ is the force acting on the test-charge $q_0$. The test charge must be vanishingly small to avoid altering the electric field being measured.

Key Concept: Quantisation, Conservation of Charge

b) Both Assertion and Reason are true, but Reason is NOT the correct explanation of Assertion.
[Solution Description]
The assertion states that the total charge of a system with charges $+2e$ and $-3e$ is $-1e$, which is correct because the net charge is calculated as the algebraic sum: $+2e - 3e = -1e$. The reason explains the additivity of electric charges by highlighting quantisation and algebraic summation. However, while both statements are true, the reason does not directly explain why the assertion is true; it instead provides a general property of charges. Thus, the reason is not the correct explanation of the assertion.

Key Concept: Coulomb's Law, Superposition Principle

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
Consider four point charges $q_A = +q$, $q_B = -q$, $q_C = +q$, and $q_D = -q$ placed at the corners of a square. A test charge $q_0$ is placed at the center. According to Coulomb’s law, the force between two charges is given by $F = k \frac{q_1 q_2}{r^2}$. Due to symmetry, the forces exerted by $q_A$ and $q_C$ on $q_0$ are equal in magnitude but opposite in direction ($\vec{F_A} = -\vec{F_C}$). Similarly, $\vec{F_B} = -\vec{F_D}$. By the superposition principle, the net force is the vector sum of all individual forces: $\vec{F} = \vec{F_A} + \vec{F_B} + \vec{F_C} + \vec{F_D} = 0$. Thus, the Assertion is true. The Reason correctly explains the Assertion because the superposition principle ensures that symmetrical forces cancel each other out.

Key Concept: Force on a charged particle in an electric field

d) 10 N
[Solution Description]
The force experienced by a test charge in an electric field is given by $\overrightarrow{F} = q \overrightarrow{E}$. Here, $q = +2 \, \text{C}$ and $E = 5 \, \text{N/C}$. So, $F = 2 \times 5 = 10 \, \text{N}$. Since the charge is positive, the direction of force is along the direction of $\overrightarrow{E}$.

Key Concept: Definition of electric field intensity

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The electric field intensity at a point is indeed defined as the force experienced by a unit positive test charge placed at that point, given by $\vec{E} = \frac{\vec{F}}{q_0}$. This definition requires the test charge $q_0$ to be negligibly small so that its placement does not disturb the original electric field, mathematically represented as $\vec{E} = \lim_{q_0 \to 0} \left( \frac{\vec{F}}{q_0} \right)$. Thus, the Assertion correctly states the definition, and the Reason explains why the test charge must be small, making it the correct explanation of the Assertion.

Key Concept: Mass change due to electron transfer

c) Mass of the comb increases.
[Solution Description]
In the given series, fur appears before plastic. Thus, when rubbed, electrons are transferred from fur to plastic. The plastic comb gains electrons and becomes negatively charged. Since electrons have mass, the mass of the plastic comb increases slightly due to the added electrons.

Key Concept: Electron Theory of Electrification

d) It gains electrons
[Solution Description]
According to the electron theory of electrification, an atom gains a negative charge when it acquires extra electrons. Conversely, losing electrons makes the atom positively charged. Therefore, a negatively charged atom has an excess of electrons.

Key Concept: Coulomb's law, Electric force in dielectric medium

a) 0.844 N
[Solution Description]
First, convert the charges to Coulombs: $q_1 = 5 \times 10^{-6} \, \text{C}$, $q_2 = -3 \times 10^{-6} \, \text{C}$. The distance $r = 20 \, \text{cm} = 0.20 \, \text{m}$. Using Coulomb's law for a dielectric medium:
$F = \frac{1}{4 \pi \varepsilon_0 K} \frac{|q_1 q_2|}{r^2}$
Substitute the given values:
$F = \frac{9 \times 10^9}{4} \times \frac{(5 \times 10^{-6})(3 \times 10^{-6})}{(0.20)^2}$
$F = 2.25 \times 10^9 \times \frac{15 \times 10^{-12}}{0.04}$
$F = 2.25 \times 10^9 \times 375 \times 10^{-12}$
$F = 843.75 \times 10^{-3} \, \text{N} = 0.844 \, \text{N}$

Key Concept: Quantization of Charge

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
Step 1: Analyze the Assertion (A). An electron has a charge of $-e$ while a positron has a charge of $+e$. Therefore, the net charge on the system is $-e + e = 0$, making Assertion (A) true.
Step 2: Analyze the Reason (R). According to the law of conservation of charge, the total charge in an isolated system remains constant. This is a fundamental principle and is correctly stated in Reason (R), making it true.
Step 3: Check if Reason (R) correctly explains Assertion (A). The assertion about the net charge being zero is indeed a consequence of the conservation of charge, as the charges cancel each other out while the total charge remains unchanged. Thus, Reason (R) is the correct explanation for Assertion (A).
Final Answer: Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.

Key Concept: Electric field intensity and force relationship

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
- The Assertion states that the electric field intensity at a point is independent of the test charge. This is true because the field intensity is a property of the source charges creating the field and not the test charge.
- The Reason provides the definition of electric field intensity, where the limit $q_0 \to 0$ ensures that the test charge does not alter the existing electric field. This is also true.
- Moreover, the Reason correctly explains why the Assertion is true, as it shows that the test charge's effect is negligible in determining the field intensity.

Key Concept: Coulomb's Law in Vector Form

b) Both Assertion and Reason are true, but Reason is NOT the correct explanation of Assertion.
[Solution Description] The Assertion states that the electrostatic force $\vec{F_{12}}$ between two charges $q_1$ and $q_2$ is always directed along the line joining them. This is true because Coulomb's law in vector form is given by $\vec{F_{12}} = \frac{1}{4 \pi \epsilon_0} \cdot \frac{q_1 q_2}{r^3} \vec{r_{12}}$, where $\vec{r_{12}}$ is the vector from $q_1$ to $q_2$. The Reason states that Coulomb's law in vector form shows that the unit vector $\hat{r_{12}}$ points from $q_1$ to $q_2$. This is also true as $\hat{r_{12}}$ is defined as $\frac{\vec{r_{12}}}{|\vec{r_{12}}|}$. However, the Reason does not correctly explain the Assertion because the direction of the force is determined by both the product of the charges ($q_1 q_2$) and the direction of $\vec{r_{12}}$. For example, if $q_1 q_2 < 0$, the force would be in the opposite direction to $\hat{r_{12}}$. Therefore, while both statements are true, the Reason is not the correct explanation of the Assertion.

Key Concept: Resultant Electric Field Magnitude

a) $93.53 \times 10^3 \, \text{V/m}$
[Solution Description]
The centroid divides each median in a 2:1 ratio, so the distance from any charge to the centroid is $\frac{\sqrt{3}}{3}$ m.
Step-by-step:
1. Magnitude of fields:
$E_1 = \frac{9 \times 10^9 \times 2 \times 10^{-6}}{(\frac{\sqrt{3}}{3})^2} = 54 \times 10^3 \, \text{V/m}$ (away from $q_1$)
$E_2 = \frac{9 \times 10^9 \times 1 \times 10^{-6}}{(\frac{\sqrt{3}}{3})^2} = 27 \times 10^3 \, \text{V/m}$ (towards $q_2$)
$E_3 = \frac{9 \times 10^9 \times 3 \times 10^{-6}}{(\frac{\sqrt{3}}{3})^2} = 81 \times 10^3 \, \text{V/m}$ (away from $q_3$)
2. Symmetry cancels out $y$-components. The $x$-components add up as follows:
$\vec{E}_{\text{net}} = (E_1 - E_2 + E_3) \cos(30^\circ) \hat{i}$
$= (54 - 27 + 81) \times 10^3 \times \frac{\sqrt{3}}{2} = 108 \times 10^3 \times \frac{\sqrt{3}}{2} = 93.53 \times 10^3 \, \text{V/m}$

Key Concept: Charge Calculation

b) $4.0 \times 10^{-7} \, \text{C}$
[Solution Description]
The charge of one electron is $-1.6 \times 10^{-19} \, \text{C}$. Losing $2.5 \times 10^{12}$ electrons means the conductor gains a positive charge equal in magnitude:
$q = n \times e = 2.5 \times 10^{12} \times 1.6 \times 10^{-19} = 4.0 \times 10^{-7} \, \text{C}.$

Key Concept: Electric Field, Dipole Moment

c) $90^\circ$
[Solution Description]
The torque $\vec{\tau}$ acting on an electric dipole in a uniform electric field is given by $\vec{\tau} = \vec{p} \times \vec{E}$. The magnitude of the torque is $\tau = pE \sin\theta$, where $\theta$ is the angle between $\vec{p}$ and $\vec{E}$. The torque will be maximum when $\sin\theta$ is maximum, i.e., when $\theta = 90^\circ$.
Calculation:
$\sin 90^\circ = 1 \Rightarrow \tau_{\text{max}} = pE \times 1 = pE$

Key Concept: Electric Lines of Force, Dipole Field

d) In the region between the two charges along the axial line.
[Solution Description]
The electric field of a dipole is strongest along the axial line near the charges, particularly in the region between the two opposite charges. This is because the field lines are most concentrated here, indicating a stronger electric field compared to other regions.

Key Concept: Charge Carriers in Conductors

c) Free electrons
[Solution Description]
In metals, only the outer electrons of atoms are free to move, known as 'free electrons'. These free electrons act as the charge carriers in metals.

Key Concept: Torque on a Dipole

c) $5\sqrt{3} \times 10^{-4}$ Nm
[Solution Description]
The torque $\tau$ on a dipole in a uniform electric field is given by:
$\tau = p E \sin \theta$
Here, $p = 2 \times 10^{-8}$ C-m, $E = 5 \times 10^4$ N/C, and $\theta = 60^\circ$. Substituting these values:
$\tau = 2 \times 10^{-8} \times 5 \times 10^4 \times \sin 60^\circ$
Since $\sin 60^\circ = \frac{\sqrt{3}}{2}$,
$\tau = 2 \times 5 \times \frac{\sqrt{3}}{2} \times 10^{-4} = 5\sqrt{3} \times 10^{-4} \text{ Nm}$

Key Concept: Electric dipole moment, torque in uniform electric field

a) $1.27 \times 10^{-4} \, N \cdot m$
[Solution Description]
The torque $\tau$ on an electric dipole in a uniform electric field is given by:
$$ \tau = pE \sin \theta $$
Here, $p = 6 \times 10^{-9} \, C \cdot m$, $E = 3 \times 10^4 \, N/C$, and $\theta = 45^\circ$.
Substituting the values:
$$ \tau = 6 \times 10^{-9} \times 3 \times 10^4 \times \sin 45^\circ $$
$$ \tau = 18 \times 10^{-5} \times \frac{\sqrt{2}}{2} $$
$$ \tau = 12.73 \times 10^{-5} \, N \cdot m $$
The closest value among the options is $1.27 \times 10^{-4} \, N \cdot m$.

Key Concept: Charge Calculation

b) $8 \times 10^{-19}$ C
[Solution Description]
Each electron has a charge of $-1.6 \times 10^{-19}$ C. A deficiency of 5 electrons means the conductor has a positive charge equal to 5 times the charge of an electron:
$Q = 5 \times 1.6 \times 10^{-19} \, \text{C} = 8 \times 10^{-19} \, \text{C}$

Key Concept: Coulomb's Law

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The Assertion states that the force between two point-charges decreases with increasing distance, which is true as per Coulomb's law: $F = k \frac{q_1 q_2}{r^2}$, where $F$ is inversely proportional to $r^2$. The Reason correctly explains the Assertion by referring to Coulomb's law, which establishes the inverse square relationship between force and distance. Therefore, both Assertion and Reason are true, and Reason is the correct explanation of Assertion.

Key Concept: Millikan’s experiment, force balance

b) 1.6 \times 10^{-19} C
[Solution Description]
At equilibrium, the electric force balances the weight of the drop:
$qE = mg \implies q = \frac{mg}{E}$
Substituting the values:
$q = \frac{(3.2 \times 10^{-15})(9.8)}{2 \times 10^5} = \frac{31.36 \times 10^{-15}}{2 \times 10^5} = 1.568 \times 10^{-19} C$
Approximating to one significant figure: $q \approx 1.6 \times 10^{-19} C$.

Key Concept: Coulomb’s Law - Force between charges and distance

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 two point charges in a dielectric medium, which includes the dielectric constant $K$ in the denominator, reducing the force compared to vacuum. The reason explains that the dielectric medium polarizes in the presence of an electric field, creating an opposing field that weakens the net electric force between the charges. Both the assertion and reason are true, and the reason correctly explains why the force is reduced in a dielectric medium. Therefore, the correct answer is option (a).

Key Concept: Coulomb's Law: Nature of Force

b) Attractive
[Solution Description]
Coulomb's law states that two unlike charges attract each other, whereas like charges repel each other. Therefore, the force between two unlike charges is attractive.

Key Concept: Direction of dipole moment

d) From negative charge to positive charge
[Solution Description]
The electric dipole moment vector (\$ \vec{p} \$) is directed from the negative charge (\$ -q \$) to the positive charge (\$ +q \$) along the line joining the two charges.

Key Concept: Electric Field due to two point charges

c) $18 \times 10^3 \, \text{N/C} \, (\text{towards} \, +x)$
[Solution Description]
The electric field due to $q_1$ at $(2, 0)$ is:
$E_1 = \frac{k|q_1|}{r^2} = \frac{9 \times 10^9 \times 5 \times 10^{-6}}{(2)^2} = 11.25 \times 10^3 \, \text{N/C} \, (\text{towards} \, +x)$
The electric field due to $q_2$ at $(2, 0)$ is:
$E_2 = \frac{k|q_2|}{r^2} = \frac{9 \times 10^9 \times 3 \times 10^{-6}}{(2)^2} = 6.75 \times 10^3 \, \text{N/C} \, (\text{towards} \, +x)$
Net electric field:
$E_{\text{net}} = E_1 + E_2 = (11.25 + 6.75) \times 10^3 = 18 \times 10^3 \, \text{N/C} \, (\text{towards} \, +x)$

Key Concept: Equilibrium of Charge