Class 12 Physics Chapter 1 Electric Charges And Fields

Key Concept: Field Line Continuity

b) Continuous
[Solution Description] In a charge-free region, electric field lines are continuous, indicating they do not have any breaks or interruptions. This continuity is due to the absence of sources or sinks of the electric field in that region.

Key Concept: Comparative Analysis

b) Volume charge density is measured in C/m³

[Solution Description]

Volume charge density $\rho$ is defined as the charge per unit volume. Its units are given by:

$\rho = \frac{\Delta Q}{\Delta V} \quad \text{with units C/m}^3$

Thus, option b accurately describes the unit of measurement for volume charge density.

Key Concept: Function Understanding

d) To restrict the flow of electricity
[Solution Description] The primary function of an insulator is to prevent the undesired flow of electricity. This helps in ensuring safety and efficiency by containing the electrical current within a desired path.

Key Concept: Historical Experimentation, Real-world Application

c) Electrometer
[Solution Description] The torsion balance allowed precise measurements of minute forces between charged objects, showing that the force was directly proportional to the product of charges and inversely proportional to the square of the distance. Modern electrometers serve similar functions by providing digital readings of electrical forces or potential difference, relying on advanced electronics rather than mechanical disturbance.

Key Concept: Charge Transfer Process

c) +1.5 units on each sphere
[Solution Description]
When two conductors touch, charges redistribute equally due to their identical nature. Total initial charge is +8 + (-5) = +3 units.

Since the spheres are identical, this total charge will divide equally:

$\text{Charge per sphere} = \frac{+3}{2} = +1.5$ units.

Thus, each sphere ends up with a charge of +1.5 units.

Key Concept: Conductors vs Insulators

b) Conductors have free-moving electrons
[Solution Description] Conductors have free-moving electrons that allow electricity to pass through them easily. In contrast, insulators do not have such free-moving electrons and resist the flow of electricity.

Key Concept: Solid Angle and Field Strength, Electromagnetic Waves

c) The flux remains unchanged since the total charge inside the sphere is constant.
[Solution Description] According to Gauss’s Law, the total electric flux $\Phi$ through a closed surface is given by $\Phi = \frac{Q}{\varepsilon_0}$, where Q is the charge enclosed and $\varepsilon_0$ is the permittivity of free space. If the charge starts to accelerate, it generates electromagnetic waves, but this motion doesn’t change the amount of charge enclosed within the sphere. Consequently, the total electric flux through the sphere remains constant regardless of the acceleration of the charge.

Key Concept: Field Line Properties

c) Crossing would imply multiple field directions at a point
[Solution Description] Electric field lines cannot cross each other because this would suggest two different directions for the electric field at the same point, which is not possible. Each point can only have one unique electric field direction.

Key Concept: Charge Transfer

c) Increases its conductivity
[Solution Description] Increasing the temperature typically increases the charge carrier concentration in a semiconductor, thus enhancing its electrical conductivity. This is contrary to metals, where increased temperature tends to decrease conductivity.

Key Concept: Basic Definition

c) $9 \times 10^9 \, Nm^2/C^2$
[Solution Description]
The constant k in Coulomb’s law, also known as Coulomb’s constant, is used to calculate the electrostatic force between two charges. It is equal to $9 \times 10^9 \, Nm^2/C^2$.

Key Concept: Field Line Interpretation, Complex Superposition

a) Toward positive x-direction
[Solution Description]
To analyze the electric field just to the right of x = 3 cm, evaluate the contributions from each charge based on their positions.

Contribution due to $+2 \, \mu C$ at x = -3 cm:
$E_1 \text{ direction toward } x > 3 \, cm \text{ since it is positively charged.}$

Contribution due to $-3 \, \mu C$ at x = 0 cm:
$E_2 \text{ direction toward } x < 3 \, cm \text{ because it is negatively charged.} $ Contribution due to $+1 \, \mu C$ at $x = 3 \, cm$: $ E_3 \text{ direction toward } x > 3 \, cm \text{ as it is positively charged.}$

Since the fields from $+2 \, \mu C$ and $+1 \, \mu C$ both direct rightward and overcome the leftward pull from $-3 \, \mu C$, the total electric field direction will be toward the right.

Strength is determined by the proximity and charge magnitudes, but qualitatively, the field lines indicate that the net electric field is indeed moving rightward due to dominance by the surrounding positive charges.

Therefore, the nature of the electric field is toward positive x-direction.

Key Concept:  Conservation of Charge

b) +2 C
[Solution Description] The conservation of charge principle states that the total charge in an isolated system remains constant. Thus, if a system initially has 5 C of positive charge and 3 C of negative charge, the net charge would be:

$\text{Net charge} = +5 \, \text{C} – 3 \, \text{C} = +2 \, \text{C}$

Therefore, the total charge of the system is +2 C.

Key Concept: Basic Concept

b) Strength of the field
[Solution Description] The relative density of electric field lines is a visual representation that indicates the strength of the electric field. Where the lines are denser, the field is stronger, and where they are spaced apart, the field is weaker.

Key Concept: Basic Formula

a) $F = \frac{1}{4\pi\varepsilon_0} \frac{q_1 q_2}{r^2}$
[Solution Description]
The formula for Coulomb’s force between two point charges $q_1$ and $q_2$ separated by a distance r, using the permittivity of free space $\varepsilon_0$, is given by:

$F = \frac{1}{4\pi\varepsilon_0} \frac{q_1 q_2}{r^2}.$

This is a direct application of Coulomb’s Law.

Key Concept: Basic Definitions

c) C/m
[Solution Description] Linear charge density $(\lambda)$ is defined as charge per unit length and its units are given in C/m.

Key Concept: Dipole Field Variation

c) Varies as $\frac{1}{r^3}$
[Solution Description] The electric field E on the axial line of a dipole at a large distance is given by the formula $E = \frac{2p}{4\pi\epsilon_0r^3}$. From this equation, it is clear that the field decreases as $1/r^3$. Therefore, the correct answer is $1/r^3$.

Key Concept: Basic Formula

a) $E = \frac{1}{4\pi \varepsilon_0} \frac{p}{(r^2 + a^2)^{3/2}}$
[Solution Description] The electric field E on the equatorial plane of a dipole can be calculated using the following formula:
$E = \frac{1}{4\pi\varepsilon_0} \frac{p}{(r^2 + a^2)^{3/2}}$
This formula is derived based on the assumption that the dipole consists of charges separated by a distance and the field is measured perpendicular to the dipole axis.

Key Concept: Charge Effects

c) The sphere gets attracted
[Solution Description] When a positively charged object is brought near a neutral metal sphere, it induces polarization within the sphere. The electrons in the sphere will move towards the side closest to the charged object, causing the nearest side to become negatively charged while the far side becomes positively charged. Thus, there will be an attractive force effect on the neutral sphere.

Key Concept: Zero Field Inside

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description] According to Gauss’s Law, for any point inside a uniformly charged spherical shell, the net charge enclosed by a Gaussian surface is zero because the shell itself does not have any charge inside it. Due to this, the electric field inside the shell must be zero. This is also supported by the fact that the distribution of charge on the shell is symmetric, leading to cancellation of electric fields at any point inside the shell. Hence, both Assertion and Reason are true, and Reason is the correct explanation of Assertion.

Key Concept: Conceptual Understanding

d) Because it is independent of distance from the sheet

[Solution Description] The electric field due to an infinite plane sheet is given by:

$E = \frac{\sigma}{2\epsilon_0}$

Since the expression for the electric field does not include the distance x, it implies that the magnitude of the electric field is independent of the distance from the sheet. This results in a uniform electric field.

Key Concept: Basic Concept

c) $\Phi = \frac{q}{\epsilon_0}$
[Solution Description] According to Gauss’s Law, the electric flux $\Phi$ through a closed surface is given by the formula $\Phi = \frac{q}{\epsilon_0}$ where q is the total charge enclosed and $\epsilon_0$ is the permittivity of free space.

Key Concept: Basic Definition

b) The force per unit charge
[Solution Description]
The electric field is defined as the force per unit charge experienced by a small positive test charge placed in the field. Mathematically, it is expressed as $E = \frac{F}{q}$ where E is the electric field, F is the force, and q is the test charge.

Key Concept: Dipole Field Formula

b) $E = \frac{2p}{4\pi\epsilon_0 r^3}$
[Solution Description]
The electric field E due to a dipole on its axis at large distances is given by the formula:

$E = \frac{2p}{4\pi\epsilon_0 r^3}$

Here, p is the dipole moment, $\epsilon_0$ is the permittivity of free space, and r is the distance from the dipole.

Key Concept: Charge Redistribution

b) 3 C

[Solution Description] When sphere A and B touch, the total charge is shared equally due to their identical nature:

$Q_{\text{total}} = Q_A + Q_B = 6 + 0 = 6 \, C$

After redistribution, each sphere has half of the total charge:

$Q_{\text{final}} = \frac{Q_{\text{total}}}{2} = \frac{6}{2} = 3 \, C$

So, the final charge on sphere A is 3 C.

Key Concept: Basic Concept

b) $4.50 \times 10^{6} \, \text{N/C}$
[Solution Description] The electric field E due to a point charge is given by Coulomb’s Law,
$E = \frac{k \cdot |q|}{r^2}$
where $k = 8.99 \times 10^{9} \, \text{Nm}^2/\text{C}^2$, $q = 5 \times 10^{-6} \, \text{C}$, and $r = 0.1 \, \text{m}$.

Substituting the values,
$E = \frac{8.99 \times 10^{9} \times 5 \times 10^{-6}}{(0.1)^2} = 4.495 \times 10^{6} \, \text{N/C}$
Therefore, the electric field is approximately $4.50 \times 10^{6} \, \text{N/C}$.

Key Concept: Basic Concept

b) Principle of Superposition
[Solution Description] The principle of superposition states that the net force on a charge due to multiple other charges is the vector sum of the forces due to each charge individually.

Key Concept: Closed Surface Convention

b) Outward normal
[Solution Description] By convention, for a closed surface, the direction of the area element vector is taken as the outward normal. This means the vector points away from the enclosed volume.

Key Concept:  Basic Concept

a) $1.13 \times 10^{6} \, N \cdot m^2/C$

[Solution Description]

According to Gauss’s law, the electric flux through a closed surface enclosing a charge q is given by $\Phi = \frac{q}{\varepsilon_0}$ where $\varepsilon_0$ is the permittivity of free space $(8.85 \times 10^{-12} \, C^2/N \cdot m^2)$.

Let’s calculate the electric flux:

$\Phi = \frac{10 \times 10^{-6} \, C}{8.85 \times 10^{-12} \, C^2/N \cdot m^2} = 1.13 \times 10^{6} \, N \cdot m^2/C$

So, the electric flux $\Phi$ is $1.13 \times 10^{6} \, N \cdot m^2/C$.

Key Concept: Electric Field Concept

d) It reduces to one-ninth

[Solution Description] According to Coulomb’s law, the electric field E due to a point charge q at a distance r is given by $E = \frac{kq}{r^2}$, where k is Coulomb’s constant. If the distance is tripled, the new distance is 3r, then:

$E_{new} = \frac{kq}{(3r)^2} = \frac{kq}{9r^2} = \frac{1}{9}E$

Therefore, the magnitude of the electric field becomes one-ninth of its original value.

Key Concept: Flux Calculation with Angle

c) $2.0784 \times 10^3 \, \text{Nm}^2/\text{C}$

[Solution Description]

To calculate the electric flux $\Phi$, we use the formula:

$\Phi = E \cdot A \cdot \cos \theta$

where $E = 200 \, \text{N/C}$, A is the area of the rectangle, and $\theta = 30^\circ$. The area A can be calculated as:

$A = 4 \, \text{m} \times 3 \, \text{m} = 12 \, \text{m}^2$

Now compute the flux:

$\Phi = 200 \times 12 \times \cos(30^\circ) = 2400 \times \frac{\sqrt{3}}{2} = 2400 \times 0.866 = 2078.4 \, \text{Nm}^2/\text{C}$

Key Concept: Electric Field Definition

d) $E = \frac{1}{4\pi\epsilon_0} \frac{Q}{r^2} \hat{r}$
[Solution Description] The electric field E due to a point charge Q at a distance r can be calculated using the formula for the electric field:
$E = \frac{1}{4\pi\epsilon_0} \frac{Q}{r^2} \hat{r}$
Where $\epsilon_0$ is the permittivity of free space, and $\hat{r}$ is the unit vector in the direction from the charge to the point where the field is being measured.

Key Concept: Basic Definition

c) Two equal and opposite charges separated by a distance

[Solution Description] An electric dipole is defined as a pair of equal and opposite charges separated by a certain distance.

Key Concept: Charge Distribution

b) Distributes evenly over the surface
[Solution Description] When an electrical charge is introduced to a conductor, it distributes evenly over the surface due to the free movement of electrons within the material.

Key Concept: Charge Neutralization

c) They neutralize each other
[Solution Description] When two oppositely charged objects are brought into contact, they undergo neutralization. The charges cancel out, making both objects lose their charge.

Key Concept: Conceptual Understanding

d) Because there is no enclosed charge

[Solution Description] According to Gauss’s law, the net electric flux through a closed surface is proportional to the enclosed charge. For any point inside a uniformly charged thin spherical shell, one can imagine a Gaussian surface that does not enclose any of the shell’s charge. As a result, the enclosed charge is zero. Consequently, the net electric field inside such a shell must also be zero because there are no charges within the Gaussian surface to produce an electric field.

Key Concept: Charge Interaction

b) They repel each other
[Solution Description] When two objects are rubbed and acquire charges of the same sign, they repel each other due to like charges repelling.

Key Concept: Basic Formula

d) $E = -\frac{1}{4\pi\varepsilon_0} \frac{p}{r^3}$
[Solution Description] For the equatorial plane of a dipole, the electric field E is given by:
$E = -\frac{1}{4\pi\varepsilon_0} \frac{p}{r^3}.$
This expression is used when the distance r from the dipole is much greater than the separation a between the charges.

Key Concept: Vector Addition

b) Parallelogram Law
[Solution Description] The net force on a charge due to multiple other charges can be calculated using vector addition principles, specifically the parallelogram law or triangle rule for adding vectors.

Key Concept: Charge Conservation in Reactions, Charge Redistribution

c) 0

[Solution Description] Initially, the neutron has no net charge (0). When it decays into a proton, electron, and antineutrino, the charges are distributed as follows: proton (+1e), electron (-1e), and antineutrino (0). The sum of these charges must be conserved based on charge conservation principles. Therefore, the total charge after the decay can be calculated as:

0 = (+1e) + (-1e) + 0

Thus, the total charge remains 0.

Key Concept: Charge Interaction

b) They repel each other
[Solution Description] According to basic electrostatic principles, like charges repel each other. So, if both balloons are negatively charged, they will repel each other.

Key Concept: Basic Concept

b) Superposition Principle
[Solution Description]
The principle that involves calculating the net force on a charge due to multiple other charges by taking the vector sum of individual forces from each charge is known as the Superposition Principle. This principle allows for the consideration of each interaction separately and then combines them using vector addition.

Key Concept: Charge Conservation

d) -2 C
[Solution Description] According to the principle of charge conservation, the total charge in an isolated system does not change. Initially, the total charge of the system is $(+3\,C) + (-5\,C) = -2\,C$. After coming into contact and separating again, the total charge remains the same at $-2\,C$.

Key Concept: Field Independence

b) Both Assertion and Reason are true, but Reason is NOT the correct explanation of Assertion.
[Solution Description]
The assertion states that the electric field produced by a uniformly charged infinite plane sheet does not change with distance from the plane. This is indeed true because, mathematically, the expression for the electric field $E = \frac{\sigma}{2\epsilon_0}$ shows no dependence on the distance variable.

The reason provided explains this phenomenon by attributing it to the uniformity in the surface charge density $\sigma$. While it is true that the uniform distribution of charge contributes to the uniformity of the field, the primary reason for the field’s independence from distance lies in the symmetrical nature of the problem and Gauss’s law considerations.

Hence, while both statements are true, the reason given is not the correct explanation of the assertion.

Key Concept:  Electrification Process

b) Negative
[Solution Description] During the process of rubbing, electrons are transferred from the glass rod to the silk. As a result, the silk becomes negatively charged.

Key Concept:  Charge Quantisation

c) q = ne
[Solution Description] According to the principle of charge quantisation, any charge q is an integral multiple of a fundamental unit of charge, denoted by e. Thus, the formula for quantisation of charge is given by q = ne, where n is an integer.

Key Concept: Formula Identification

d) $F = k \frac{q_1 q_2}{r^2}$
[Solution Description] The formula for Coulomb’s Law is given by $F = k \frac{q_1 q_2}{r^2}$ where F is the force between two charges, k is Coulomb’s constant, $q_1$ and $q_2$ are the magnitudes of the charges, and r is the distance between them.

Key Concept: Field Direction

b) Inward toward the wire

[Solution Description] Since the linear charge density $\lambda = -5 \, \mu C/m$ is negative, the electric field direction is inward toward the wire. This is because, by convention, if $\lambda < 0$, the electric field points inward.

Key Concept: Torque Calculation

c) $12 \, \text{N} \cdot \text{m}$

[Solution Description] The torque $\tau$ on the dipole is given by the formula $\tau = pE\sin\theta$, where p is the dipole moment, E is the electric field, and $\theta$ is the angle between them.

Given:

$p = 4 \, \text{C} \cdot \text{m}, \quad E = 6 \, \text{N/C}, \quad \theta = 30^\circ$

Calculate $\tau$

$\tau = 4 \times 6 \times \sin(30^\circ) = 24 \times 0.5 = 12 \, \text{N} \cdot \text{m}$

Hence, the torque experienced by the dipole is $12 \, \text{N} \cdot \text{m}$

Key Concept: Basic Definition

d) It is proportional to the charge enclosed.

[Solution Description]

Gauss’s Law states that the electric flux $\Phi$ through a closed surface is equal to the charge q enclosed divided by the permittivity of free space $\epsilon_0$. Mathematically, this can be expressed as:

$\Phi = \frac{q}{\epsilon_0}$

Therefore, according to Gauss’s Law, the correct answer is that the electric flux through a closed surface is proportional to the charge enclosed.

Key Concept: Application of Concepts

b) 12 C
[Solution Description]
To find the total charge on the rod, we use the formula for linear charge density: $\lambda = \Delta Q / \Delta l$. Here, $\Delta Q$ is the total charge, and $\Delta l$ is the length of the rod.

Rearranging the formula to solve for $\Delta Q$, we have:
$\Delta Q = \lambda \cdot \Delta l = 4 \, \text{C/m} \times 3 \, \text{m} = 12 \, \text{C}$

Therefore, the total charge on the rod is $12 \, \text{C}$.

Key Concept: Basic Formula

c) $36 \times 10^9 \, \text{N}$
[Solution Description]
To calculate the force between the charges using Coulomb’s Law, apply the formula:
$F = k \frac{q_1 q_2}{r^2}.$
Substitute $q_1 = 2 \, \text{C}, q_2 = 2 \, \text{C}, r = 1 \, \text{m}$, and $ k = 9 \times 10^9 \, \text{N m}^2/\text{C}^2 $.

\begin{align*}
F &= 9 \times 10^9 \, \frac{2 \times 2}{1^2} \\
&= 9 \times 10^9 \, \times 4 \\
&= 36 \times 10^9 \, \text{N}.
\end{align*}
Therefore, the force is $ 36 \times 10^9 \, \text{N} $.

Key Concept: Field Line Density

a) Electric field lines are closest together.
[Solution Description] The strength of the electric field can be determined by the density of electric field lines. The closer the field lines are to each other, the stronger the electric field at that location.

Key Concept: Basic Concept

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description] The assertion states the behaviour of an electric field along the axis of a dipole, which indeed decreases as the distance increases. This decrease is specifically described by a $1/r^3$ dependence, as given in the reason. Thus, both statements are true, and the reason correctly explains why the assertion is true.

Key Concept: Positive vs Negative

a) +2 C
[Solution Description]
Add the charges by considering the positive and negative signs:

$+6 \, \text{C} + (-4) \, \text{C} = +2 \, \text{C}$

Therefore, the net charge is +2 C.

Key Concept: Basic Definition

b) Coulomb
[Solution Description] The fundamental unit of electric charge is defined as the elementary charge, denoted by e.

Key Concept: Force Comparison

c) $4.17 \times 10^{42}$
[Solution Description]
To find the ratio of electric force to gravitational force between two electrons, we use the formulas for both forces and their constants. The electric force $F_e$ can be calculated using:

$F_e = k \frac{q_1 q_2}{r^2}$

where $q_1 = q_2 = -1.6 \times 10^{-19} \, \text{C}$ and $k = 9 \times 10^9 \, \text{N m}^2 \text{C}^{-2}$.

The gravitational force $F_g$ is given by:

$F_g = G \frac{m_1 m_2}{r^2}$

with $m_1 = m_2 = 9.11 \times 10^{-31} \, \text{kg}$ and $G = 6.674 \times 10^{-11} \, \text{N m}^2 \text{kg}^{-2}$.

The ratio is:

$\text{Ratio} = \frac{k \cdot (-1.6 \times 10^{-19})^2}{G \cdot (9.11 \times 10^{-31})^2}$

Substituting the known values, we get:

$\text{Ratio} = \frac{9 \times 10^9 \cdot (1.6 \times 10^{-19})^2}{6.674 \times 10^{-11} \cdot (9.11 \times 10^{-31})^2} \approx 4.17 \times 10^{42}$

Thus, the ratio is approximately $4.17 \times 10^{42}$.

Key Concept: Field Magnitude Comparison

a) Greater

[Solution Description]

For large distances r such that r >> a, we have:

– On the axis:

$E_{\text{axis}} \approx \frac{1}{4\pi\epsilon_0} \frac{2p}{r^3}$

– In the equatorial plane:

$E_{\text{equator}} \approx \frac{1}{4\pi\epsilon_0} \frac{p}{r^3}$

Comparing the two expressions, $E_{\text{axis}}$ is exactly twice $E_{\text{equator}}$.

Thus, at large distances, the electric field on the axis is greater than in the equatorial plane.

Key Concept: Angle Effect

c) It becomes zero
[Solution Description] When the area is parallel to the electric field, the angle $\theta$ between the area vector DS and the electric field E is 90 degrees. The electric flux $\Phi$ is given by:
$\Phi = E DS \cos \theta$
Since $\cos 90^\circ = 0$, the electric flux becomes zero.

Key Concept: Charge Interaction

b) They repel each other.
[Solution Description] Both plastic rods become negatively charged after being rubbed with wool. According to the principle that like charges repel, the two similarly charged rods will repel each other.