Class 12 NCERT Physics Chapter 1: Electric Charges and Fields – Detailed Notes

 

Class 12 NCERT Physics Chapter 1: Electric Charges and Fields – Detailed Notes

Introduction to Electrostatics

Electrostatics is the branch of physics that deals with the study of electric charges at rest, the forces between them, and the electric fields they produce.

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1. Electric Charge

Definition:

Electric charge is a fundamental property of matter that causes it to experience a force when placed in an electric or magnetic field.

Properties of Electric Charge:

1. Charge is Quantized: The charge on any object is always an integer multiple of the elementary charge ($e$).

   $$q = \pm ne$$

   where $n$ is an integer and $e = 1.6 \times 10^{-19} C$.

2. Charge is Conserved: The total charge in an isolated system remains constant.

3. Charge is Additive: The total charge of a system is the algebraic sum of all individual charges.

4. Charge is Invariant: The charge of a particle does not change with motion or frame of reference.

5. Charge can be Transferred: Charge can move from one body to another via conduction, induction, or friction.

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2. Conductors and Insulators

Definition:

• Conductors: Materials that allow free movement of charges. Example: Metals, human body.

• Insulators (Dielectrics): Materials that do not allow free movement of charges. Example: Glass, rubber.

Semiconductors:

Materials that have conductivity between conductors and insulators (e.g., Silicon, Germanium).

Coulomb’s Law – Detailed Derivation

Introduction

Coulomb’s law describes the force between two stationary electric charges. It was formulated by Charles-Augustin de Coulomb in 1785 using a torsion balance experiment.

 The law states that the electrostatic force between two point charges is directly proportional to the product of their magnitudes and inversely proportional to the square of the distance between them.

The force $F$ is directly proportional to the product of the two charges:

$$F \propto q_1 q_2$$

1. The force $F$ is inversely proportional to the square of the distance between them:

   $$F \propto \frac{1}{r^2}$$

Combining both relations:

$$F \propto \frac{q_1 q_2}{r^2}$$

Introducing a proportionality constant $k$:

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

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Step 3: Determining the Value of $k$

The constant $k$ depends on the medium in which the charges are placed. In free space (vacuum), it is given by:

$$k = \frac{1}{4\pi\varepsilon_0}$$

Substituting this into the equation:

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

 where 

$\varepsilon_0$ (permittivity of free space) has the value:

$$\varepsilon_0 = 8.854 \times 10^{-12} \, C^2 N^{-1} m^{-2}$$

Thus, the numerical value of $k$ is:

$$k = 9.0 \times 10^9 \, N m^2 C^{-2}$$

Vector Form of Coulomb’s Law

Since force is a vector quantity, we express Coulomb’s law in vector form.

Let $\mathbf{r}_{12}$ be the displacement vector from charge $q_1$ to $q_2$

$$\mathbf{r}_{12} = \mathbf{r}_2 - \mathbf{r}_1$$

The unit vector along this direction is:

$$\hat{\mathbf{r}}_{12} = \frac{\mathbf{r}_{12}}{|\mathbf{r}_{12}|}$$

The force exerted by charge $q_1$ on charge $q_2$ is:

$$\mathbf{F}_{12} = \frac{1}{4\pi\varepsilon_0} \frac{q_1 q_2}{r^2} \hat{\mathbf{r}}_{12}$$

Similarly, the force on charge $q_1$ due to $q_2$ (by Newton’s Third Law) is:

$$\mathbf{F}_{21} = -\mathbf{F}_{12}$$

This confirms that the forces between two charges are equal in magnitude and opposite in direction

Derivation of the Superposition Principle

Statement of Superposition Principle

The Superposition Principle states that the net force (or electric field) on a charge due to multiple other charges is the vector sum of the forces (or fields) due to each individual charge, independently calculated as if the other charges were not present.

Mathematically, for a system of $n$ charges $q_1, q_2, q_3, ... q_n$ exerting forces on a test charge $q$ the net force is:

$$\mathbf{F}_{\text{net}} = \mathbf{F}_1 + \mathbf{F}_2 + \mathbf{F}_3 + ... + \mathbf{F}_n = \sum_{i=1}^{n} \mathbf{F}_i$$

 Consider a System of Point Charges

• Let there be n point charges $q_1, q_2, q_3, ... q_n$ placed at positions $\mathbf{r}_1, \mathbf{r}_2, ... \mathbf{r}_n$
  

• A test charge $q_0$ is placed at $\mathbf{r}_0$.

• Each charge $q_i$ exerts a force on $q_0$ given by Coulomb’s Law:

$$\mathbf{F}_i = \frac{1}{4\pi \epsilon_0} \frac{q_0 q_i}{r_{i0}^2} \hat{\mathbf{r}}_{i0}$$

where:

• $r_{i0}$ is the distance between $q_i$ and $q_0$.

• $\hat{\mathbf{r}}_{i0}$ is the unit vector from $q_i$ to $q_0$.

• Derivation of the Electric Field Due to a Point Charge

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  Definition of Electric Field

  The electric field at a point in space is defined as the force experienced per unit positive test charge placed at that point:

  $$\mathbf{E} = \frac{\mathbf{F}}{q_0}$$

  where:

  • $\mathbf{E}$ is the electric field vector (N/C or V/m).
  • $\mathbf{F}$ is the electrostatic force acting on the test charge $q_0$.
  • $q_0$ is a small test charge used to measure the field (without disturbing the source charge).

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  Derivation of the Electric Field Due to a Point Charge

  Step 1: Consider a Point Charge

  • Let a source charge $Q$ be placed at the origin $O$.
  • We need to determine the electric field at a point $P$ located at a distance $r$ from $Q$.
  • A small test charge $q_0$ is placed at $P$ to experience the force.

  Step 2: Apply Coulomb’s Law

  From Coulomb’s law, the electrostatic force on $q_0$ due to $Q$ is:

  $$\mathbf{F} = \frac{1}{4\pi \epsilon_0} \frac{|Q q_0|}{r^2} \hat{\mathbf{r}}$$

  where:

  • $\epsilon_0 = 8.854 \times 10^{-12} \ C^2 N^{-1} m^{-2}$ (permittivity of free space).
  • $\hat{\mathbf{r}}$ is the unit vector directed outward if $Q$ is positive, and inward if $Q$ is negative.

  Step 3: Find the Electric Field

  Since the electric field is defined as force per unit charge:

  $$\mathbf{E} = \frac{\mathbf{F}}{q_0}$$

  Substituting the value of $\mathbf{F}$:

  $$\mathbf{E} = \frac{1}{4\pi \epsilon_0} \frac{|Q q_0|}{r^2 q_0} \hat{\mathbf{r}}$$ $$\mathbf{E} = \frac{1}{4\pi \epsilon_0} \frac{Q}{r^2} \hat{\mathbf{r}}$$

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  Final Expression for the Electric Field

  $$E = \frac{1}{4\pi \epsilon_0} \frac{|Q|}{r^2}$$

  • The electric field due to a positive charge points radially outward.
  • The electric field due to a negative charge points radially inward.

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  Vector Form of Electric Field

  The electric field vector at a point $P$ located at position $\mathbf{r}$ (relative to charge $Q$) is:

  $$\mathbf{E} = \frac{1}{4\pi \epsilon_0} \frac{Q}{r^3} \mathbf{r}$$

  where:

  • $\mathbf{r}$ is the position vector from charge $Q$ to the point $P$.
  • $r = |\mathbf{r}|$ is the distance between charge $Q$ and $P$.

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  Graphical Representation

  • The electric field around a positive charge forms radial outward lines.
  • The electric field around a negative charge forms radial inward lines.

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  Important Points

  1. Electric field follows an inverse-square law: $E \propto \frac{1}{r^2}$.
  2. The field is strongest near the charge and weakens with distance.
  3. The direction of the field depends on the sign of the charge.
  4. Electric field is a vector quantity, so it follows vector addition (superposition principle).

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  Example Calculation

  Problem: Find the electric field due to a charge $Q = 5 \mu C$ at a distance of 2 m in vacuum.

  Solution:
  Using the formula:

  $$E = \frac{1}{4\pi \epsilon_0} \frac{Q}{r^2}$$ $$E = \frac{9 \times 10^9}{(2)^2} \times 5 \times 10^{-6}$$ $$E = \frac{9 \times 10^9 \times 5 \times 10^{-6}}{4}$$ $$E = \frac{45 \times 10^3}{4} = 11250 \text{ N/C}$$

  So, the electric field is 11250 N/C outward (since $Q$ is positive).

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