Electrostatics Notes

Introduction

Electrostatics is a branch of physics that deals with the study of forces, fields, and potentials arising from static charges. It is a fundamental topic in the JEE Advanced Physics syllabus and is pivotal for understanding more complex concepts in electromagnetism. This study note will break down the key concepts of electrostatics, providing detailed explanations and examples to aid comprehension.

Basic Concepts of Electrostatics

Electric Charge

• Definition: Electric charge is a fundamental property of matter that causes it to experience a force when placed in an electric field.
• Types: There are two types of charges: positive and negative.
• Quantization of Charge: Charge is quantized, meaning it exists in discrete amounts. The smallest unit of charge is the elementary charge, $e$, where $e = 1.6 \times 10^{-19}$ C.

Coulomb's Law

• Statement: Coulomb's Law describes the force between two point charges. It states that the magnitude of the electrostatic force $F$ between two point charges $q_1$ and $q_2$ separated by a distance $r$ is given by:

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

where $k_e$ is the Coulomb constant, $k_e = 8.99 \times 10^9 , \text{Nm}^2/\text{C}^2$.

Electric Field

• Definition: The electric field $\mathbf{E}$ at a point in space is the force $\mathbf{F}$ experienced by a positive test charge $q_0$ placed at that point, divided by the magnitude of the test charge:

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

• Field Due to Point Charge: The electric field due to a point charge $q$ at a distance $r$ is given by:

$$ \mathbf{E} = k_e \frac{q}{r^2} \hat{r} $$

where $\hat{r}$ is the unit vector from the charge to the point of interest.

Electric Potential

• Definition: Electric potential $V$ at a point is the work done in bringing a unit positive charge from infinity to that point:

$$ V = k_e \frac{q}{r} $$

• Potential Difference: The potential difference between two points is the work done in moving a unit charge from one point to another:

$$ \Delta V = V_B - V_A $$

Gauss's Law

Statement

Gauss's Law relates the electric flux through a closed surface to the charge enclosed by that surface. Mathematically, it is expressed as:

$$ \oint \mathbf{E} \cdot d\mathbf{A} = \frac{q_{\text{enc}}}{\epsilon_0} $$

where $q_{\text{enc}}$ is the total charge enclosed within the surface, and $\epsilon_0$ is the permittivity of free space, $\epsilon_0 = 8.85 \times 10^{-12} , \text{C}^2/\text{Nm}^2$.

Applications of Gauss's Law

Spherical Symmetry

For a point charge $q$ at the center of a spherical surface of radius $r$:

$$ E \cdot 4\pi r^2 = \frac{q}{\epsilon_0} $$

Thus,

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

Cylindrical Symmetry

For an infinite line charge with linear charge density $\lambda$:

$$ E \cdot 2\pi rL = \frac{\lambda L}{\epsilon_0} $$

Thus,

$$ E = \frac{\lambda}{2\pi \epsilon_0 r} $$

Planar Symmetry

For an infinite plane sheet of charge with surface charge density $\sigma$:

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

Thus,

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

Capacitance

Definition

Capacitance $C$ is the ability of a system to store charge per unit potential difference. It is defined as:

$$ C = \frac{Q}{V} $$

where $Q$ is the charge stored, and $V$ is the potential difference.

Parallel Plate Capacitor

For a parallel plate capacitor with plate area $A$ and separation $d$:

$$ C = \epsilon_0 \frac{A}{d} $$

Energy Stored in a Capacitor

The energy $U$ stored in a capacitor is given by:

$$ U = \frac{1}{2} CV^2 $$

Conclusion

Electrostatics forms the foundation for understanding various phenomena in electromagnetism. Mastering the concepts of electric charge, electric field, electric potential, Gauss's Law, and capacitance is crucial for excelling in JEE Advanced Physics. Practice problems and real-world applications will further solidify these concepts and prepare you for the examination.