Chapter 1: Electric Charges and Fields - Comprehensive NEET Physics Notes

1. Fundamental Concepts

1.1 Electric Charge

Formula:
$q=ne$

Explanation:
Electric charge ($q$) is the physical property of matter that causes it to experience a force when placed in an electromagnetic field. The charge is quantized, meaning it is always an integral multiple of a basic unit of charge ($e=1.602\times 10^{-19}$ C).

Common Mistake:

Students often confuse the sign of the charge. Remember, electrons have a negative charge, while protons have a positive charge.

NEET Tip:

Charges add up algebraically. Always consider the signs when adding charges.

1.2 Coulomb’s Law

Formula:
$F=\frac{1}{4\pi {ϵ}_0}\cdot \frac{q_1{q}_2}{r^2}$

Explanation:
Coulomb's Law describes the electrostatic force between two point charges. The force ($F$) is directly proportional to the product of the charges ($q_1$ and $q_2$) and inversely proportional to the square of the distance ($r$) between them. ${ϵ}_0$ is the permittivity of free space.

Example Application:
Two charges, $q_1=2\times 10^{-6}$ C and $q_2=3\times 10^{-6}$ C, are placed 10 cm apart. The force between them is calculated as:

$F=\frac{9\times 10^9\cdot 2\times 10^{-6}\cdot 3\times 10^{-6}}{(0.1{)}^2}=5.4,\text{N}$

Common Mistake:

Students often forget to convert distances to meters when applying the formula.

1.3 Electric Field

Formula:
$E=\frac{F}{q}=\frac{1}{4\pi {ϵ}_0}\cdot \frac{Q}{r^2}$

Explanation:
The electric field ($E$) at a point in space is defined as the force ($F$) experienced by a unit positive charge placed at that point. It is a vector quantity and its direction is along the force experienced by a positive test charge.

Real-life Application:

Electric fields are used in capacitors, which store energy in the form of an electric field between two conductive plates.

Mnemonic:

"FEAR": Force equals Electric field times Another Radius (distance squared).

1.4 Electric Potential Energy

Formula:
$U=\frac{1}{4\pi {ϵ}_0}\cdot \frac{q_1{q}_2}{r}$

Explanation:
Electric potential energy ($U$) is the energy a charge possesses due to its position in an electric field. It's similar to gravitational potential energy but in the context of electric fields.

Common Misconception:

Electric potential energy is not the same as electric potential. Electric potential is energy per unit charge.

2. Electrostatics Principles

2.1 Principle of Superposition

Formula:
${\vec{F}}_{\text{total}}=\sum {\vec{F}}_i$

Explanation:
The principle of superposition states that the total force on any charge is the vector sum of the forces exerted by other individual charges.

NEET Problem-Solving Strategy:

Break down complex charge systems into individual pairs and apply Coulomb’s Law to each pair, then sum the forces vectorially.

2.2 Gauss’s Law

Formula:
$\oint \vec{E}\cdot d\vec{A}=\frac{q_{\text{enc}}}{{ϵ}_0}$

Explanation:
Gauss's Law relates the electric flux through a closed surface to the charge enclosed within the surface. It is particularly useful for calculating electric fields in symmetric charge distributions.

Real-life Application:

Gauss's Law is used to determine the electric field in configurations like spherical shells and infinite planes.

Common Mistake:

Confusing the total charge with the charge enclosed within the Gaussian surface.

Quick Recap

• Electric charge is quantized and follows Coulomb's Law.
• Electric field is the force per unit charge and is calculated using $E=\frac{F}{q}$.
• Electric potential energy depends on the position of the charge in an electric field.
• Superposition principle helps in calculating forces in systems with multiple charges.
• Gauss's Law is essential for simplifying electric field calculations in symmetric cases.

Practice Questions

1. Question: Two charges, $+3\times 10^{-6}$ C and $-3\times 10^{-6}$ C, are placed 5 cm apart. Calculate the force between them. Solution: Use Coulomb's Law to find $F=\frac{9\times 10^9\times (3\times 10^{-6}{)}^2}{(0.05{)}^2}=32.4$ N.
2. Question: Find the electric field at a point 0.2 m away from a charge of $5\times 10^{-9}$ C. Solution: Apply $E=\frac{1}{4\pi {ϵ}_0}\cdot \frac{Q}{r^2}$ to get $E=\frac{9\times 10^9\times 5\times 10^{-9}}{(0.2{)}^2}=1125$ N/C.
3. Question: What is the potential energy of a system of two charges, $4\times 10^{-9}$ C and $-4\times 10^{-9}$ C, separated by 3 cm? Solution: $U=\frac{9\times 10^9\times (4\times 10^{-9})\times (-4\times 10^{-9})}{0.03}=-4.8\times 10^{-6}$ J.
4. Question: Calculate the flux through a spherical surface that encloses a charge of $2\times 10^{-6}$ C. Solution: Using Gauss's Law, $\oint \vec{E}\cdot d\vec{A}=\frac{q_{\text{enc}}}{{ϵ}_0}=\frac{2\times 10^{-6}}{8.854\times 10^{-12}}=2.26\times 10^5$ Nm²/C.