Class 12 Physics Chapter 1: Electric Charges and Fields – Notes, Formulas & Board Questions

Class 12 Physics Chapter 1: Electric Charges and Fields – Notes, Formulas & Questions

1. Chapter Summary

The first chapter of Class 12 Physics, Electric Charges and Fields, introduces the fundamental concept of electrostatics. It deals with electric charges, their properties, interactions, and the fields they produce. The chapter builds the foundation for the entire Electrostatics Unit, which is highly important for both CBSE Board Exams and competitive exams like JEE/NEET.

Key highlights of this chapter include:

• The basic nature and quantization of electric charge.
• Coulomb’s law and its vector form.
• Superposition principle of charges.
• Continuous charge distributions and charge density.
• Electric field and field lines.
• Electric flux and Gauss’s law with applications.

This chapter emphasizes understanding the laws governing charges and the mathematical tools required to calculate electric fields.

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2. Important Concepts

(a) Electric Charge

• Fundamental property of matter responsible for electric interactions.
• Two types: positive and negative.
• Charges are quantized: 

• $q = \pm ne$, where $e = 1.6 \times 10^{-19} C$.

• Charges are conserved – cannot be created or destroyed, only transferred.

(b) Coulomb’s Law

• Force between two point charges:

$$F = \frac{1}{4\pi\epsilon_0} \cdot \frac{q_1 q_2}{r^2}$$

• $\epsilon_0 = 8.85 \times 10^{-12} C^2/Nm^2$.

• Force is along the line joining the charges and obeys the inverse square law.

(c) Principle of Superposition

• The net force on a charge due to a system of charges is the vector sum of individual forces.

(d) Electric Field (E)

• Electric field due to a point charge:

$$E = \frac{1}{4\pi\epsilon_0} \cdot \frac{q}{r^2}$$

• Direction: outward for +ve charge, inward for –ve charge.

(e) Electric Field Lines

• Imaginary lines representing field direction.

• Properties:

  Originate from positive charges and terminate on negative charges.
• Never intersect.
• Density of lines ∝ magnitude of field.

(f) Electric Dipole

• System of two equal and opposite charges separated by distance $2a$.

• Dipole moment:

$$\vec{p} = q \cdot 2a \, (\text{along negative to positive charge})$$

• Field due to dipole on axial line:

$$E = \frac{1}{4\pi\epsilon_0} \cdot \frac{2p}{r^3}$$

• Field due to dipole on equatorial line:

$$E = \frac{1}{4\pi\epsilon_0} \cdot \frac{p}{r^3}$$

(g) Gauss’s Law

$$\oint \vec{E} \cdot d\vec{A} = \frac{q_{\text{enclosed}}}{\epsilon_0}$$

• Useful for symmetric charge distributions (sphere, cylinder, sheet).

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3. Important Formulas

1. Coulomb’s Law:

   $$F = \frac{1}{4\pi\epsilon_0} \cdot \frac{q_1 q_2}{r^2}$$

2. Electric Field (point charge):

   $$E = \frac{1}{4\pi\epsilon_0} \cdot \frac{q}{r^2}$$

3. Electric Dipole Moment:

   $$p = q \times 2a$$

4. Field due to Dipole (Axial line):

   $$E = \frac{1}{4\pi\epsilon_0} \cdot \frac{2p}{r^3}$$

5. Field due to Dipole (Equatorial line):

   $$E = \frac{1}{4\pi\epsilon_0} \cdot \frac{p}{r^3}$$

6. Torque on Dipole in Electric Field:

   $$\tau = pE \sin\theta$$

7. Gauss’s Law:

   $$\oint \vec{E} \cdot d\vec{A} = \frac{q}{\epsilon_0}$$

8. Field due to Charged Sphere:

   • Outside: $E = \frac{1}{4\pi\epsilon_0} \cdot \frac{Q}{r^2}$

   • Inside: $E = 0$ (for conductor)

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4. Board Pattern Questions & Answers

Q1. State Coulomb’s law in vector form. (1 mark)

Answer:

$$\vec{F} = \frac{1}{4\pi\epsilon_0} \cdot \frac{q_1 q_2}{r^2} \hat{r}$$

Where $\hat{r}$ is the unit vector along the line joining the two charges.

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Q2. Define electric flux. State its SI unit. (2 marks)

Answer:
Electric flux through a surface is the number of electric field lines passing normally through that surface.

$$\phi_E = \vec{E} \cdot \vec{A}$$

SI unit: Newton meter² per Coulomb (Nm²/C).

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Q3. Derive expression for electric field on axial line of an electric dipole. (3 marks)

Answer:

• Consider dipole with charges $+q$ and $-q$ separated by $2a$.

• At point $P$ on axial line, distance from center = $r$.

• Net field:

$$E = \frac{1}{4\pi\epsilon_0} \cdot \frac{2p}{r^3}$$

where $p = q \times 2a$.

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Q4. Explain Gauss’s law. Derive expression for field due to uniformly charged spherical shell. (5 marks)

Answer:
Gauss’s law states:

$$\oint \vec{E} \cdot d\vec{A} = \frac{q_{\text{enclosed}}}{\epsilon_0}$$

• For a spherical shell of charge $Q$:

  • Outside (r > R): $E = \frac{1}{4\pi\epsilon_0} \cdot \frac{Q}{r^2}$

  • Inside (r < R): $E = 0$.

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Q5. Previous Year Question (CBSE 2024):

“A charge $+Q$ is uniformly distributed on a spherical shell. Draw electric field lines and explain the field pattern inside and outside the shell.” (3 marks)

Answer:

• Inside the shell: E = 0 (no field lines inside).
• Outside the shell: Field is similar to that due to a point charge Q at center.
• Field lines are radially outward.

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5. Quick Revision Notes

• Charges are quantized and conserved.
• Coulomb’s law follows inverse square law.
• Superposition principle is vector addition of forces/fields.
• Dipole has axial and equatorial fields.
• Gauss’s law simplifies calculation of electric field for symmetric distributions.

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This completes a Class 12 Physics Chapter 1 blog post with summary, key concepts, all formulas, and board-style Q&A including PYQs.

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