Class 12 Physics MCQ Test – Electric charges and fields (Chapter 1) | Full Chapter Test
Class 12 Physics – Chapter 1: Electric Charges and Fields (Detailed Notes)
These notes are designed for CBSE, GSEB, JEE & NEET students. Includes theory, formulas, derivation concepts, and numerical-based understanding.
1. Electric Charge
Electric charge is a fundamental property of matter due to which it experiences force in an electric field.
- Types: Positive (+) and Negative (−)
- SI Unit: Coulomb (C)
- Symbol: \( q \)
Properties of Charge
- Additivity: Total charge is algebraic sum
- Conservation: Charge cannot be created or destroyed
- Quantization: \( q = ne \), where \( e = 1.6 \times 10^{-19} C \)
2. Coulomb’s Law
The electrostatic force between two point charges is directly proportional to product of charges and inversely proportional to square of distance.
\( F = \frac{1}{4\pi \varepsilon_0} \cdot \frac{q_1 q_2}{r^2} \)
- \( \varepsilon_0 \) = Permittivity of free space
- Value: \( \frac{1}{4\pi \varepsilon_0} = 9 \times 10^9 \, Nm^2/C^2 \)
Nature of Force:
- Like charges → Repulsion
- Unlike charges → Attraction
3. Principle of Superposition
Net force on a charge is vector sum of forces due to all other charges.
\( \vec{F}_{net} = \vec{F}_1 + \vec{F}_2 + \vec{F}_3 + ... \)
4. Electric Field
Electric field at a point is the force experienced by unit positive charge.
\( E = \frac{F}{q} \)
For point charge:
\( E = \frac{1}{4\pi \varepsilon_0} \cdot \frac{q}{r^2} \)
- Unit: N/C
- Direction: Along force on positive charge
5. Electric Field Lines
- Originate from positive and end at negative charges
- Never intersect
- Density represents strength
- Closer lines → Strong field
6. Electric Dipole
A pair of equal and opposite charges separated by small distance.
Dipole moment: \( \vec{p} = q \cdot \vec{2a} \)
- Direction: From negative to positive
- Unit: C·m
Electric Field of Dipole
Axial Line:
\( E_{axial} = \frac{1}{4\pi \varepsilon_0} \cdot \frac{2p}{r^3} \)
Equatorial Line:
\( E_{equatorial} = \frac{1}{4\pi \varepsilon_0} \cdot \frac{p}{r^3} \)
7. Electric Flux
Electric flux is measure of electric field passing through a surface.
\( \Phi = \vec{E} \cdot \vec{A} = EA \cos \theta \)
- Unit: \( Nm^2/C \)
8. Gauss’s Law
Total electric flux through closed surface is equal to charge enclosed divided by permittivity.
\( \Phi = \frac{Q_{enc}}{\varepsilon_0} \)
Applications
- Infinite line charge
- Infinite plane sheet
- Spherical shell
9. Electric Field due to Different Charge Distributions
1. Infinite Line Charge
\( E = \frac{\lambda}{2\pi \varepsilon_0 r} \)
2. Infinite Plane Sheet
\( E = \frac{\sigma}{2\varepsilon_0} \)
3. Charged Spherical Shell
- Outside: \( E = \frac{1}{4\pi \varepsilon_0} \frac{Q}{r^2} \)
- Inside: \( E = 0 \)
10. Important Exam Points
- Coulomb’s law is inverse square law
- Field is vector quantity
- Dipole field varies as \( \frac{1}{r^3} \)
- Flux depends on area orientation
- Gauss law depends only on enclosed charge
⬇️ Attempt the test given below after revising these notes
VEDANT CLASSES
Class 12 Physics MCQ Test – Electric Charges & Fields (Full Chapter Test)
Class 12 Physics Chapter 1: Electric Charges and Fields
This chapter is one of the most fundamental topics in Class 12 Physics. It builds the base for understanding electrostatics, which is further used in chapters like Electrostatic Potential, Current Electricity, and Magnetism. A strong grasp of this chapter is essential for board exams as well as competitive exams like JEE and NEET.
Learning Outcomes
After completing this chapter, students will be able to:
- Understand the concept of electric charge and its properties (conservation and quantization).
- Differentiate between conductors and insulators.
- Understand methods of charging:
- Charging by friction
- Charging by conduction
- Charging by induction
- Apply Coulomb’s Law to calculate electrostatic force between charges.
- Understand the principle of superposition of charges.
- Define and calculate electric field and electric field intensity.
- Draw and interpret electric field lines and their properties.
- Understand electric dipole and calculate dipole moment.
- Apply Gauss’s Law for symmetrical charge distributions.
Real Life Applications
Concepts from this chapter are widely used in daily life and technology:
- Electrostatic Precipitators: Used in industries to remove dust and smoke particles from exhaust gases.
- Photocopiers and Printers: Work on the principle of electrostatic charges.
- Lightning Protection: Lightning conductors protect buildings by safely transferring charge to the ground.
- Capacitive Touchscreens: Used in smartphones and tablets.
- Painting Industry: Electrostatic spray painting ensures uniform coating.
Common Mistakes Students Make
- Forgetting that electric charge is a scalar quantity.
- Wrong application of Coulomb’s Law (especially sign conventions).
- Ignoring vector nature of electric field.
- Confusing electric field with electric force.
- Incorrect direction of electric field lines.
- Applying Gauss’s Law in non-symmetrical situations incorrectly.
Conclusion
The chapter “Electric Charges and Fields” is highly scoring if concepts are clear and regularly practiced. Focus on diagrams, vector concepts, and numerical problems. Consistent revision and solving previous year questions will help you secure excellent marks.
Class 12 Physics Chapter 1: Electric Charges and Fields
10 Detailed Solved Examples
Calculate the force between two charges 2 µC and 3 µC placed 0.5 m apart in air. (k = 9 × 10⁹ N m²/C²)
Solution:
Step 1: Write formula
F = k q₁q₂ / r²
Step 2: Substitute values
F = (9×10⁹ × 2×10⁻⁶ × 3×10⁻⁶) / (0.5)²
Step 3: Calculate
F = (54×10⁻³) / 0.25 = 0.216 N
Find electric field at a point 0.2 m away from a charge of 5 µC.
Solution:
E = kq / r²
E = (9×10⁹ × 5×10⁻⁶) / (0.2)²
E = (45×10³) / 0.04 = 1.125 × 10⁶ N/C
A charge of 2 C is placed in an electric field of 10 N/C. Find the force on it.
Solution:
F = qE = 2 × 10 = 20 N
Two charges +2 C and +3 C act on a point with forces 4 N and 6 N in same direction. Find net force.
Solution:
Net force = 4 + 6 = 10 N
Two equal charges of 2 µC are separated by 0.1 m. Calculate dipole moment.
Solution:
p = q × d = 2×10⁻⁶ × 0.1 = 2 × 10⁻⁷ C·m
Find electric field at a point on axial line of dipole. (Use p = 2×10⁻⁷ C·m, r = 0.5 m)
Solution:
E = (1/4πε₀) × (2p / r³)
E = (9×10⁹ × 2×2×10⁻⁷) / (0.5)³
E = (36×10²) / 0.125 = 2.88 × 10⁴ N/C
Find electric flux through a closed surface enclosing charge 5 C. (ε₀ = 8.85×10⁻¹²)
Solution:
Φ = q / ε₀ = 5 / (8.85×10⁻¹²)
Φ = 5.65 × 10¹¹ N·m²/C
Find electric field at distance r from infinite line charge λ.
Solution:
E = λ / (2πε₀ r)
(Formula based answer)
Find electric field outside a charged sphere.
Solution:
E = kQ / r² (same as point charge)
Find number of electrons in charge 1 C.
Solution:
n = Q / e = 1 / (1.6×10⁻¹⁹)
n = 6.25 × 10¹⁸ electrons
10 Detailed FAQs
It is a fundamental property of matter responsible for electric force.
Charge is conserved, quantized, and additive in nature.
It gives force between two charges and depends on distance and magnitude.
Region around a charge where another charge experiences force.
Imaginary line representing direction of electric field.
Net force is vector sum of individual forces.
Pair of equal and opposite charges separated by small distance.
Electric flux is proportional to enclosed charge.
Measure of electric field passing through a surface.
It forms the base of entire electrostatics and modern physics applications.
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