Electric Charges and Fields — Class 12 Physics NCERT Solutions (Free)
Free step-by-step NCERT solutions for Class 12 Physics chapter "Electric Charges and Fields" — 8 important questions with detailed answers for CBSE board exam preparation.
TL;DR: Free step-by-step NCERT solutions for Class 12 Physics chapter "Electric Charges and Fields" — 8 important questions with detailed answers for CBSE bo…
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NCERT Solutions for Class 12 Physics — Electric Charges and Fields. Step-by-step answers to all 8 textbook questions from this chapter, written for CBSE board preparation and free to use.
Electric Charges and Fields — All 8 Questions Solved
Q1. Define electric charge. State the SI unit and symbol for electric charge.
Electric charge is the fundamental property of matter responsible for electromagnetic interactions. It is conserved and quantized.
SI Unit: Coulomb (C)
Symbol: q or Q
Quantization: q = ne, where n is an integer and e = 1.6 × 10⁻¹⁹ C (elementary charge)
Q2. State Coulomb's law. Two charges of 2 μC and 3 μC are separated by 30 cm. Calculate the electrostatic force between them.
Coulomb's Law: F = k(q₁q₂)/r²
where k = 9 × 10⁹ N⋅m²/C² (Coulomb's constant)
Given: q₁ = 2 μC = 2 × 10⁻⁶ C
q₂ = 3 μC = 3 × 10⁻⁶ C
r = 30 cm = 0.3 m
Step 1: F = (9 × 10⁹) × (2 × 10⁻⁶) × (3 × 10⁻⁶) / (0.3)²
Step 2: F = (9 × 10⁹) × (6 × 10⁻¹²) / 0.09
Step 3: F = 54 × 10⁻³ / 0.09
Step 4: F = 0.6 N
Final Answer: 0.6 N (repulsive if charges are like, attractive if opposite)
Q3. What is electric field? Write its definition and SI unit.
Electric Field (E) is the region around a charge where another charge experiences an electrostatic force.
Definition: E = F/q₀ (force per unit positive test charge)
SI Unit: N/C or V/m (Newton per Coulomb or Volt per meter)
Electric field due to point charge q at distance r:
E = kq/r² = q/(4πε₀r²)
where ε₀ = 8.85 × 10⁻¹² F/m (permittivity of free space)
Q4. A point charge of +5 μC is placed at a point. What is the electric field at a distance of 10 cm from this charge?
Given: q = +5 μC = 5 × 10⁻⁶ C
r = 10 cm = 0.1 m
k = 9 × 10⁹ N⋅m²/C²
Formula: E = kq/r²
Step 1: E = (9 × 10⁹) × (5 × 10⁻⁶) / (0.1)²
Step 2: E = (45 × 10³) / 0.01
Step 3: E = 4.5 × 10⁶ N/C
Final Answer: 4.5 × 10⁶ N/C (directed radially outward from positive charge)
Q5. What is an electric dipole? Define dipole moment and give its SI unit.
Electric Dipole: System of two equal and opposite charges separated by a small distance.
Dipole Moment (p): Product of charge magnitude and separation distance.
p = q × d
where q = magnitude of each charge
d = separation between charges (vector from negative to positive charge)
SI Unit: C⋅m (Coulomb-meter)
Direction: From negative charge toward positive charge
Alternatively: Can be expressed in terms of electron charge as Debye unit: 1 D = 3.33 × 10⁻³⁰ C⋅m
Q6. Derive the expression for electric field on the axial line of a dipole at distance r from center.
Consider dipole with charges +q and -q separated by distance 2a. Point P on axial line at distance r from center.
Field due to +q at distance (r - a):
E₊ = kq/(r - a)²
Field due to -q at distance (r + a):
E₋ = -kq/(r + a)²
Net field (E = E₊ + E₋):
E = kq/(r - a)² - kq/(r + a)²
E = kq[(r + a)² - (r - a)²] / [(r - a)²(r + a)²]
E = kq[4ar] / [(r - a)²(r + a)²]
For r >> a (dipole approximation):
E = 4kqa/r³ = 2kp/r³ = p/(2πε₀r³)
where p = 2qa (dipole moment)
Q7. State Gauss's law. Calculate the electric flux through a closed sphere of radius R with a point charge +q at its center.
Gauss's Law: Total electric flux through closed surface = enclosed charge / ε₀
Φ = ∮ E⋅dA = q_enclosed/ε₀
For sphere with +q at center:
Step 1: Due to spherical symmetry, E is radial and uniform on surface
Step 2: E = kq/R² = q/(4πε₀R²) at every point on surface
Step 3: Flux = E × Area = E × 4πR²
Step 4: Φ = [q/(4πε₀R²)] × 4πR²
Step 5: Φ = q/ε₀
Alternatively using Gauss directly: Φ = q_enclosed/ε₀ = q/ε₀
Final Answer: Φ = q/ε₀ (independent of radius R)
Q8. Find the electric field due to uniformly charged infinite plane sheet with surface charge density σ using Gauss's law.
Consider infinite plane sheet with uniform surface charge density σ. Use cylindrical Gaussian surface (pillbox) perpendicular to sheet.
Approach using Gauss's Law:
Step 1: By symmetry, field is perpendicular to sheet and same magnitude on both sides
Step 2: Choose cylindrical Gaussian surface with area A on each flat face
Step 3: Flux through curved surface = 0 (E parallel to surface)
Step 4: Total flux = E⋅A (top) + E⋅A (bottom) = 2EA
Step 5: Enclosed charge = σ⋅A
Step 6: By Gauss's law: 2EA = σA/ε₀
Step 7: E = σ/(2ε₀)
Final Answer: E = σ/(2ε₀) (perpendicular to sheet, independent of distance)
Direction: Away from sheet if σ > 0, toward sheet if σ < 0
Electric Charges and Fields — Practice MCQs with Answers
Attempt these 20 multiple-choice questions after working through the solutions above. Each carries the correct option and the reasoning behind it, so a wrong answer tells you which idea to revisit.
MCQ 1. Which of the following is NOT a fundamental property of electric charge?
- A. Quantization
- B. Conservation
- C. Additivity
- D. Polarity
Answer: D. Polarity — Electric charges exhibit quantization (e.g., e), conservation, and additivity. While charges have polarity (positive/negative), polarity itself is not listed as a fundamental property in the same category as quantization, conservation, or additivity.
MCQ 2. What is the SI unit of the permittivity of free space (ε₀)?
- A. N m² C⁻²
- B. N C⁻² m⁻²
- C. C² N⁻¹ m⁻²
- D. C N⁻¹ m⁻¹
Answer: C. C² N⁻¹ m⁻² — From Coulomb's Law, F = (1 / 4πε₀) (q₁q₂ / r²). Rearranging for ε₀, we get ε₀ = (1 / 4πF) (q₁q₂ / r²). Thus, the unit is C² N⁻¹ m⁻².
MCQ 3. Which statement about electric field lines is incorrect?
- A. They originate from positive charges and terminate on negative charges.
- B. They never intersect each other.
- C. They can form closed loops.
- D. The tangent to a field line at any point gives the direction of the electric field at that point.
Answer: C. They can form closed loops. — Electric field lines do not form closed loops; they start from positive charges and end on negative charges (or extend to infinity). Closed loops are characteristic of magnetic field lines.
MCQ 4. Two point charges, q₁ = +2 µC and q₂ = -4 µC, are placed 3 cm apart in air. What is the magnitude of the electrostatic force between them?
- A. 80 N
- B. 800 N
- C. 8 N
- D. 8000 N
Answer: A. 80 N — Using Coulomb's Law, F = k |q₁q₂| / r². F = (9 × 10⁹ N m²/C²) × ((2 × 10⁻⁶ C) × (4 × 10⁻⁶ C)) / (0.03 m)² = 80 N.
MCQ 5. A point charge +q is placed at the center of a cube of side 'a'. What is the electric flux passing through one face of the cube?
- A. q / ε₀
- B. q / (6ε₀)
- C. q / (4πε₀)
- D. q / (12ε₀)
Answer: B. q / (6ε₀) — By Gauss's Law, the total flux through the closed surface (the cube) is q / ε₀. Since the charge is at the center, the flux is distributed symmetrically among the six identical faces, so the flux through one face is q / (6ε₀).
MCQ 6. An electric dipole is placed in a uniform electric field. What is the net force experienced by the dipole?
- A. Zero
- B. Depends on the orientation of the dipole
- C. Always directed opposite to the field
- D. Always directed along the field
Answer: A. Zero — In a uniform electric field, the forces on the positive and negative charges of the dipole are equal in magnitude and opposite in direction. Therefore, the net force on the dipole is zero, although it may experience a torque.
MCQ 7. The electric field at a point is defined as the force experienced by a unit __________ kept at that point.
- A. Negative charge
- B. Positive charge
- C. Neutral particle
- D. Mass
Answer: B. Positive charge — The electric field strength at a point is defined as the electrostatic force per unit positive test charge (E = F/q₀) acting on a vanishingly small positive test charge placed at that point.
MCQ 8. A point charge of +45 nC is placed at the origin. What is the magnitude of the electric field at a point P located at (3 m, 0)?
- A. 5 N/C
- B. 15 N/C
- C. 45 N/C
- D. 9 N/C
Answer: C. 45 N/C — The electric field due to a point charge is given by E = k |q| / r². Here, q = 45 × 10⁻⁹ C and r = 3 m. So, E = (9 × 10⁹) × (45 × 10⁻⁹) / (3²) = 45 N/C.
MCQ 9. Three charges, +q, +q, and -q, are placed at the vertices of an equilateral triangle of side 'a'. What is the magnitude of the net force on the charge -q?
- A. k q² / a²
- B. √3 k q² / a²
- C. 2 k q² / a²
- D. k q² / (2a²)
Answer: B. √3 k q² / a² — Each +q charge exerts an attractive force of magnitude F = k q² / a² on the -q charge. Since the triangle is equilateral, the angle between these two force vectors is 60°. The resultant force is F_net = √(F² + F² + 2F² cos(60°)) = √(3F²) = √3 F = √3 k q² / a².
MCQ 10. When a glass rod is rubbed with silk, it acquires a positive charge. This is because:
- A. Protons are transferred from the silk to the glass rod.
- B. Electrons are transferred from the silk to the glass rod.
- C. Electrons are transferred from the glass rod to the silk.
- D. Protons are transferred from the glass rod to the silk.
Answer: C. Electrons are transferred from the glass rod to the silk. — When a glass rod is rubbed with silk, the glass rod loses electrons to the silk, making the glass rod positively charged and the silk negatively charged.
MCQ 11. An electric dipole of moment 'p' is placed in a uniform electric field 'E'. The dipole is initially aligned parallel to the field. How much work is required to rotate it by 90 degrees?
- A. pE
- B. pE/2
- C. 2pE
- D. 0
Answer: A. pE — The work done in rotating an electric dipole in a uniform electric field is given by W = U_f - U_i = -pE(cosθ_f - cosθ_i). Initially, θ_i = 0° (parallel), so cosθ_i = 1. Finally, θ_f = 90°, so cosθ_f = 0. Thus, W = -pE(0 - 1) = pE.
MCQ 12. Electric field lines provide information about: I. Field strength II. Direction III. Nature of charge
- A. I only
- B. I and II only
- C. II and III only
- D. I, II and III
Answer: D. I, II and III — The density of field lines indicates the field strength, the tangent to a field line gives the direction of the field, and the origin/termination of lines indicates the nature of the charges (originate from positive, terminate on negative).
MCQ 13. Which of the following values of charge is NOT possible?
- A. 1.6 × 10⁻¹⁹ C
- B. 3.2 × 10⁻¹⁹ C
- C. 4.8 × 10⁻¹⁹ C
- D. 2.4 × 10⁻¹⁹ C
Answer: D. 2.4 × 10⁻¹⁹ C — Charge is quantized, meaning it must be an integral multiple of the elementary charge, e = 1.6 × 10⁻¹⁹ C. 2.4 × 10⁻¹⁹ C is not an integral multiple of e.
MCQ 14. If the net electric flux through a closed surface is zero, then:
- A. The electric field everywhere on the surface is zero.
- B. The net charge enclosed by the surface is zero.
- C. The electric potential everywhere on the surface is zero.
- D. The charge distribution is uniform inside the surface.
Answer: B. The net charge enclosed by the surface is zero. — According to Gauss's Law, the net electric flux through a closed surface is equal to the net charge enclosed divided by ε₀. If the net flux is zero, it implies that the net charge enclosed by the surface is zero.
MCQ 15. Two point charges +Q and -Q are placed at points (0, d) and (0, -d) respectively. What is the electric field at a point (x, 0) on the x-axis?
- A. 1/(4πε₀) * (2Qd / (x²+d²)^³/²) along x-axis
- B. 1/(4πε₀) * (2Qx / (x²+d²)^³/²) along y-axis
- C. 1/(4πε₀) * (2Qd / (x²+d²)^³/²) along negative y-axis
- D. 1/(4πε₀) * (2Qd / (x²+d²)^³/²) along positive y-axis
Answer: C. 1/(4πε₀) * (2Qd / (x²+d²)^³/²) along negative y-axis — This setup forms an electric dipole along the y-axis. At point (x,0), the x-components of the fields due to +Q and -Q cancel. The y-components add up, both pointing along the negative y-axis, resulting in E_net = (1 / 4πε₀) * (2Qd / (x²+d²)^³/²) along the negative y-axis.
MCQ 16. The direction of the electric field intensity at a point is:
- A. Always perpendicular to the force acting on a test charge.
- B. Always opposite to the force acting on a test charge.
- C. In the direction of the force acting on a positive test charge.
- D. In the direction of the force acting on a negative test charge.
Answer: C. In the direction of the force acting on a positive test charge. — The electric field vector E at a point is defined as the force F experienced by a positive test charge q₀ at that point, i.e., E = F/q₀. Therefore, the direction of E is the same as the direction of F for a positive test charge.
MCQ 17. An electric dipole consists of two opposite charges of magnitude 1 µC separated by a distance of 2 cm. What is the magnitude of its electric dipole moment?
- A. 2 × 10⁻⁸ C m
- B. 2 × 10⁻⁶ C m
- C. 1 × 10⁻⁸ C m
- D. 1 × 10⁻⁶ C m
Answer: A. 2 × 10⁻⁸ C m — The electric dipole moment 'p' is given by p = q × 2a, where q is the magnitude of the charge and 2a is the separation distance. Here, q = 1 × 10⁻⁶ C and 2a = 2 × 10⁻² m. So, p = (1 × 10⁻⁶ C) × (2 × 10⁻² m) = 2 × 10⁻⁸ C m.
MCQ 18. A conductor with a cavity has a charge 'Q' placed inside the cavity. The surface of the conductor is uncharged. Which of the following statements is true?
- A. The electric field inside the cavity is zero.
- B. An equal and opposite charge (-Q) appears on the outer surface of the conductor.
- C. An equal and opposite charge (-Q) appears on the inner surface of the conductor.
- D. The electric field outside the conductor is zero.
Answer: C. An equal and opposite charge (-Q) appears on the inner surface of the conductor. — When a charge Q is placed in a cavity of an isolated conductor, by induction, an equal and opposite charge -Q is induced on the inner surface of the conductor, and an equal charge +Q is induced on the outer surface of the conductor. The electric field inside the conductor material is zero.
MCQ 19. Two point charges +q and +4q are kept at a distance 'r' from each other. At what point on the line joining the charges is the electric field zero?
- A. At a distance r/3 from +q
- B. At a distance 2r/3 from +q
- C. At a distance r/2 from +q
- D. At a distance r/5 from +q
Answer: A. At a distance r/3 from +q — Let the point be at a distance x from +q. For the field to be zero, k q/x² = k (4q)/(r-x)². This simplifies to 1/x = 2/(r-x), which gives r - x = 2x, so r = 3x or x = r/3.
MCQ 20. Electric flux is a measure of:
- A. The total number of electric field lines passing through a given area.
- B. The electric potential at a point.
- C. The strength of the electric field.
- D. The charge enclosed by a surface.
Answer: A. The total number of electric field lines passing through a given area. — Electric flux is defined as the total number of electric field lines passing normally through a given surface. It is a measure of the 'flow' of the electric field through an area.
How to Use These Solutions
Attempt each question yourself first and only then compare with the worked answer. Marks in the CBSE board exam are awarded for the METHOD as much as the final result, so reproduce the steps rather than memorising the last line. Where a solution states a law, a formula or a definition, learn that wording — examiners look for it.
Related practice for this chapter: chapter MCQs, previous-year questions, revision notes and sample papers.