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Samacheer Kalvi 12th Physics Chapter 1 Exercise Solutions

Tamil Nadu Class 12 Physics Chapter 1 Electrostatics exercise solutions with Coulomb law, electric field, potential, capacitance, formulas, numerical patterns, and exam tips.

Updated 25 May 202615 min read

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Step-by-step answers for all textbook questions in this chapter

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Chapter 1 — Electrostatics Exercise Solutions

Electrostatics questions usually test formula selection, sign convention, and unit conversion. Write the formula first before substituting values.

How to Use This Page

  • Read the Key Concepts and Formula Recap tables first to refresh the basics before attempting questions.
  • Try each MCQ and short-answer question on your own, then check the answer and explanation.
  • For numerical problems, follow the Numerical Answer Pattern step by step — always write the formula before substituting.
  • Use the Common Mistakes to Avoid section as a final review before your exam to catch frequent errors.
  • Revisit the Exam Checklist the night before the exam to make sure you have covered every essential point.

Key Concepts (Quick Recap)

TermDefinition
Coulomb's lawThe force between two point charges is directly proportional to the product of the charges and inversely proportional to the square of the distance between them.
Electric fieldThe region around a charge where another charge experiences a force; defined as force per unit positive test charge.
Electric field linesImaginary lines that show the direction of the electric field; they start on positive charges and end on negative charges and never cross.
Electric potentialThe work done per unit charge in bringing a test charge from infinity to a point in the electric field.
Equipotential surfaceA surface where every point has the same electric potential; work done to move a charge on it is zero.
Gauss's lawThe total electric flux through a closed surface equals the net charge enclosed divided by the permittivity of free space.
Electric dipoleA pair of equal and opposite charges separated by a small distance; characterised by dipole moment p = q × 2a.
CapacitanceThe ability of a conductor to store charge; defined as C = Q / V and measured in farads.
DielectricAn insulating material placed between capacitor plates that increases capacitance by reducing the electric field.
Van de Graaff generatorA device that uses a moving belt to accumulate charge on a large metal dome, producing very high voltages.
Electric fluxThe number of electric field lines passing through a given area; denoted by φ and measured in N·m²/C.
PermittivityA measure of how easily a medium allows electric field lines to pass through it; ε₀ for free space, ε = ε₀εᵣ for a medium.
Superposition principleThe net force or field due to multiple charges is the vector sum of the forces or fields due to each charge individually.

I. Choose the Best Answer (MCQs)

1. Two point charges +q and +4q are separated by a distance r. The point where the electric field is zero lies (a) at r/2 from +q (b) at r/3 from +q (c) at r/5 from +q (d) at 2r/3 from +q
Answer(c) — The field is zero where kq/x² = k(4q)/(r − x)², giving x = r/5 from +q.
2. The SI unit of electric flux is (a) N/C (b) N·m²/C (c) C/m² (d) V/m
Answer(b) — Electric flux = E × A, so its unit is (N/C)(m²) = N·m²/C.
3. If the distance between two point charges is doubled, the Coulomb force between them becomes (a) half (b) one-fourth (c) double (d) four times
Answer(b) — Coulomb force is inversely proportional to r², so doubling r reduces force to 1/4.
4. A parallel plate capacitor with a dielectric of constant K has capacitance (a) C/K (b) KC (c) C/K² (d) K²C
Answer(b) — A dielectric increases capacitance by factor K, so C' = KC.
5. The work done in moving a charge along an equipotential surface is (a) always positive (b) always negative (c) zero (d) depends on the path
Answer(c) — Potential difference between any two points on an equipotential surface is zero, so work done is zero.
6. Gauss's law is useful for calculating electric field when the charge distribution has (a) arbitrary shape (b) spherical, cylindrical, or planar symmetry (c) only point charges (d) only dipole arrangements
Answer(b) — Gauss's law simplifies field calculations only when a Gaussian surface can exploit symmetry.
7. The electric field inside a charged conductor is (a) maximum at the centre (b) zero everywhere (c) proportional to the charge (d) proportional to the radius
Answer(b) — In electrostatic equilibrium, charges reside on the surface; the field inside a conductor is zero.
8. The energy stored in a capacitor charged to voltage V is (a) CV (b) ½CV² (c) ½C²V (d) QV
Answer(b) — The formula for energy stored is U = ½CV² = ½QV = Q²/(2C).
9. The dipole moment of an electric dipole with charges ±q separated by distance 2a is (a) qa (b) 2qa (c) q²a (d) q/2a
Answer(b) — Dipole moment p = q × 2a, the product of charge and the full separation.
10. If the permittivity of a medium is ε, the relative permittivity (dielectric constant) is (a) ε/ε₀ (b) ε₀/ε (c) εε₀ (d) 1/(εε₀)
Answer(a) — Relative permittivity εᵣ = ε/ε₀ by definition.

II. Fill in the Blanks

1. The value of Coulomb's constant k is ________ N·m²/C². Answer: 9 × 10⁹

2. The SI unit of electric field intensity is ________. Answer: N/C (or V/m)

3. The electric potential at infinity is taken as ________. Answer: zero

4. The relation between electric field E and electric potential V is E = ________. Answer: −dV/dr

5. The capacitance of a parallel plate capacitor with area A, separation d, and no dielectric is C = ________. Answer: ε₀A/d

6. According to Gauss's law, the total flux through a closed surface equals Q/________. Answer: ε₀

7. The torque on an electric dipole in a uniform field is τ = ________. Answer: pE sin θ

8. Two capacitors C₁ and C₂ in series have equivalent capacitance C = ________. Answer: C₁C₂ / (C₁ + C₂)

III. True or False (with Corrections)

1. Electric field lines can cross each other. False. Electric field lines never cross because at the point of crossing there would be two directions of the field, which is impossible.

2. The electric potential due to a positive charge is negative. False. The electric potential due to a positive charge is positive because V = kq/r and q is positive.

3. The capacitance of a capacitor depends on the charge stored. False. Capacitance depends only on geometry (plate area, separation) and the dielectric medium, not on the charge or voltage.

4. An equipotential surface is always perpendicular to electric field lines. True. No work is done moving a charge along an equipotential surface, so the field must be perpendicular to it.

5. Gauss's law applies only to symmetrical charge distributions. False. Gauss's law is valid for any closed surface and any charge distribution, but it is useful for calculation only when symmetry exists.

6. The electric field inside a uniformly charged spherical shell is zero. True. By Gauss's law, the charge enclosed by a Gaussian surface inside the shell is zero, so the field is zero.

7. The energy stored in a capacitor is proportional to the charge. False. Energy U = Q²/(2C), so it is proportional to the square of the charge, not the charge itself.

IV. Match the Following

Column AColumn B
1. Coulomb's lawA. Energy stored in capacitor
2. Gauss's lawB. Force between point charges
3. CapacitanceC. C = ε₀A/d
4. Electric potentialD. Q = φε₀
5. Dipole momentE. Work done per unit charge
6. Parallel plate capacitorF. p = q × 2a
Answer1—B, 2—D, 3—C (capacitance formula for parallel plate), 4—E, 5—F, 6—A (energy relates to capacitor storage).

Note: Match 3 with C and 6 with A; each item in Column B is used once.

V. Answer Briefly (2-mark)

1. Define electric field at a point.
Answer

The electric field at a point is the force experienced by a unit positive test charge placed at that point, without disturbing the source charge. Mathematically, E = F/q₀, where F is the force and q₀ is the test charge. Its direction is the direction of force on a positive charge.

2. State Gauss's law.
Answer

Gauss's law states that the total electric flux through any closed surface is equal to the net charge enclosed divided by the permittivity of free space. In equation form:

φ = Q_enclosed / ε₀

This law relates the distribution of electric charge to the resulting electric field.

3. What is an equipotential surface?
Answer

An equipotential surface is a surface where every point has the same electric potential. No work is done in moving a test charge along such a surface. Equipotential surfaces are always perpendicular to electric field lines and never intersect each other.

4. What is a dielectric? Give an example.
Answer

A dielectric is an insulating material that does not conduct electricity but gets polarised when placed in an electric field, thereby reducing the effective field inside it. Common examples include glass, mica, and paper. A dielectric increases the capacitance of a capacitor by a factor equal to its dielectric constant K.

5. State the superposition principle for electric fields.
Answer

The superposition principle states that the net electric field at a point due to a group of charges is the vector sum of the individual electric fields produced by each charge at that point alone. This means:

E_net = E₁ + E₂ + E₃ + ... (vector sum)

Each field is calculated independently as if the other charges were absent.

VI. Answer in Detail (5-mark)

1. Derive Coulomb's law and explain its vector form.
Answer

Coulomb's law describes the electrostatic force between two point charges. Consider two charges q₁ and q₂ separated by a distance r in vacuum. The magnitude of the force is directly proportional to the product of the charges and inversely proportional to the square of the distance:

F = k |q₁||q₂| / r²

where k = 1/(4πε₀) = 9 × 10⁹ N·m²/C² is Coulomb's constant. In vector form, the force on q₂ due to q₁ is:

F₂₁ = k q₁q₂ / r² · r̂₂₁

where r̂₂₁ is the unit vector pointing from q₁ to q₂. If both charges are like charges, the force is repulsive; if they are unlike, it is attractive. The force obeys Newton's third law — both charges experience equal and opposite forces. Coulomb's law is valid only for point charges at rest and in vacuum or air as an approximation.

2. State Gauss's law and apply it to find the electric field due to a uniformly charged infinite plane sheet.
Answer

Gauss's law states that the total electric flux through a closed surface equals the net charge enclosed divided by ε₀. For a uniformly charged infinite plane sheet with surface charge density σ, we choose a Gaussian surface in the form of a pillbox (cylinder) of cross-section A that pierces the sheet symmetrically. By symmetry, the electric field is perpendicular to the sheet and has the same magnitude on both sides. The flux through the curved surface is zero because E is parallel to the surface. The flux through each flat face is EA, so total flux = 2EA. The charge enclosed = σA. Applying Gauss's law:

2EA = σA / ε₀ E = σ / (2ε₀)

The field is uniform and independent of the distance from the sheet, pointing away from the sheet if σ is positive.

3. Derive the expression for capacitance of a parallel plate capacitor with a dielectric slab between the plates.
Answer

A parallel plate capacitor has two plates of area A separated by distance d. Without a dielectric, the field between the plates is E₀ = V/d and capacitance is C₀ = ε₀A/d. When a dielectric slab of dielectric constant K and thickness d (fully filling the gap) is inserted, the electric field inside the dielectric reduces to E = E₀/K. The potential difference becomes V = Ed = (E₀/K) × d = V₀/K. The new capacitance is:

C = Q/V = Q/(V₀/K) = K × Q/V₀ = K C₀ C = K ε₀ A / d

Thus, inserting a dielectric increases the capacitance by factor K. If the slab is of thickness t < d, the effective capacitance can be calculated treating it as two capacitors in series.

4. Derive the expression for electric potential due to a point charge.
Answer

Consider a point charge q at the origin. The electric potential at a point P at distance r from q is defined as the work done per unit charge in bringing a test charge from infinity to P. The electric field due to q at distance x is:

E = kq / x²

The work done in moving a test charge q₀ through a small distance dx against the field is dW = q₀ E dx. The total work from infinity to r is:

W = ∫(∞ to r) q₀ E dx = q₀ kq ∫(∞ to r) dx / x² = q₀ kq [-1/x](∞ to r) = q₀ kq / r

Therefore, the electric potential is:

V = W / q₀ = kq / r

For a positive charge, V is positive; for a negative charge, V is negative. Potential is a scalar quantity and follows the superposition principle.

Formula Recap

ConceptFormula
Coulomb forceF = (1 / 4 pi epsilon0) q1q2 / r²
Electric fieldE = F / q
Field due to point chargeE = kq / r²
Electric potentialV = kq / r
CapacitanceC = Q / V
Energy in capacitorU = 1/2 CV²

Numerical Answer Pattern

1. Write the given values with SI units. 2. Choose Coulomb law, electric field, potential, or capacitance formula. 3. Convert microcoulomb or nanocoulomb values to coulomb. 4. Substitute and simplify. 5. Write the unit in the final answer.

Short Answer Points

  • Electric field is the force experienced by a unit positive charge.
  • Electric potential is work done per unit charge.
  • Equipotential surfaces have zero work done along them.
  • Capacitance depends on plate area, separation, and medium.

Worked Examples

Example 1 — Coulomb's law

Find the force between charges of 2 μC and 3 μC placed 0.3 m apart in air.

F = k q₁q₂ / r² = (9×10⁹)(2×10⁻⁶)(3×10⁻⁶) / (0.3)² = (9×10⁹)(6×10⁻¹²) / 0.09 = 0.054 / 0.09 = 0.6 N (repulsive)

Example 2 — Electric field due to a point charge

Find E at 0.2 m from a 4 μC charge.

E = k q / r² = (9×10⁹)(4×10⁻⁶) / (0.2)² = 36000 / 0.04 = 9×10⁵ N/C

Example 3 — Capacitance of a parallel plate capacitor

A capacitor stores 6 μC at 3 V. Find its capacitance.

C = Q / V = (6×10⁻⁶) / 3 = 2×10⁻⁶ F = 2 μF

Common Mistakes to Avoid

1. Forgetting to convert micro/nano prefixes ❌ Writing q = 5 in Coulomb's law directly as q = 5 ✅ Convert first: 5 μC = 5 × 10⁻⁶ C, then substitute

2. Using distance in cm instead of metres ❌ Substituting r = 30 cm as r = 30 in Coulomb's law ✅ Convert to SI: r = 30 cm = 0.30 m before using the formula

3. Missing the square in Coulomb's law ❌ Writing F = kq₁q₂/r instead of F = kq₁q₂/r² ✅ Remember: force depends on inverse square of distance, not just the distance

4. Confusing electric field and potential formulas ❌ Using V = kq/r² (this is the field formula, not potential) ✅ E = kq/r² (field), V = kq/r (potential) — potential has one power of r less

5. Ignoring vector nature of electric field ❌ Adding fields of multiple charges algebraically without considering direction ✅ Use the superposition principle with vector addition; resolve into components when needed

6. Misapplying Gauss's law ❌ Choosing a Gaussian surface that does not match the symmetry of the charge distribution ✅ Choose a surface where E is either constant or zero on each face — sphere for point charge, cylinder for line charge, pillbox for plane sheet

Exam Checklist

  • Formula is written before substituting values — never substitute into a blank line.
  • SI units are used throughout — convert cm to m, μC to C, nC to 10⁻⁹ C.
  • Direction or sign of the answer is mentioned (attractive/repulsive, towards/away).
  • Final answer includes the correct unit (N, N/C, V, F, J, N·m²/C).
  • Coulomb's law: remember the r² in the denominator.
  • Electric field vs potential: E ∝ 1/r², V ∝ 1/r — do not mix them up.
  • Capacitors: series adds as reciprocals, parallel adds directly.
  • Gauss's law: choose the right Gaussian surface based on symmetry.
  • Dipole: remember the torque formula τ = pE sin θ and the factor of 2 in p = 2qa.
  • Energy in capacitor: U = ½CV² = ½QV = Q²/(2C) — know all three forms.

Frequently Asked Questions

Coulomb's law, electric field, electric potential, Gauss law, capacitors in series and parallel, and energy stored in a capacitor are very important.