Class 12 Physics Chapter 1 – Electric Charges and Fields | MP Board (MPBSE) Important Questions & Notes
Electric Charges and Fields is the opening chapter of the Class 12 MP Board (MPBSE) Physics syllabus, and it lays the conceptual groundwork for almost every unit that follows — from capacitance and current electricity to magnetism and electromagnetic induction. For MP Board students, this chapter is examination gold: it consistently contributes questions across every mark category, from quick one-mark conceptual checks to full five-mark derivations such as Gauss's theorem and the electric field due to a dipole.
This chapter introduces the idea of electric charge as a fundamental property of matter, explains the two types of charge and the principle of quantisation and conservation of charge, and builds up to Coulomb's law — the force law that governs interactions between static charges. From there, the chapter develops the vector concept of the electric field, superposition of fields due to multiple charges, and the elegant idea of electric field lines as a way to visualise invisible forces. A major portion of the chapter is devoted to electric dipoles, their field along the axial and equatorial lines, and the torque they experience in a uniform field — all frequent five-mark derivation questions in MP Board exams.
The second half of the chapter shifts to continuous charge distributions (linear, surface, and volume charge density) and culminates in Gauss's law, one of the most important and most-tested results in the entire syllabus. Students are expected to not only state Gauss's law but apply it to derive electric fields due to an infinite line charge, an infinite plane sheet of charge, and a uniformly charged spherical shell — each a classic long-answer question.
Below you will find a complete, board-focused question bank for this chapter: 1-mark, 2-mark, 3-mark, and 5-mark questions, solved numericals, previous-year-style questions, a quick formula sheet, key definitions, comparison tables, assertion-reason questions, HOTS questions, common mistakes to avoid, and expert exam tips — everything organised so you can revise the entire chapter in one sitting before your MP Board exam.
Table of Contents
1 Mark Questions
Question 1
Question: Define electric charge.
Answer:
Electric charge is a fundamental property of matter that causes it to experience a force when placed in an electric field. It exists in two kinds — positive and negative — and is measured in coulombs (C).
Question 2
Question: State the SI unit of electric charge.
Answer: The SI unit of electric charge is the coulomb (C).
Question 3
Question: What is meant by quantisation of charge?
Answer: Quantisation of charge means that any charge on a body always exists as an integral multiple of the elementary charge e, i.e. q = ne, where n is an integer.
Question 4
Question: State the law of conservation of charge.
Answer: The total electric charge of an isolated system remains constant; charge can neither be created nor destroyed, only transferred from one body to another.
Question 5
Question: What is the value of the elementary charge e?
Answer: e = 1.6 × 10⁻¹⁹ C.
Question 6
Question: Define electric field intensity.
Answer: Electric field intensity at a point is the force experienced per unit positive test charge placed at that point: E = F/q₀.
Question 7
Question: Give the SI unit of electric field intensity.
Answer: Newton per coulomb (N/C), equivalent to volt per metre (V/m).
Question 8
Question: What is an electric dipole?
Answer: A pair of equal and opposite point charges separated by a small distance is called an electric dipole.
Question 9
Question: Define electric dipole moment.
Answer: Electric dipole moment is the product of either charge and the separation between the charges; it is a vector directed from the negative to the positive charge, p = q × 2a.
Question 10
Question: What is the SI unit of dipole moment?
Answer: Coulomb-metre (C·m).
Question 11
Question: Define electric flux.
Answer: Electric flux through a surface is the total number of electric field lines passing normally through that surface: Φ = E·A cosθ.
Question 12
Question: State the SI unit of electric flux.
Answer: N·m²/C (equivalently V·m).
Question 13
Question: State Gauss's law in electrostatics.
Answer: The total electric flux through a closed surface equals 1/ε₀ times the net charge enclosed by that surface: Φ = q/ε₀.
Question 14
Question: What is a Gaussian surface?
Answer: A Gaussian surface is an imaginary closed surface chosen conveniently to apply Gauss's law for calculating electric field.
Question 15
Question: Define linear charge density.
Answer: Charge per unit length of a charged object, λ = q/l, measured in C/m.
Question 16
Question: Define surface charge density.
Answer: Charge per unit area of a charged surface, σ = q/A, measured in C/m².
Question 17
Question: Define volume charge density.
Answer: Charge per unit volume of a charged body, ρ = q/V, measured in C/m³.
Question 18
Question: What is the value of permittivity of free space ε₀?
Answer: ε₀ = 8.85 × 10⁻¹² C²N⁻¹m⁻².
Question 19
Question: State Coulomb's law.
Answer: The 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, acting along the line joining them.
Question 20
Question: What is the value of the constant k (1/4πε₀) in Coulomb's law?
Answer: k = 9 × 10⁹ N·m²/C².
Question 21
Question: Are electric field lines real?
Answer: No, electric field lines are imaginary lines used only to visualise the direction and relative strength of an electric field.
Question 22
Question: Can two electric field lines intersect each other? Why?
Answer: No, because at the point of intersection there would be two directions of the field simultaneously, which is not possible.
Question 23
Question: What is meant by a uniform electric field?
Answer: A field in which the magnitude and direction of E are the same at every point in the region.
Question 24
Question: Is electric charge a scalar or a vector quantity?
Answer: Electric charge is a scalar quantity, though it can be positive or negative.
Question 25
Question: What happens to the force between two charges if the distance between them is doubled?
Answer: The force becomes one-fourth of its original value, since F ∝ 1/r².
2 Marks Questions
Question 1
Question: State and explain the principle of superposition of electric forces.
Answer: The principle states that the total force on a charge due to several other charges is the vector sum of the individual forces exerted by each charge, calculated as if the others were absent. Mathematically, F = F₁ + F₂ + F₃ + ... where each force is found independently using Coulomb's law.
Question 2
Question: Why does a charged body attract a light uncharged body?
Answer: The electric field of the charged body induces opposite charges on the near side of the neutral body (polarisation). The attraction between the charged body and the induced opposite charge, being closer, exceeds the repulsion from the like induced charge on the far side, resulting in a net attractive force.
Question 3
Question: Two point charges repel each other with a force F. What happens to F if a third charge is brought nearby?
Answer: The force between the original two charges remains exactly F, unchanged, because electric forces obey the superposition principle — each pair of charges interacts independently of the presence of any other charge.
Question 4
Question: Why can electric field lines never form closed loops (in electrostatics)?
Answer: Electrostatic field lines start on positive charges and end on negative charges (or at infinity). If they formed a closed loop, work done in moving a charge around the loop would be non-zero, violating the conservative nature of the electrostatic field.
Question 5
Question: Define electric flux and write the factors on which it depends.
Answer: Electric flux Φ through a surface is the measure of the number of field lines crossing it, given by Φ = E·A cosθ. It depends on the magnitude of the field E, the area of the surface A, and the angle θ between E and the area vector.
Question 6
Question: Why is the electric field inside a conductor zero in electrostatic equilibrium?
Answer: In electrostatic equilibrium, free charges inside a conductor rearrange themselves on the surface until the internal field due to induced charges exactly cancels any external field, making the net field inside zero; otherwise charges would keep moving.
Question 7
Question: What is the significance of Gauss's law?
Answer: Gauss's law provides a quick method to calculate the electric field for charge distributions possessing symmetry (spherical, cylindrical, planar) without directly integrating Coulomb's law, saving considerable calculation effort.
Question 8
Question: An electric dipole is placed in a non-uniform electric field. What is the net effect on it?
Answer: In a non-uniform field, the two forces on the dipole charges are unequal, so the dipole experiences both a net translational force and a torque, causing it to move as well as rotate.
Question 9
Question: Why is a small test charge used to define the electric field?
Answer: A small test charge is used so that it does not significantly disturb or redistribute the source charges creating the field, ensuring the measured field represents the original, undisturbed field.
Question 10
Question: Compare the strength of the gravitational and electrostatic forces between two electrons.
Answer: The electrostatic force between two electrons is enormously stronger than the gravitational force between them — approximately 10⁴² times greater — showing that gravity is negligible at atomic scales compared to electric forces.
Question 11
Question: Why does the electric field due to a dipole fall off faster (as 1/r³) than that of a single point charge (1/r²) at large distances?
Answer: Because the fields of the two equal and opposite charges of the dipole almost cancel each other at large distances, leaving only a small residual field that depends on the charge separation, causing a faster fall-off with distance.
Question 12
Question: State two properties of electric field lines.
Answer: (i) Field lines start from positive charges and end on negative charges. (ii) Two field lines never intersect each other, since the field at any point has a single unique direction.
Question 13
Question: Why is charge said to be additive in nature?
Answer: The total charge of a system of charges is the algebraic sum of all individual charges, taking sign into account, similar to how masses add up — this property is called additivity of charge.
Question 14
Question: Why is charge considered invariant?
Answer: The magnitude of an electric charge is independent of the speed of the charge or the frame of reference from which it is observed — it remains the same in every inertial frame.
Question 15
Question: What is the physical significance of the area vector in electric flux calculations?
Answer: The area vector represents the magnitude of a surface element along with a direction normal to it, allowing the flux formula Φ = E·A cosθ to correctly account for how the field is oriented relative to the surface.
Question 16
Question: Why do field lines crowd closer together near a highly charged region?
Answer: The density of field lines represents field strength; a stronger field near a highly charged region is represented by field lines drawn closer together, while a weaker field is shown by widely spaced lines.
Question 17
Question: A charge q is placed at the centre of a cube. What is the flux through one face of the cube?
Answer: By symmetry, the total flux q/ε₀ is equally distributed among the six faces, so the flux through one face is q/6ε₀.
Question 18
Question: Why does Coulomb's force obey Newton's third law?
Answer: The force exerted by charge 1 on charge 2 is equal in magnitude and opposite in direction to the force exerted by charge 2 on charge 1, satisfying the action-reaction pair required by Newton's third law.
Question 19
Question: Why is the electrostatic force called a central force?
Answer: Because the force between two point charges always acts along the straight line joining their centres, depending only on the distance between them.
Question 20
Question: What is meant by dielectric constant, and why does Coulomb's law include it?
Answer: The dielectric constant K indicates how much a medium reduces the electric force compared to vacuum. Coulomb's law in a medium becomes F = q₁q₂/4πε₀Kr², since the presence of a dielectric medium weakens the force between the charges.
3 Marks Questions
Question 1
Question: State the properties of electric charge.
Answer: (i) Additivity: total charge of a system is the algebraic sum of individual charges. (ii) Conservation: charge can neither be created nor destroyed, only transferred. (iii) Quantisation: charge always exists as an integral multiple of e. (iv) Invariance: the value of charge does not change with the speed of the charged body or the observer's frame of reference.
Question 2
Question: Derive the expression for electric field due to a point charge.
Answer: Consider a point charge Q at the origin. Place a small test charge q₀ at a distance r from Q. By Coulomb's law, the force on q₀ is F = (1/4πε₀)(Qq₀/r²), directed along the line joining them. The electric field is defined as E = F/q₀, so E = (1/4πε₀)(Q/r²), directed radially outward from Q if Q is positive, and radially inward if Q is negative.
Question 3
Question: Explain why the tangent drawn at any point on an electric field line gives the direction of the electric field at that point.
Answer: Field lines are drawn such that at every point their direction coincides with the direction the electric field would push a positive test charge. Since a field line is continuous and curved, only the tangent at any given point represents the instantaneous direction of the field there, just as the tangent to a path gives instantaneous velocity direction.
Question 4
Question: What is meant by continuous charge distribution? Explain the three types with formulas.
Answer: When charge is distributed continuously over a body rather than existing at discrete points, it is described using charge densities: (i) Linear charge density λ = dq/dl (charge per unit length, for wires). (ii) Surface charge density σ = dq/dA (charge per unit area, for sheets/plates). (iii) Volume charge density ρ = dq/dV (charge per unit volume, for solid charged bodies).
Question 5
Question: Explain electric polarisation of a dielectric.
Answer: When a dielectric (insulator) is placed in an external electric field, its molecules develop induced dipole moments aligned with the field, or its permanent dipoles get oriented along the field. This creates a net dipole moment per unit volume in the material, called polarisation, which produces an internal field opposing the external one and weakens the net field inside the dielectric.
Question 6
Question: State Gauss's law and explain the meaning of each term.
Answer: Gauss's law states that the total electric flux through any closed surface equals 1/ε₀ times the total charge enclosed: ∮E·dA = q_enc/ε₀. Here E is the electric field, dA is the vector area element of the closed (Gaussian) surface, and q_enc is the net charge inside that surface. Charges outside the surface contribute zero net flux.
Question 7
Question: Why is Gauss's law useful only for symmetric charge distributions?
Answer: Gauss's law is always true, but it is only easy to solve for E when a Gaussian surface can be chosen on which E is either constant in magnitude or zero, and E is either parallel or perpendicular to dA everywhere. This is only possible for spherically, cylindrically, or planar symmetric charge distributions, making the surface integral simple.
Question 8
Question: Explain the behaviour of an electric dipole placed in a uniform electric field.
Answer: In a uniform field, the two equal and opposite charges of the dipole experience equal and opposite forces, so the net translational force on the dipole is zero. However, since the forces act at different points, they form a couple that produces a torque τ = pE sinθ, which tends to align the dipole moment p with the field E.
Question 9
Question: Explain why a conductor placed in an external electric field has no field inside it, using the concept of induced charges.
Answer: When a conductor is placed in an external field, free electrons experience a force and redistribute themselves, accumulating negative charge on one face and leaving positive charge on the opposite face. These induced surface charges create their own field inside the conductor, opposite to the external field. This continues until the induced field exactly cancels the external field inside the conductor, making the net field zero at equilibrium.
Question 10
Question: Compare Coulomb's law with the universal law of gravitation.
Answer: Both laws follow an inverse square relationship with distance. However, Coulomb's force can be attractive or repulsive depending on the sign of charges, while gravitational force is always attractive. Coulomb's force depends on the medium between charges (via ε₀ or K), whereas gravitational force is medium-independent. Also, the electrostatic force is far stronger in magnitude than gravitational force for elementary particles.
Question 11
Question: Explain why field lines are denser near a small charged sphere of large charge and sparse near a large sphere of small charge.
Answer: Field line density represents field strength. A small sphere with a large charge produces a strong field near its surface, so more lines are drawn close together. A large sphere with a small charge has a weaker field per unit area, so fewer, more widely spaced lines are drawn, correctly representing the relative field strengths.
Question 12
Question: What is electrostatic shielding? Give one application.
Answer: Electrostatic shielding is the phenomenon by which the interior of a hollow conductor remains free of electric field even when the conductor is placed in an external electric field, because induced charges on the conductor's outer surface cancel the field inside. Application: sensitive electronic equipment is enclosed in metallic cases (Faraday cages) to protect it from external electric disturbances.
Question 13
Question: Derive the relation between electric field and electric flux for a uniform field through a flat surface.
Answer: For a flat surface of area A placed in a uniform field E, with the area vector making angle θ with E, the flux is defined as the dot product Φ = E·A = EA cosθ. When the surface is perpendicular to the field (θ = 0°), flux is maximum, Φ = EA; when the surface is parallel to the field (θ = 90°), flux is zero.
Question 14
Question: Why can a charged comb attract small bits of paper even though the paper is neutral?
Answer: The non-uniform field of the charged comb polarises the neutral paper bits, inducing opposite charge on the near side. Since the induced opposite charge is closer to the comb than the induced like charge on the far side, the attractive force dominates over the repulsive force, resulting in net attraction and the paper bits being pulled towards the comb.
Question 15
Question: Explain why the electric field due to an infinite plane sheet of charge does not depend on the distance from the sheet.
Answer: For an infinite sheet, the charge distribution extends infinitely in all directions parallel to the sheet, so as you move away from the sheet, no new geometric factor of distance enters the symmetry of the problem — using a cylindrical Gaussian pillbox, the enclosed charge and flux relation gives E = σ/2ε₀, a constant independent of the perpendicular distance from the sheet.
5 Marks Questions
Question 1
Question:
State Coulomb's law in vector form and explain each term. Discuss how it obeys Newton's Third Law of Motion.
Question 2
Question: Derive an expression for the electric field due to an electric dipole at a point on its axial line.
Question 3
Question: Derive an expression for the electric field due to an electric dipole at a point on its equatorial line.
Question 4
Question: Derive an expression for the torque acting on an electric dipole placed in a uniform electric field.
Answer: Consider a dipole with charges +q and −q, separated by 2a, placed in a uniform field E, making angle θ with the field direction. The force on +q is qE (along E), and on −q is qE (opposite to E). These two equal, opposite, and parallel forces are separated by a perpendicular distance = 2a sinθ, forming a couple.
Torque = Force × perpendicular distance = qE × 2a sinθ = (q × 2a) × E sinθ = pE sinθ
In vector form: τ = p × E
This torque tends to rotate the dipole so as to align p along E. The torque is maximum when θ = 90° (τ = pE) and zero when the dipole is aligned (θ = 0°) or anti-aligned (θ = 180°) with the field, both being equilibrium positions — stable and unstable respectively.
Question 5
Question: State Gauss's law and prove it for a point charge (derive Coulomb's law from Gauss's law).
Answer: Gauss's law states ∮E·dA = q/ε₀ over a closed surface enclosing charge q. To verify this for a point charge, take a spherical Gaussian surface of radius r centred on the charge q. By symmetry, E has the same magnitude at every point on the sphere and is directed radially outward, parallel to dA everywhere.
∮E·dA = E∮dA = E × (4πr²)
By Gauss's law, this equals q/ε₀:
E × 4πr² = q/ε₀ ⟹ E = q/(4πε₀r²)
This is exactly Coulomb's law expression for the field of a point charge, confirming that Gauss's law and Coulomb's law are consistent, and Coulomb's law can be derived as a special case of Gauss's law.
Question 6
Question: Using Gauss's law, derive the expression for the electric field due to an infinitely long, thin, uniformly charged straight wire.
Answer: Consider an infinite line charge with linear charge density λ. By symmetry, the field E is radial and has the same magnitude at every point at a perpendicular distance r from the wire. Choose a cylindrical Gaussian surface of radius r and length l, coaxial with the wire.
Flux through the curved surface: E × 2πrl (flux through the flat circular ends is zero since E is parallel to them, not perpendicular).
Charge enclosed: q = λl
By Gauss's law: E × 2πrl = λl/ε₀
E = λ/(2πε₀r)
The field is directed radially outward (for positive λ) and decreases as 1/r with distance from the wire.
Question 7
Question: Using Gauss's law, derive the expression for the electric field due to an infinite plane sheet of charge.
Answer: Consider an infinite plane sheet with uniform surface charge density σ. By symmetry, the field is perpendicular to the sheet and has equal magnitude at equal distances on both sides. Choose a cylindrical (pillbox) Gaussian surface with its flat faces of area A parallel to the sheet, one on each side.
Flux through each flat face = EA, and there are two faces, so total flux = 2EA (flux through the curved surface is zero since E is parallel to it).
Charge enclosed = σA
By Gauss's law: 2EA = σA/ε₀
E = σ/2ε₀
The field is uniform, directed away from the sheet (for positive σ) on both sides, and independent of the distance from the sheet.
Question 8
Question: Using Gauss's law, derive the expression for the electric field due to a uniformly charged thin spherical shell, both outside and inside the shell.
Answer: Let a thin spherical shell of radius R carry total charge q uniformly distributed on its surface.
Case 1: Point outside the shell (r > R). Choose a spherical Gaussian surface of radius r > R, concentric with the shell. By symmetry, E is radial and constant in magnitude on this surface.
E × 4πr² = q/ε₀ ⟹ E_out = q/(4πε₀r²)
This shows the shell behaves as if all its charge were concentrated at the centre, for points outside.
Case 2: Point on the surface (r = R). E_surface = q/(4πε₀R²)
Case 3: Point inside the shell (r < R). Choose a Gaussian sphere of radius r < R. Since all the charge resides on the shell's surface, the charge enclosed by this inner Gaussian surface is zero.
E × 4πr² = 0/ε₀ ⟹ E_in = 0
Thus the electric field is zero everywhere inside a uniformly charged spherical shell, and behaves like a point charge concentration at the centre for all points outside it.
Question 9
Question: Derive an expression for the electric field due to a system of n point charges using the superposition principle, and explain with a suitable example.
Answer: Consider n point charges q₁, q₂, ..., qₙ located at position vectors r₁, r₂, ..., rₙ from an origin O. Let P be the point where the field is to be calculated, at position vector r. The field due to each charge qᵢ alone at P is:
Eᵢ = (1/4πε₀) × qᵢ/rᵢP² × r̂ᵢP
where rᵢP is the distance from qᵢ to P and r̂ᵢP is the unit vector from qᵢ towards P. According to the superposition principle, the resultant field at P due to the entire system is the vector sum of the individual fields:
E = E₁ + E₂ + ... + Eₙ = Σ (1/4πε₀) × qᵢ/rᵢP² × r̂ᵢP
Example: For two charges +q at A and −q at B (a dipole) with P on the perpendicular bisector, the individual fields E_A and E_B are calculated separately using the above formula and then added vectorially, with their components along the axis reinforcing and components perpendicular to the axis cancelling — this is exactly the method used to derive the equatorial field of a dipole.
Question 10
Question: What are electric field lines? State their important properties and explain, with the help of diagrams described in words, how field lines look for (a) an isolated positive charge (b) two equal positive charges (c) an electric dipole.
Answer: Electric field lines are imaginary curves drawn such that the tangent at any point gives the direction of the electric field at that point, and the density of lines represents the field's relative strength.
Properties: (i) Field lines originate from positive charges and terminate on negative charges (or extend to infinity for isolated positive charge). (ii) They never intersect, since the field has a unique direction at every point. (iii) They are denser where the field is stronger. (iv) They are always normal to the surface of a conductor. (v) They contract lengthwise and repel sideways in general, representing attraction between unlike charges and repulsion between like charges.
(a) Isolated positive charge: Field lines radiate straight outward symmetrically in all directions, like spokes of a wheel, extending to infinity.
(b) Two equal positive charges: Lines radiate outward from each charge, curving away from each other, with a neutral point exactly midway between them where lines from both charges meet and cancel, so no field line passes through that point.
(c) Electric dipole: Lines emerge from the positive charge and curve around to terminate on the negative charge, forming closed-looking arcs between the two charges, densest along the line joining them and progressively curving outward on either side.
Important Numericals
Numerical 1
Given: Two point charges q₁ = 2 μC and q₂ = 3 μC are placed 30 cm apart.
Formula: F = (1/4πε₀) q₁q₂/r²
Solution: F = 9×10⁹ × (2×10⁻⁶ × 3×10⁻⁶)/(0.3)² = 9×10⁹ × 6×10⁻¹²/0.09 = 0.6 N
Final Answer: F = 0.6 N (repulsive)
Numerical 2
Given: A charge of 5 μC experiences a force of 0.2 N at a point.
Formula: E = F/q
Solution: E = 0.2/(5×10⁻⁶) = 4×10⁴ N/C
Final Answer: E = 4 × 10⁴ N/C
Numerical 3
Given: Two charges +4 μC and −4 μC are separated by 2 cm, forming a dipole.
Formula: p = q × 2a
Solution: p = 4×10⁻⁶ × 0.02 = 8×10⁻⁸ C·m
Final Answer: p = 8 × 10⁻⁸ C·m
Numerical 4
Given: Find the electric field at a distance of 0.5 m from a point charge of 1 μC.
Formula: E = (1/4πε₀) q/r²
Solution: E = 9×10⁹ × 10⁻⁶/(0.5)² = 9×10⁹ × 10⁻⁶/0.25 = 3.6×10⁴ N/C
Final Answer: E = 3.6 × 10⁴ N/C
Numerical 5
Given: A dipole of moment 4×10⁻⁸ C·m is placed in a uniform field of 10⁵ N/C at 30° to the field.
Formula: τ = pE sinθ
Solution: τ = 4×10⁻⁸ × 10⁵ × sin30° = 4×10⁻³ × 0.5 = 2×10⁻³ N·m
Final Answer: τ = 2 × 10⁻³ N·m
Numerical 6
Given: Find the flux through a surface of area 2 m² held perpendicular to a uniform field of 100 N/C.
Formula: Φ = EA cosθ
Solution: Since the surface is perpendicular to the field, θ = 0°, so Φ = 100 × 2 × cos0° = 200
Final Answer: Φ = 200 N·m²/C
Numerical 7
Given: A sphere encloses a net charge of 2 μC. Find the total electric flux through the sphere.
Formula: Φ = q/ε₀
Solution: Φ = 2×10⁻⁶/8.85×10⁻¹² = 2.26×10⁵ N·m²/C
Final Answer: Φ ≈ 2.26 × 10⁵ N·m²/C
Numerical 8
Given: An infinite plane sheet has a surface charge density of 17.7×10⁻¹² C/m². Find the electric field near the sheet.
Formula: E = σ/2ε₀
Solution: E = 17.7×10⁻¹²/(2×8.85×10⁻¹²) = 17.7/17.7 = 1 N/C
Final Answer: E = 1 N/C
Numerical 9
Given: Find the force between two charges of 1 C each placed 1 m apart in vacuum.
Formula: F = kq₁q₂/r²
Solution: F = 9×10⁹ × 1 × 1/1² = 9×10⁹ N
Final Answer: F = 9 × 10⁹ N (illustrating how large 1 coulomb actually is)
Numerical 10
Given: Charge on an object is measured as 3.2×10⁻¹⁸ C. Find the number of electrons removed or added.
Formula: n = q/e
Solution: n = 3.2×10⁻¹⁸/1.6×10⁻¹⁹ = 20
Final Answer: n = 20 electrons
Numerical 11
Given: Find the electric field at a point on the axial line of a dipole (p = 2×10⁻⁷ C·m) at a distance of 20 cm from the centre.
Formula: E_axial = (1/4πε₀) × 2p/r³
Solution: E = 9×10⁹ × 2×2×10⁻⁷/(0.2)³ = 9×10⁹ × 4×10⁻⁷/0.008 = 4.5×10⁵ N/C
Final Answer: E ≈ 4.5 × 10⁵ N/C
Numerical 12
Given: A charged wire has linear charge density 2×10⁻⁶ C/m. Find the field at a perpendicular distance of 10 cm.
Formula: E = λ/2πε₀r
Solution: E = 2×10⁻⁶/(2π × 8.85×10⁻¹² × 0.1) = 2×10⁻⁶/(5.56×10⁻¹²) ≈ 3.6×10⁵ N/C
Final Answer: E ≈ 3.6 × 10⁵ N/C
Numerical 13
Given: Three equal charges of 2 μC each are placed at the vertices of an equilateral triangle of side 10 cm. Find the force on any one charge.
Formula: F = kq²/r² (for each pair), resultant using vector addition at 60°
Solution: Force due to each neighbour: F = 9×10⁹ × (2×10⁻⁶)²/(0.1)² = 9×10⁹×4×10⁻¹²/0.01 = 3.6 N. Since the two forces on any charge act at 60° to each other, resultant = 2F cos30° = 2×3.6×0.866 ≈ 6.24 N
Final Answer: Resultant force ≈ 6.24 N, directed away from the triangle's centre
Numerical 14
Given: A solid sphere of radius 10 cm has a total charge of 4 μC uniformly distributed in its volume. Find the field at a point 5 cm from the centre (inside the sphere).
Formula: E_in = (1/4πε₀) × qr/R³ (for uniformly charged solid sphere)
Solution: E = 9×10⁹ × 4×10⁻⁶ × 0.05/(0.1)³ = 9×10⁹ × 2×10⁻⁷/0.001 = 1.8×10⁶ N/C
Final Answer: E = 1.8 × 10⁶ N/C, directed radially outward
Numerical 15
Given: Two point charges of +6 μC and −6 μC are 12 cm apart. Find the field at the midpoint of the line joining them.
Formula: E = 2 × (1/4πε₀) q/r² (fields add, both directed same way at midpoint of unlike charges)
Solution: r = 6 cm = 0.06 m. E_each = 9×10⁹ × 6×10⁻⁶/(0.06)² = 9×10⁹×6×10⁻⁶/0.0036 = 1.5×10⁷ N/C. Total E = 2 × 1.5×10⁷ = 3×10⁷ N/C
Final Answer: E = 3 × 10⁷ N/C, directed from +q towards −q
Previous Year Type Questions
Question 1
Question: State Coulomb's law and express it in vector form.
Answer: Coulomb's law states that the force between two stationary point charges is directly proportional to the product of the charges and inversely proportional to the square of the distance between them, acting along the line joining them. In vector form, F₂₁ = (1/4πε₀)(q₁q₂/r²) r̂₂₁.
Question 2
Question: Derive the expression for torque on an electric dipole in a uniform electric field.
Answer: See the full derivation under 5-Marks Question 4 above: τ = pE sinθ, or in vector form τ = p × E, arising from the couple formed by equal and opposite forces on the dipole's charges.
Question 3
Question: Using Gauss's theorem, derive the expression for the electric field due to a uniformly charged infinite plane sheet.
Answer: See the full derivation under 5-Marks Question 7 above, resulting in E = σ/2ε₀, independent of distance from the sheet.
Question 4
Question: Two point charges of magnitude q are placed at the two ends of a diameter of a circle of radius r. What is the electric field at the centre?
Answer: If the charges are equal and of the same sign, the two fields at the centre are equal in magnitude but opposite in direction, so the net field is zero. If the charges are equal and opposite, the fields add up, giving a net field of 2 × (1/4πε₀)(q/r²) directed from the positive towards the negative charge.
Question 5
Question: Define electric field intensity and derive its expression due to a point charge, showing it is a vector quantity.
Answer: Electric field intensity is force per unit positive test charge, E = F/q₀. For a point charge Q, E = (1/4πε₀)(Q/r²) r̂, where r̂ is the unit vector from the charge to the field point. Since E has both magnitude and direction (along r̂), it is a vector quantity.
Question 6
Question: State Gauss's theorem and use it to find the electric field due to a long uniformly charged straight wire.
Answer: See 5-Marks Question 6 above for the full statement and derivation, giving E = λ/2πε₀r for an infinite line charge.
Question 7
Question: Explain why the electric field is discontinuous across a charged surface but continuous across an uncharged surface.
Answer: Applying Gauss's law with a thin pillbox straddling a charged surface with surface density σ shows that the field just outside minus the field just inside equals σ/ε₀ — a jump proportional to the surface charge. If there is no charge on the surface (σ = 0), this jump is zero, so the field remains continuous across it.
Question 8
Question: Derive the expression for the field due to a dipole at a point on its equatorial line and show it is anti-parallel to the dipole moment.
Answer: See 5-Marks Question 3 above. The perpendicular components of the fields due to +q and −q cancel, while the parallel components add, and because the −q field component dominates in direction along the axis, the resultant field E_equatorial points opposite to p.
Question 9
Question: A charge q is placed at the centre of a cube of side a. Find the electric flux through each face of the cube.
Answer: Total flux through the cube (a closed surface) = q/ε₀. Since the cube has 6 identical faces and the charge is at the centre (symmetric position), the flux is equally divided: flux per face = q/6ε₀.
Question 10
Question: Explain, with reasoning, why a hollow charged conductor has no electric field inside it, regardless of the shape of the conductor.
Answer: For any closed Gaussian surface drawn just inside the conductor's cavity, the enclosed charge is zero (all charge resides on the outer surface of a conductor in equilibrium). By Gauss's law, since q_enc = 0, the flux and hence the field inside the cavity must be zero, irrespective of the conductor's shape.
Question 11
Question: State the principle of superposition of electric fields and apply it to find the field due to two point charges at a general point.
Answer: The principle states the resultant field due to several charges is the vector sum of fields due to each charge individually. For two charges q₁ and q₂ at points A and B, the field at point P is found by calculating E₁ (due to q₁) and E₂ (due to q₂) separately using Coulomb's law, then adding them vectorially using the parallelogram law: E = E₁ + E₂.
Question 12
Question: Two charges of +5 μC and +10 μC are placed 20 cm apart. Find the point on the line joining them where the electric field is zero.
Answer: The null point lies between the charges, closer to the smaller charge. Let it be at distance x from the +5 μC charge. Setting fields equal: 5/x² = 10/(20−x)², solving gives x ≈ 8.28 cm from the 5 μC charge.
Question 13
Question: Explain why the electric field due to a dipole varies as 1/r³ while that of a point charge varies as 1/r².
Answer: A single point charge produces a field that falls off as 1/r² by Coulomb's law directly. A dipole consists of two opposite charges whose fields nearly cancel at large distances, leaving only a small net field that depends additionally on the small charge separation 2a, introducing an extra factor of 1/r and making the net dependence 1/r³.
Question 14
Question: Derive Gauss's law from Coulomb's law for a point charge enclosed by an arbitrary closed surface (qualitative reasoning).
Answer: For a point charge q, the flux through any small solid angle dΩ of an arbitrary closed surface equals E·dA cosθ, which can be shown (using the definition of solid angle) to equal (q/4πε₀) dΩ, independent of the surface's shape or the distance r. Integrating over the full solid angle 4π around the charge gives total flux = q/ε₀, proving Gauss's law holds for any shape of closed surface enclosing the charge.
Question 15
Question: A dipole is placed in a uniform electric field with its dipole moment perpendicular to the field. Find the torque and discuss the nature of equilibrium.
Answer: When θ = 90°, torque τ = pE sin90° = pE, which is the maximum possible torque. This is not an equilibrium position since the torque is non-zero; the dipole will rotate towards alignment with the field (θ = 0°), which is the stable equilibrium position, while θ = 180° (anti-parallel) is the unstable equilibrium position.
Important Formula Sheet
| Quantity | Formula |
|---|---|
| Coulomb's Law | F = (1/4πε₀) × q₁q₂/r² |
| Coulomb's Law (vector form) | F₂₁ = (1/4πε₀)(q₁q₂/r²) r̂₂₁ |
| Value of 1/4πε₀ | 9 × 10⁹ N·m²/C² |
| Electric Field | E = F/q₀ |
| Electric Field due to a point charge | E = (1/4πε₀) × q/r² |
| Dipole Moment | p = q × 2a |
| Field on axial line of dipole | E_axial = (1/4πε₀) × 2p/r³ |
| Field on equatorial line of dipole | E_equatorial = (1/4πε₀) × p/r³ |
| Torque on dipole | τ = pE sinθ = p × E |
| Electric Flux | Φ = E·A = EA cosθ |
| Gauss's Law | Φ = q_enc/ε₀ |
| Field due to infinite line charge | E = λ/2πε₀r |
| Field due to infinite plane sheet | E = σ/2ε₀ |
| Field due to charged spherical shell (outside) | E = q/(4πε₀r²) |
| Field due to charged spherical shell (inside) | E = 0 |
| Field inside uniformly charged solid sphere | E = qr/(4πε₀R³) |
| Linear charge density | λ = q/l |
| Surface charge density | σ = q/A |
| Volume charge density | ρ = q/V |
| Permittivity of free space | ε₀ = 8.85 × 10⁻¹² C²N⁻¹m⁻² |
Important Definitions
Differences
| Coulomb's Law | Newton's Law of Gravitation |
|---|---|
| Force can be attractive or repulsive | Force is always attractive |
| Depends on the medium between charges | Independent of the medium |
| Constant k = 9 × 10⁹ N·m²/C² | Constant G = 6.67 × 10⁻¹¹ N·m²/kg² |
| Acts between charges | Acts between masses |
| Conductor | Insulator (Dielectric) |
|---|---|
| Contains free electrons that can move | Electrons are bound to atoms, cannot move freely |
| Electric field inside is zero in equilibrium | Electric field can exist inside |
| Charge resides only on the outer surface | Charge can exist throughout the volume |
| Example: copper, silver | Example: rubber, glass |
| Field on Axial Line of Dipole | Field on Equatorial Line of Dipole |
|---|---|
| E_axial = (1/4πε₀) × 2p/r³ | E_equatorial = (1/4πε₀) × p/r³ |
| Direction same as dipole moment p | Direction opposite to dipole moment p |
| Magnitude is twice that of equatorial field at same r | Magnitude is half that of axial field at same r |
Diagram Based Questions
Draw the electric field lines for an isolated positive point charge and describe their pattern.
Draw the electric field lines of an electric dipole and label the positive and negative charges.
Draw a labelled diagram of the cylindrical Gaussian surface used to find the field due to an infinite line charge.
Draw a diagram showing an electric dipole placed at an angle θ in a uniform electric field, and mark the forces and torque.
Assertion and Reason Questions
Directions: Each question has an Assertion (A) and a Reason (R). Choose whether both are true and R correctly explains A, both are true but R does not explain A, A is true but R is false, or both are false.
Question 1
Assertion (A): The electric field inside a charged conductor is zero.
Reason (R): All the charge on a conductor resides only on its outer surface.
Answer: Both A and R are true, and R correctly explains A — induced charges rearrange on the surface, cancelling the internal field.
Question 2
Assertion (A): Electric field lines never cross each other.
Reason (R): The electric field at a point has a unique direction.
Answer: Both A and R are true, and R correctly explains A.
Question 3
Assertion (A): Electric charge is quantised.
Reason (R): Charge always exists in integral multiples of the elementary charge e.
Answer: Both A and R are true, and R correctly explains A.
Question 4
Assertion (A): The electric field due to a dipole at a large distance falls off faster than that due to a point charge.
Reason (R): The net charge of a dipole is zero.
Answer: Both A and R are true, and R correctly explains A — since net charge is zero, only the residual effect of separation causes the 1/r³ dependence.
Question 5
Assertion (A): Gauss's law is valid only for symmetric charge distributions.
Reason (R): Gauss's law itself, ∮E·dA = q/ε₀, holds true for any closed surface and any charge distribution.
Answer: A is false, R is true. Gauss's law is universally valid; it is only easy to solve for E when the distribution is symmetric.
Question 6
Assertion (A): A hollow charged sphere behaves like the entire charge is concentrated at its centre for external points.
Reason (R): Electric field inside a uniformly charged spherical shell is always zero.
Answer: Both A and R are true, but R is not the direct explanation for A (they are two separate consequences of applying Gauss's law to a shell).
Question 7
Assertion (A): Two charges of equal magnitude but opposite sign are placed close together to form a dipole; the net force on the dipole in a uniform field is zero.
Reason (R): Equal and opposite forces act on the two charges of a dipole in a uniform field.
Answer: Both A and R are true, and R correctly explains A.
Question 8
Assertion (A): A charged comb attracts small uncharged pieces of paper.
Reason (R): The comb induces polarisation in the paper, and the closer induced opposite charge dominates the attraction.
Answer: Both A and R are true, and R correctly explains A.
Question 9
Assertion (A): The electric flux through a closed surface depends only on the net charge enclosed.
Reason (R): Charges outside the closed surface contribute zero net flux through it.
Answer: Both A and R are true, and R correctly explains A.
Question 10
Assertion (A): The electric field due to an infinite plane sheet of charge is independent of the distance from the sheet.
Reason (R): The sheet is assumed to be infinitely large, so no new distance-dependent geometry enters the problem.
Answer: Both A and R are true, and R correctly explains A.
HOTS Questions
Question 1
Question: Two identical charged spheres are suspended from the same point by strings of equal length and make an angle of 30° with each other due to mutual repulsion. If the spheres are immersed in a liquid of dielectric constant K, and the angle remains the same, what does this tell you about the density relationship of the spheres and the liquid?
Answer: For the angle to remain unchanged after immersion, the reduced electrostatic force (due to the dielectric medium weakening Coulomb's force by factor 1/K) must be compensated by a reduced effective weight due to buoyancy. This requires the density of the sphere material to be K times the density of the liquid, so that the ratio of effective forces remains the same.
Question 2
Question: A point charge is placed at the centre of a hollow, uncharged conducting spherical shell. Describe the charge distribution on the shell and the field in all regions.
Answer: The point charge +q induces a charge −q on the inner surface of the shell and, since the shell is neutral overall, a charge +q appears on the outer surface. Inside the cavity (between the point charge and inner surface), the field follows Coulomb's law for the point charge. Inside the conductor material itself, the field is zero. Outside the shell, the field is as if a point charge +q were located at the centre.
Question 3
Question: Two dipoles of equal dipole moment p are placed at the two ends of a line, both pointing in the same direction, separated by a large distance r. Estimate how the interaction energy between them varies with r.
Answer: Since each dipole creates a field falling off as 1/r³ at the location of the other, and the interaction energy involves the product of one dipole's moment with the other's field, the interaction energy between two dipoles varies as 1/r³.
Question 4
Question: A charge Q is distributed uniformly over a ring of radius R. Explain, without detailed integration, why the field at the centre of the ring is zero but non-zero at a point on its axis.
Answer: At the centre, contributions from diametrically opposite elements of the ring are equal in magnitude but point in opposite directions, so by symmetry the net field cancels to zero. At a point on the axis (not at the centre), the components of the field perpendicular to the axis still cancel by symmetry, but the components along the axis from every element point in the same direction and add up, giving a non-zero net field along the axis.
Question 5
Question: If Coulomb's law were an inverse cube law instead of an inverse square law, would Gauss's law in its present form (Φ = q/ε₀) still hold? Justify.
Answer: No. Gauss's law in its simple form relies specifically on the inverse-square nature of the force, since the 1/r² dependence exactly cancels the r² growth of a sphere's surface area, making flux independent of the Gaussian sphere's radius. If the force followed an inverse-cube law, the flux would depend on r, and the simple statement Φ = q/ε₀ would no longer hold.
Question 6
Question: A small metal sphere carrying charge q is placed inside (but not touching) a larger hollow uncharged metal sphere, and then the two are connected by a wire. What happens to the charge?
Answer: When connected, all the charge q flows from the inner sphere to the outer surface of the larger sphere, because charge on a conductor system always resides on the outermost surface — this is the principle behind the Van de Graaff generator.
Question 7
Question: An electric dipole is released from rest in a uniform electric field with its axis perpendicular to the field. Describe its subsequent motion (ignoring gravity).
Answer: Since the net force is zero but the torque τ = pE sinθ is maximum at θ = 90°, the dipole begins to rotate towards alignment with the field. As it rotates, it gains rotational kinetic energy and, without any resistive forces, it will oscillate back and forth about the stable equilibrium position (θ = 0°) like a torsional pendulum, since it overshoots due to inertia.
Question 8
Question: Why can Gauss's law not be used directly to find the electric field due to a finite line charge or a finite charged disc?
Answer: Gauss's law requires a Gaussian surface on which the field is constant in magnitude and has a simple angular relationship with the area vector everywhere. A finite line or disc lacks the necessary translational or full symmetry (the field varies in both magnitude and direction along such a surface), so no simple Gaussian surface can be chosen, making direct integration of Coulomb's law necessary instead.
Question 9
Question: Two point charges +q and +4q are placed a distance d apart. At what point on the line joining them (other than infinity) is the net electric field zero?
Answer: The null point must lie between the charges, closer to the smaller charge +q. Setting fields equal: q/x² = 4q/(d−x)², which gives (d−x)/x = 2, so x = d/3. The field is zero at a distance d/3 from the charge +q.
Question 10
Question: Explain why, in practice, it is impossible to have an "isolated" magnetic-like point charge analog in electrostatics, i.e., why electric field lines must always begin and end on charges rather than forming closed loops.
Answer: The electrostatic field is conservative, meaning the work done in moving a charge around any closed path is zero. If field lines formed closed loops, a positive test charge moved along such a loop in the direction of the field would gain net kinetic energy for free, violating energy conservation. Hence electrostatic field lines must always originate on positive charges and terminate on negative charges, never forming closed loops on their own.
Common Mistakes
- Forgetting to square the distance r in Coulomb's law and Gauss's law calculations.
- Confusing the direction of the electric field on the equatorial line of a dipole — remember it is opposite to the dipole moment p, not the same direction.
- Applying Gauss's law to non-symmetric charge distributions and expecting a simple closed-form answer for E.
- Forgetting that the field inside a uniformly charged spherical shell is zero, not equal to the surface field.
- Mixing up units — charge density formulas (λ, σ, ρ) have different units (C/m, C/m², C/m³) and must not be interchanged.
- Writing Coulomb's law without the 1/4πε₀ constant, or using the wrong value of k in numerical problems.
- Assuming field lines can cross or that they represent real physical entities — they are only a visualisation tool.
- Forgetting to convert all given quantities to SI units (especially distances in cm to m) before substituting into formulas.
Board Exam Tips
- Always draw a clean, labelled diagram before starting Gauss's law derivations — MP Board examiners award marks specifically for correct diagrams.
- Memorise the standard derivations (dipole field on axial/equatorial line, infinite sheet, infinite line, spherical shell) word for word, since these repeat almost every year as 5-mark questions.
- In numerical problems, write down Given, Formula, Substitution, and Final Answer with correct units clearly — MP Board follows step-wise marking.
- Revise all SI units and standard constants (ε₀, k = 9×10⁹) a day before the exam; unit errors are the most common reason for lost marks.
- Practice vector-form questions separately, as many students lose marks by giving only scalar magnitude without direction.
- For assertion-reason questions, read both statements carefully — many students mark "both true" without checking if R actually explains A.
Frequently Asked Questions
Q1. What is the weightage of Electric Charges and Fields in MP Board Class 12 Physics?
This chapter typically contributes questions across 1-mark, 3-mark, and 5-mark categories, forming a core part of the Electrostatics unit, which carries significant weightage in the MP Board exam.
Q2. Which derivations from this chapter are most important for the board exam?
The field due to a dipole (axial and equatorial), torque on a dipole, and Gauss's law applications (infinite line charge, infinite sheet, spherical shell) are the most frequently asked long-answer derivations.
Q3. Is Gauss's law difficult for board exam students?
Gauss's law itself is a simple statement, but its application requires practice in choosing the correct Gaussian surface; with repeated practice of the three standard derivations, it becomes very manageable.
Q4. What is the difference between electric field and electric flux?
Electric field is a vector quantity representing force per unit charge at a point, while electric flux is a scalar quantity representing the total field lines passing through a given surface.
Q5. How many numerical questions should I practice from this chapter?
At least 15-20 well-varied numericals covering Coulomb's law, field calculations, dipole problems, and Gauss's law applications are recommended for thorough board exam preparation.
Q6. Why is the electric field inside a conductor always zero?
Because free charges redistribute themselves on the conductor's surface until their induced field exactly cancels any external field inside, a state called electrostatic equilibrium.
Q7. What is the difference between axial and equatorial fields of a dipole?
The axial field is twice as strong as the equatorial field at the same distance, and the two fields point in opposite directions relative to the dipole moment.
Q8. Are field lines and flux the same concept?
No, field lines are a visualisation tool showing direction and relative strength, whereas flux is a specific, quantifiable scalar measure of field lines passing through a chosen surface.
Q9. What should I revise last before the exam from this chapter?
Focus on the formula sheet, the three Gauss's law derivations, and the dipole torque formula — these consistently appear in MP Board question papers.
Q10. Does this chapter connect to later chapters in the syllabus?
Yes, the concepts of electric field and Gauss's law form the foundation for Electrostatic Potential and Capacitance (Chapter 2) and are also used later in Electromagnetic Induction and Electromagnetic Waves.
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