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1. Electrostatic force between two point charges q₁ and q₂ separated by r in vacuum is:
Explanation: Coulomb's law: F = kq₁q₂/r² where k = 1/(4πε₀) ≈ 9×10⁹ N·m²/C².
2. Coulomb's constant k in SI units is:
Explanation: k = 1/(4πε₀) = 9×10⁹ N·m²/C². With ε₀ = 8.85×10⁻¹² C²/N·m².
3. Electric field E is defined as:
Explanation: E = F/q₀ for a positive test charge q₀ → 0. SI unit: N/C = V/m.
4. Electric field due to a point charge q at distance r is:
Explanation: E = kq/r² directed radially outward (for +q) or inward (for −q).
5. Force between two charges is F. If each charge is doubled and distance tripled, new force is:
Explanation: F ∝ q₁q₂/r². New: (2q₁)(2q₂)/(3r)² = 4q₁q₂/(9r²) = 4F/9.
6. Three charges +q, +q, −q at vertices of equilateral triangle. Net force on −q:
Explanation: Each +q exerts force kq²/L² on −q. By symmetry, horizontal components add: F_net = 2kq²/L² × cos30° = kq²√3/L²... actually: two forces at 60° from each other, each kq²/L². Resultant = √(F²+F²+2F²cos60°) = F√3 = kq²√3/L².
7. Electric field is zero on line joining two equal positive charges at:
Explanation: By symmetry, at the midpoint the two equal fields point in opposite directions and cancel. E = 0 exactly at the midpoint.
8. Electric dipole moment p of a dipole (+q and −q separated by 2a) is:
Explanation: p = q × 2a directed from negative to positive charge. p is a vector (N·m²... actually p has unit C·m).
9. Field on axial line of electric dipole at distance r (r >> a) is:
Explanation: E_axial = 2kp/r³ (directed along p). E_equatorial = kp/r³ (anti-parallel to p). Axial field is twice equatorial.
10. Electric field lines:
Explanation: Field lines represent direction of E at each point. Two lines cannot cross — that would mean E has two directions at one point, which is impossible.
11. Two identical conducting spheres (charges Q and −Q) touch and separate. Force between them after separation compared to before:
Explanation: When they touch, charge redistributes: equal and opposite charges neutralise. Total charge = 0. Each sphere gets 0 charge → force = 0.
12. Torque on an electric dipole p in uniform field E is:
Explanation: τ = p × E. Magnitude τ = pE sinθ where θ is angle between p and E.
13. A uniformly charged ring (charge Q, radius R) — field at centre is:
Explanation: By symmetry, every element contributes a field that is exactly cancelled by the diametrically opposite element. Net E = 0 at centre.
14. Force on charge q in field E is:
Explanation: F = qE. This is the definition of electric field: E = F/q → F = qE.
15. Electric field at distance r from infinite line charge (linear charge density λ) is:
Explanation: E = λ/(2πε₀r) = 2kλ/r. Perpendicular to the line, decreasing as 1/r (not 1/r² like a point charge).
16. Work done in rotating a dipole from θ₁ to θ₂ in a uniform electric field:
Explanation: U = −pE cosθ. Work done = ΔU = U₂ − U₁ = −pE cosθ₂ + pE cosθ₁ = pE(cosθ₁ − cosθ₂).
17. Field inside a uniformly charged spherical shell:
Explanation: By Gauss's law (or shell theorem): E = 0 everywhere inside a uniformly charged shell. Outside: E = kQ/r².
18. Electrostatic force is:
Explanation: Like charges repel, unlike charges attract. The sign of qq' in Coulomb's law determines direction.
19. Surface charge density σ on an infinite plane gives electric field:
Explanation: Infinite plane with surface charge σ: E = σ/(2ε₀) on each side, perpendicular to the surface. Between two plates of opposite σ: E = σ/ε₀.
20. Permittivity ε₀ has SI unit:
Explanation: From F = q₁q₂/(4πε₀r²): ε₀ = q₁q₂/(4πFr²) → C²/(N·m²) = C²·N⁻¹·m⁻².
21. Gauss's law states that electric flux through a closed surface is:
Explanation: ΦE = ∮E·dA = q_enc/ε₀. The flux depends only on the charge enclosed, not on the shape of the surface.
22. Electric flux through a surface is:
Explanation: Flux Φ = E · A = EA cosθ where θ is angle between E and the outward normal to the surface.
23. Electric field inside a conductor in electrostatic equilibrium is:
Explanation: Free charges in conductor rearrange to cancel any internal field. E = 0 inside conductor in electrostatics.
24. By Gauss's law, field just outside a conductor surface with surface charge density σ is:
Explanation: Using Gauss's law with a pillbox: E_out = σ/ε₀ (perpendicular to surface). Inside conductor: E = 0. The surface charge creates field σ/ε₀ outside.
25. Using Gauss's law, field inside a solid uniformly charged sphere (radius R) at distance r < R:
Explanation: Enclosed charge = Q(r/R)³. E×4πr² = Q(r/R)³/ε₀. E = kQr/R³ ∝ r inside the sphere.
26. Gauss's law is most useful for calculating E when:
Explanation: With high symmetry, E is constant over a Gaussian surface, allowing it to be factored out of the integral.
27. A charge q is at centre of a cube. Flux through each face is:
Explanation: Total flux = q/ε₀ by Gauss's law. By symmetry, each of 6 faces gets equal flux: q/(6ε₀).
28. If a closed surface encloses no net charge, then flux through the surface is:
Explanation: By Gauss's law, ΦE = q_enc/ε₀ = 0 when q_enc = 0. All field lines that enter must also exit.
29. Field at distance r from a long straight uniformly charged rod (linear density λ, r < rod length) is:
Explanation: Cylindrical Gaussian surface: E×2πrL = λL/ε₀ → E = λ/(2πε₀r). Same as 2kλ/r.
30. Excess charge on a conductor:
Explanation: Charge redistributes to outer surface. If any charge were inside, Gauss's law would require an E field there — but E = 0 inside conductor, so no charge can reside inside.
31. A point charge +q is placed at one corner of a cube of side L. Flux through one face not adjacent to the charge:
Explanation: Use 8-cube argument: +q at corner means 8 identical cubes share this corner. Total flux from 8 cubes = q/ε₀. Each cube gets q/(8ε₀). Cube has 3 faces adjacent to charge (zero flux by symmetry) and 3 non-adjacent. By symmetry, each non-adjacent face gets q/(8ε₀)/3 = q/(24ε₀).
32. Field inside a spherical cavity (radius a) carved from a uniformly charged solid sphere (radius R, charge density ρ), cavity centre displaced by d from sphere centre:
Explanation: By superposition: field = (field of full sphere) − (field of material removed). The cavity has uniform E = ρd/(3ε₀) directed from sphere centre to cavity centre.
33. A conducting shell shields the interior from:
Explanation: External field: charge redistributes on shell surface → E = 0 inside (shielding). But a charge inside the cavity induces surface charges and creates field inside — not shielded from internal sources.
34. Electric flux through a hemisphere of radius R placed in a uniform field E (field along axis of hemisphere):
Explanation: Flux through curved part = flux through flat base (by Gauss's law, no charge enclosed, total = 0 for closed surface). Flat base: area = πR², flux = E×πR². Curved hemisphere gets same but opposite sign. So flux through curved hemisphere = πR²E (outward).
35. Field at distance r from centre of infinite flat plane of charge (volume density ρ, thickness t):
Explanation: Total surface charge density σ = ρt. E = σ/(2ε₀) = ρt/(2ε₀) on each side.
36. Electric field E and displacement D in a dielectric medium (relative permittivity ε_r) relate as:
Explanation: D = ε₀ε_rE = εE where ε is absolute permittivity. D (displacement vector) accounts for free charges; polarization P accounts for bound charges.
37. For a non-uniform electric field, flux through a surface element dA is:
Explanation: Flux = ∮E·dA = surface integral. For non-uniform E, must integrate over the entire surface element by element.
38. Charge density on a conductor surface is greatest where:
Explanation: Field and surface charge density σ are highest at regions of greatest curvature (sharp points). Lightning rods use this — intense field at tip ionises air, providing discharge path.
39. Two concentric shells (radii R₁ < R₂) carry charges Q₁ and Q₂. Field at R₁ < r < R₂:
Explanation: Gauss's law: enclosed charge at R₁
40. Differential form of Gauss's law is:
Explanation: Maxwell's first equation: ∇·E = ρ/ε₀ where ρ is the free charge volume density. This is the differential (local) form of Gauss's law.
41. Electric potential V at a point is defined as:
Explanation: V = W/q₀ = work done by external agent per unit charge to bring a positive test charge from infinity to that point, with no kinetic energy gained.
42. Electric potential due to point charge q at distance r is:
Explanation: V = kq/r. Unlike E (∝ 1/r²), potential falls off as 1/r.
43. Potential difference between points A and B is V_A − V_B. Work done to move charge q from A to B is:
Explanation: W_AB = q(V_A − V_B). Positive when moving from high V to low V (charge naturally flows this way).
44. Equipotential surfaces are:
Explanation: No work is done moving a charge along an equipotential surface (ΔV = 0). Since W = F·d, E must be perpendicular to the surface.
45. Potential energy of two point charges q₁ and q₂ at distance r is:
Explanation: U = kq₁q₂/r. This is the work done to assemble the pair from infinity. Positive for like charges (repulsion stores energy).
46. Relation between electric field and potential is:
Explanation: E = −∇V. In one dimension: Ex = −dV/dx. Electric field points from high to low potential (negative gradient).
47. Inside a conductor, potential is:
Explanation: E = 0 inside conductor → E = −dV/dx = 0 → V = constant. This constant equals the surface potential.
48. For a uniform electric field E in +x direction, equipotential surfaces are:
Explanation: V = −Ex + const. V = const means x = const — planes perpendicular to E.
49. Potential energy of three charges +q each placed at corners of equilateral triangle (side L) is:
Explanation: U_total = U₁₂ + U₁₃ + U₂₃ = 3×(kq×q/L) = 3kq²/L.
50. Potential at centre of a square with charges +q at all four corners (side L) is:
Explanation: Distance from corner to centre = L√2/2. V = 4 × kq/(L√2/2) = 4 × 2kq/(L√2) = 8kq/(L√2) = 4√2 kq/L.
51. V = 3x² + 4y² (in SI units). Electric field at (1,0) m is:
Explanation: E_x = −∂V/∂x = −6x. E_y = −∂V/∂y = −8y. At (1,0): E = −6x̂ − 0ŷ = −6x̂ N/C.
52. Work done to assemble 4 charges +q each at corners of square (side L) is:
Explanation: 6 pairs: 4 pairs at distance L (sides), 2 pairs at distance L√2 (diagonals). U = 4kq²/L + 2kq²/(L√2) = kq²(4 + √2)/L.
53. Potential on axis of a uniformly charged disc (surface charge density σ, radius R) at distance x:
Explanation: V = σ/(2ε₀)[√(R²+x²) − x]. For x >> R: V → kQ/x (point charge limit). At x=0 (centre): V = σR/(2ε₀).
54. If V is constant throughout a region, the electric field in that region is:
Explanation: E = −∇V. If V = constant, ∇V = 0, so E = 0 everywhere in the region.
55. No two equipotential surfaces can intersect because:
Explanation: Potential V is a single-valued function at each point. Intersection of two equipotentials of different V values would mean two potentials at one point — a contradiction.
56. A charge q is released from rest in a field E. It moves a distance d. Kinetic energy gained:
Explanation: Work done by field = force × distance = qE × d = qEd. By work-energy theorem, this equals kinetic energy gained.
57. Potential at a point due to an electric dipole (p) at distance r, angle θ from dipole axis:
Explanation: V_dipole = kp cosθ/r². At θ=0 (axial): V = kp/r². At θ=90° (equatorial): V = 0.
58. Potential energy of an electric dipole in a uniform field E at angle θ:
Explanation: U = −p·E = −pE cosθ. Minimum energy (stable) at θ = 0 (aligned with field). Maximum (unstable) at θ = π.
59. Near a conductor surface, which quantity changes abruptly (has a discontinuity)?
Explanation: E_normal = σ/ε₀ just outside, 0 inside — discontinuous jump. V is continuous across the surface. Tangential E is continuous (no work done along surface).
60. Electrostatic potential energy stored in an electric field E occupying volume V is:
Explanation: Energy density = ½ε₀E². Total energy = ½ε₀E² × Volume (for uniform field). This is analogous to ½μ₀H² for magnetic energy density.
61. Force between charges of +4 μC and −2 μC placed 0.3 m apart is (k = 9×10⁹):
Explanation: F = kq₁q₂/r² = 9×10⁹ × 4×10⁻⁶ × 2×10⁻⁶ / (0.09) = 9×10⁹ × 8×10⁻¹² / 0.09 = 72×10⁻³/0.09 = 0.8 N. Unlike charges → attractive.
62. A charge of 2 μC is placed in a field of 1000 N/C. Force on it is:
Explanation: F = qE = 2×10⁻⁶ × 1000 = 2×10⁻³ N = 0.002 N.
63. Work done to move 3 C from A (V=20V) to B (V=5V) is:
Explanation: W = q(V_A−V_B) = 3×(20−5) = 45 J (work done by external agent). Work done by field = −45 J... Wait: W_external = q(V_A − V_B) — moving charge from A(20V) to B(5V), high to low potential. W_field = q(V_A−V_B) = +45 J by field. External work = −45 J. Test context: W = q×ΔV = 3×(5−20) = −45 J for work done by external force going to lower potential. Choose based on convention asked.
64. Charge +Q is placed at centre of a spherical shell of radius R. Charge on inner surface of shell is:
Explanation: Gauss's law with surface inside shell conductor: E = 0 inside conductor → charge on inner surface = −Q to cancel the field.
65. An electric dipole in a non-uniform field experiences:
Explanation: In uniform field: torque exists but net force = 0. In non-uniform field: both net force (since field varies across dipole extent) and torque exist.
66. At which point is potential zero for two charges +q at (0,0) and −q at (d,0)?
Explanation: V = kq/r₁ + k(−q)/r₂. V = 0 where r₁ = r₂, which is the perpendicular bisector plane (equatorial plane of the dipole). Any point equidistant from both charges.
67. Two large parallel plates with charges +σ and −σ. Field between them is:
Explanation: Each plate contributes σ/(2ε₀). Between plates, both contributions add: E = σ/(2ε₀) + σ/(2ε₀) = σ/ε₀. Outside: they cancel, E = 0.
68. Three charges +q, +q, −q are placed at (0,0), (d,0), (d/2,d√3/2) — equilateral triangle. Net force on −q:
Explanation: Each +q exerts kq²/d² on −q. By vector addition (120° between them): F_net = √3×kq²/d² directed toward midpoint of the two +q charges.
69. If E = 5x² ĵ N/C, work done to move 2 C from (0,0) to (0,3) m:
Explanation: W = ∫F·dl = q∫E_y dy = 2∫₀³ 5x² dy. But E = 5x²ĵ — along x = 0 (path along y-axis), x = 0, so E = 0. W = 0 J. If the field were 5y² ĵ: W = 2∫₀³ 5y² dy = 10[y³/3]₀³ = 10×9 = 90 J.
70. Electric field on axis of uniformly charged ring (charge Q, radius R) at distance x from centre:
Explanation: E_axial = kQx/(R²+x²)^(3/2). Maximum at x = R/√2. Integrating ring elements, components perpendicular to axis cancel by symmetry.
71. An infinite line charge (λ = 2 μC/m) is at origin. Field at 0.1 m:
Explanation: E = λ/(2πε₀r) = 2kλ/r = 2×9×10⁹×2×10⁻⁶/0.1 = 36×10³/0.1... = 2×9×10⁹×2×10⁻⁶/0.1 = 360,000 N/C = 360 kN/C.
72. Two protons (charge e, mass m) are released from rest at separation r₀. Speed of each when separation doubles:
Explanation: Energy conservation: ke²/r₀ = ke²/(2r₀) + 2×½mv². ke²/(2r₀) = mv². v = √(ke²/(2mr₀)).
73. Non-conducting sphere (radius R, uniform volume charge density ρ): ratio of fields at r = R/2 and r = 2R:
Explanation: Inside (r=R/2): E_in = ρr/(3ε₀) = ρR/(6ε₀). Outside (r=2R): E_out = kQ/(2R)² = ρR³/(3ε₀×4R²) = ρR/(12ε₀). Ratio = (ρR/6ε₀)/(ρR/12ε₀) = 2:1. Hmm: E_in:E_out = 2:1. So answer is 2:1.
74. Number of field lines emerging from a charge q is proportional to:
Explanation: By Gauss's law, total flux = q/ε₀ ∝ q. In a field line diagram convention, number of lines ∝ q.
75. A hollow conducting sphere (inner radius R₁, outer R₂) with charge Q on it. Potential at r = R₁ (inner surface) is:
Explanation: All charge resides on outer surface (R₂). Potential is constant throughout conductor = kQ/R₂. So V at inner surface = kQ/R₂.
76. Gauss's law in a dielectric medium: ε₀∇·E = ρ_free + ρ_bound. With displacement field D: ∇·D equals:
Explanation: D = ε₀E + P. ∇·D = ρ_free. This is the macroscopic Gauss's law in matter — D responds only to free charges.
77. Equipotential surface through a point in free space must be:
Explanation: By definition: E = −∇V. On an equipotential surface, V = const → ∇V along surface = 0 → E is perpendicular to the surface at every point.
78. Field between two conducting concentric spheres (radii R₁ and R₂, potential V₁ and V₂) at radius r:
Explanation: V(r) = A + B/r with B.C.s V(R₁)=V₁, V(R₂)=V₂. Solving: E = −dV/dr = B/r² = R₁R₂(V₁−V₂)/((R₂−R₁)r²).
79. Charge Q is uniformly distributed on surface of a sphere of radius R. Electric field at the surface:
Explanation: By Gauss's law: E at surface = kQ/R² — same as if all charge were at centre. This is the shell theorem result.
80. Method of images: a point charge q at distance d from an infinite grounded conductor can be replaced by:
Explanation: Method of images: grounded conductor (V=0) replaced by image charge −q at the mirror position (d behind the surface). The induced surface charge creates the same field as this image charge.
81. Force on charge q at distance d from a grounded conducting plane is:
Explanation: Image charge −q is at distance 2d away. F = kq²/(2d)² = kq²/(4d²), directed toward the plane (attractive).
82. Polarisation P in a dielectric is:
Explanation: P = dp/dV = electric dipole moment per unit volume. ∇·P = −ρ_bound (bound charge comes from divergence of polarisation).
83. Uniqueness theorem in electrostatics states:
Explanation: Given the potential (or its normal derivative) on all boundaries, the solution to Laplace's equation (∇²V = 0) in a region is unique. Guarantees conductor arrangements have one unique field solution.
84. Earnshaw's theorem states that:
Explanation: Earnshaw's theorem: ∇²V = 0 (Laplace's equation) in free space means no local maximum/minimum of V, hence no stable equilibrium position for a charge in free space.
85. Total energy stored in the field of a uniformly charged sphere (charge Q, radius R):
Explanation: U = (3/5)kQ²/R for a uniformly charged sphere (volume charge). For surface-charged shell: U = kQ²/(2R).
86. At large distances, the dominant term in the potential of an arbitrary charge distribution is (if total charge ≠ 0):
Explanation: Multipole expansion: V = V_mono + V_dipole + V_quadrupole +... If total charge Q ≠ 0, monopole term kQ/r dominates at large r. If Q = 0, dipole term dominates.
87. Laplace's equation ∇²V = 0 applies in regions with:
Explanation: In charge-free regions, ρ = 0, so Poisson's equation ∇²V = −ρ/ε₀ reduces to Laplace's equation ∇²V = 0.
88. Why is a conducting shell an ideal Faraday cage?
Explanation: External field causes charge redistribution on the outer surface. This induced charge creates a field inside that exactly cancels the external field. E = 0 inside regardless of external field strength.
89. Two conducting spheres of radii R₁ and R₂ connected by a long wire. In equilibrium, surface charge densities σ₁ and σ₂ satisfy:
Explanation: Connected spheres: same potential V = kQ₁/R₁ = kQ₂/R₂. σ = Q/(4πR²). kσ×4πR²/R = const → σR = const → σ₁R₁ = σ₂R₂. Smaller sphere has higher σ (and higher E at surface).
90. A charge distribution has zero monopole moment (total charge = 0) and zero dipole moment. The dominant far-field potential is:
Explanation: With Q = 0 and p = 0, the quadrupole term dominates: V_quadrupole ∝ 1/r² and E_quadrupole ∝ 1/r³. Example: two dipoles back-to-back.
91. At an interface between two dielectrics, which component of field is continuous?
Explanation: Boundary conditions: E_tangential is continuous (from ∮E·dl = 0). D_normal is continuous if no surface free charge. E_normal is discontinuous (D_n continuous + different ε_r).
92. When a +Q charged object is brought near an isolated neutral conductor, the conductor's net charge is:
Explanation: Induction redistributes charge but doesn't change net charge. Near side: −Q induced. Far side: +Q induced. Total remains zero (conductor was neutral).
93. Dielectric breakdown (sparking) occurs when electric field exceeds:
Explanation: Every insulator has a dielectric strength — maximum field before breakdown (ionisation/conduction). For air: ≈ 3×10⁶ V/m. Lightning occurs when charge buildup exceeds this threshold.
94. The self-energy of a point charge (formal calculation) is:
Explanation: Self-energy = energy to assemble a point charge from infinity = ∫₀^∞ field energy = (½ε₀)(kq/r)² × 4πr²dr → diverges as r → 0. This UV divergence is a foundational problem in classical electrodynamics, resolved only in quantum field theory (renormalisation).