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NATIONAL INSTITUTE OF OPEN SCHOOLING

Senior Secondary Course - Physics (312)

Answer Key

Lessons 15: Electric Charge and Electric Field and 16: Electric Potential and Capacitors

Question Paper: 80 Marks Answer Format: Concise Model Key
Use this as a marking key. Equivalent wording, correct formulas, and correct units should also be accepted.

Section A - Multiple Choice Answers

  1. Answer: C
  2. Answer: B
  3. Answer: C
  4. Answer: B
  5. Answer: A
  6. Answer: C
  7. Answer: B
  8. Answer: A
  9. Answer: D
  10. Answer: B
  11. Answer: B
  12. Answer: C
  13. Answer: B
  14. Answer: C
  15. Answer: B
  16. Answer: D

Section B - Short Answer Key

  1. Any two: charge is conserved, charge is quantised, like charges repel, unlike charges attract, charge is additive/scalar in nature.
  2. Conservation of charge means total charge of an isolated system remains constant. When a glass rod is rubbed with silk, electrons transfer from glass to silk; both acquire equal and opposite charges, so net charge remains unchanged.
  3. Quantisation means charge exists in integral multiples of electronic charge: Q = ne, where n is an integer. A body cannot have 2.5e because charge cannot exist in fractional multiples of e.
  4. F = (1 / 4πε0) (q1q2 / r2). The force depends on the magnitudes of the charges, the distance between them, and the nature of the medium.
  5. Electric field intensity at a point is the force acting on a unit positive test charge placed there. For a point charge q at distance r, E = (1 / 4πε0) (q / r2).
  6. An electric dipole is a pair of equal and opposite charges separated by a small distance. Its dipole moment is p = q(2l) and is directed from negative to positive charge.
  7. Any two: field lines start from positive and end at negative charges; tangent at any point gives field direction; no two field lines intersect; closer lines indicate stronger field.
  8. Gauss's law states that the total electric flux through any closed surface equals the enclosed charge divided by permittivity of free space: ΦE = qenclosed / ε0.
  9. Electric potential at a point is the work done in bringing a unit positive charge from infinity to that point. It is scalar because work is a scalar quantity and potentials add algebraically.
  10. VB - VA = WAB / q0, where WAB is the work done in moving the test charge q0 from A to B.
  11. Capacitance is the charge stored per unit potential difference: C = Q / V. Its SI unit is farad (F).
  12. Dielectric constant or relative permittivity is the ratio of permittivity of the medium to permittivity of vacuum. One relation is K = ε / ε0 = C / C0.

Section C - Very Short Answer Key

  1. n = Q / e = (4.8 × 10-19) / (1.6 × 10-19) = 3. The body possesses 3 electronic charges.
  2. F = 9 × 109 [(2 × 10-6)(2 × 10-6) / 32] = 4 × 10-3 N. Since both charges are like, the force is repulsive.
  3. E = 9 × 109 (5 × 10-9 / 0.102) = 4.5 × 103 N C-1.
  4. p = qd = (4 × 10-6)(5 × 10-3) = 2 × 10-8 C m.
  5. The electric field inside a uniformly charged spherical shell is zero.
  6. V = 9 × 109 (8 × 10-9 / 0.20) = 360 V.
  7. C = Q / V = (6 × 10-4) / 200 = 3 × 10-6 F = 3 μF.
  8. For parallel combination, Ceq = 2 + 3 + 5 = 10 μF.
  9. U = 1/2 CV2 = (1/2)(4 × 10-6)(122) = 2.88 × 10-4 J.

Section D - 3 Mark Answer Key

  1. Consider a short dipole of moment p = q(2l). At a point on the axial line at distance r, the two charges produce fields in the same direction. Using superposition and for r >> l, Eaxial = (1 / 4πε0)(2pr / r4) = (1 / 4πε0)(2p / r3), directed along the dipole axis.
  2. Take a cylindrical Gaussian surface of radius r and length l around the wire. By symmetry, field is radial and constant on the curved surface, and flux through the end caps is zero. Hence Φ = E(2πrl). If charge per unit length is λ, enclosed charge is λl. By Gauss's law, E(2πrl) = λl / ε0, so E = λ / (2πε0r).
  3. Electric potential due to a point charge is the work done per unit positive charge in bringing it from infinity to the point. Since V = -∫∞r E dr and E = (1 / 4πε0)(q / r2), integration gives V = (1 / 4πε0)(q / r).
  4. For two capacitors in series, the charge on each capacitor is the same: Q. Total potential difference is V = V1 + V2 = Q/C1 + Q/C2. If equivalent capacitance is Cs, then V = Q/Cs. Therefore 1/Cs = 1/C1 + 1/C2. In series grouping, charge is same on each capacitor while potential difference is divided.

Section E - 5 Mark Answer Key

  1. Gauss's law: ΦE = qenclosed / ε0. For a uniformly charged spherical shell of radius R carrying charge Q, choose a concentric spherical Gaussian surface. For an external point (r > R), symmetry gives E(4πr2) = Q / ε0, hence E = (1 / 4πε0)(Q / r2). On the surface (r = R), E = (1 / 4πε0)(Q / R2). For an internal point (r < R), enclosed charge is zero, so E = 0. Thus outside the shell it behaves like a point charge at the centre, while inside the shell the field is zero.
  2. A parallel plate capacitor consists of two large conducting plates of area A separated by distance d, carrying equal and opposite charges. The electric field between the plates is approximately uniform: E = σ / ε0, where σ = Q/A. The potential difference is V = Ed = Qd / (ε0A). Therefore capacitance is C = Q / V = ε0A / d. If a dielectric slab of constant K is inserted fully between the plates, capacitance becomes C = Kε0A / d = KC0, so the capacitance increases by the factor K.