Torque Dipole

Answer: B. The forces qE and −qE cancel in a uniform field, but they act at different points and form a couple of magnitude pE sin θ (NCERT Class 12 Physics Part I, Chapter 1, Section 1.11, Eq. 1.22, page 27).

Why A is wrong: A is wrong because it is the trap this lesson names: the net force is zero, but the two forces act at different points, so a torque pE sin θ remains unless θ is 0° or 180°.

Why C is wrong: C is wrong because it counts only the force on +q; the force −qE on −q cancels it, and the couple they form gives a non-zero torque.

Why D is wrong: D is wrong because it adds the two force magnitudes instead of cancelling opposite forces, and it uses cos θ where the torque needs sin θ.

Answer: C. NCERT Class 12 Physics Part I, Chapter 1, Section 1.11 (page 27): the torque tends to align the dipole with the field E, and when p is aligned with E the torque is zero.

Why A is wrong: A is wrong because NCERT states the torque tends to align p with E, not to point p opposite to E.

Why B is wrong: B is wrong because the perpendicular position is where the torque is largest (sin 90° = 1), so the dipole is still being turned there, not brought to rest.

Why D is wrong: D is wrong because the net force in a uniform field is zero, so there is no push along the field; the effect is a rotation.

Answer: B. NCERT Class 12 Physics Part I, Chapter 1, Section 1.11 (page 27): when p is parallel to E in a non-uniform field, the net torque is zero and the net force is in the direction of increasing field.

Why A is wrong: A is wrong because a net force toward decreasing field is NCERT's result for p antiparallel to E, not parallel.

Why C is wrong: C is wrong because it applies the uniform-field result; in a non-uniform field the forces on +q and −q differ, so the net force is not zero, and with p parallel to E the torque is zero.

Why D is wrong: D is wrong because with p parallel to E both charge forces lie along the field direction, so their resultant cannot be perpendicular to E.

Answer: A. τ = pE sin θ = 3.0 × 10⁻⁹ × 4.0 × 10⁴ × sin 30° = 1.2 × 10⁻⁴ × 0.5 = 6.0 × 10⁻⁵ N m (NCERT Class 12 Physics Part I, Chapter 1, Eq. 1.22, page 27).

Why B is wrong: B is wrong because 1.2 × 10⁻⁴ N m is pE with sin θ dropped; that is the maximum torque, which applies only at 90°.

Why C is wrong: C is wrong because 1.0 × 10⁻⁴ N m comes from pE cos 30° = 1.2 × 10⁻⁴ × 0.866; the torque needs sin θ, not cos θ.

Why D is wrong: D is wrong because 2.4 × 10⁻⁴ N m comes from dividing pE by sin 30° (1.2 × 10⁻⁴ / 0.5) instead of multiplying.

Answer: D. Each charge feels qE = 2.0 × 10⁻⁶ × 3.0 × 10⁵ = 0.60 N; the forces are equal and opposite, so in a uniform field the net force is zero (NCERT Class 12 Physics Part I, Chapter 1, Section 1.11, page 27).

Why A is wrong: A is wrong because 1.2 N adds the two 0.60 N magnitudes (0.60 + 0.60) as if both forces pointed along the field; the force on −q points opposite to E.

Why B is wrong: B is wrong because 0.60 N counts only the force on +q and ignores the equal and opposite force on −q.

Why C is wrong: C is wrong because 3.0 × 10⁻³ comes from multiplying qE by the 5.0 × 10⁻³ m separation (0.60 × 5.0 × 10⁻³); that product belongs to the torque (qE × 2a, in N m), not to the force on a charge.

Answer: C. The torque pE sin θ is maximum at θ = 90°, where it equals pE. So p = τ_max / E = 8.0 × 10⁻³ / 2.0 × 10⁵ = 4.0 × 10⁻⁸ C m (NCERT Class 12 Physics Part I, Chapter 1, Eq. 1.22, page 27).

Why A is wrong: A is wrong because 1.6 × 10³ comes from multiplying τ_max by E (8.0 × 10⁻³ × 2.0 × 10⁵) instead of dividing.

Why B is wrong: B is wrong because 2.5 × 10⁷ comes from E / τ_max (2.0 × 10⁵ / 8.0 × 10⁻³), the inverted ratio.

Why D is wrong: D is wrong because 8.0 × 10⁻⁸ comes from p = τ / (E sin 30°) = 8.0 × 10⁻³ / (2.0 × 10⁵ × 0.5); the maximum torque occurs at 90°, where sin θ = 1.

Answer: D. p = q × 2a = 5.0 × 10⁻⁶ × 2.0 × 10⁻³ = 1.0 × 10⁻⁸ C m. Then τ = pE sin 150° = 1.0 × 10⁻⁸ × 1.5 × 10⁵ × 0.5 = 7.5 × 10⁻⁴ N m (NCERT Class 12 Physics Part I, Chapter 1, Section 1.11, page 27).

Why A is wrong: A is wrong because 1.5 × 10⁻³ N m is pE with sin θ dropped; sin 150° = 0.5, so the torque is half of pE.

Why B is wrong: B is wrong because 3.8 × 10⁻⁴ N m uses half the separation (1.0 mm) for p, giving p = 5.0 × 10⁻⁹ C m and τ = 3.75 × 10⁻⁴ N m; p is q times the full separation 2a.

Why C is wrong: C is wrong because 1.3 × 10⁻³ N m uses the size of cos 150° (0.866) instead of sin 150° (0.5): 1.5 × 10⁻³ × 0.866 ≈ 1.3 × 10⁻³.

Answer: A. Maximum torque is pE (at 90°). Setting pE sin θ = pE/2 gives sin θ = 1/2, so the acute angle is θ = 30° (NCERT Class 12 Physics Part I, Chapter 1, Eq. 1.22, page 27).

Why B is wrong: B is wrong because 60° solves cos θ = 1/2; the torque depends on sin θ, and sin 60° ≈ 0.87, not 0.5.

Why C is wrong: C is wrong because at 90° sin θ = 1, so the torque is at its maximum, not half of it.

Why D is wrong: D is wrong because 45° halves the angle of maximum torque (90°/2) instead of halving the torque; sin 45° ≈ 0.71, not 0.5.