JEE Physics Electric Charges And Fields MCQs Set 26

Answer: (b) electric dipole moment

Explanation:
1. 1 debye = 3.34 × 10⁻³⁰ C m, a unit of dipole moment.

Teacher's Note:
1. Debye measures dipole moment.

Answer: (a) zero

Explanation:
1. The field is parallel to the curved surface everywhere, so there is no flux through it.

Teacher's Note:
1. E is tangent to the curved surface.

Answer: (c) there is a minimum permissible charge on a particle

Explanation:
1. Every charge is an integral multiple of e.
2. So there is a minimum possible charge, e = 1.6 × 10⁻¹⁹ C.

Teacher's Note:
1. q = ne.

Answer: (d) are imaginary

Explanation:
1. Lines of force are a visual device for representing the field.
2. They are imaginary, not physical.

Teacher's Note:
1. They are only a representation.

Answer: (d) directly proportional to E

Explanation:
1. For a linear dielectric P = χε₀E.
2. So P is directly proportional to E.

Teacher's Note:
1. P = χε₀E.

Answer: (a) 90°

Explanation:
1. τ = pE sin θ is maximum when sin θ = 1, that is θ = 90°.

Teacher's Note:
1. τ is maximum at 90°.

Answer: (b) A-3, B-4, C-2, D-1

Explanation:
1. The dipole moment points from −q towards +q (A-3).
2. The field points away from a positive charge (B-4).
3. Flux is the scalar product E·A (C-2).
4. Torque is the vector product p × E (D-1).

Teacher's Note:
1. Direction and product type.

Answer: (c) \( \frac1r \)

Explanation:
1. By Gauss's law E = λ/(2πε₀r).

Teacher's Note:
1. E ∝ 1/r.

Answer: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. Rubbing only moves electrons between the bodies.
2. The total charge stays the same because charge is conserved.

Teacher's Note:
1. Conservation of charge.

Answer: (d) A-3, B-1, C-2, D-4

Explanation:
1. Coulomb's law gives an inverse square force (A-3).
2. Gauss's law concerns the total flux through a closed surface (B-1).
3. Superposition gives the vector sum of forces (C-2).
4. Quantisation is the discrete nature of charge (D-4).

Teacher's Note:
1. Name the idea behind each law.

Answer: (b) conductor, insulator

Explanation:
1. Free charges move easily on a conductor and spread over its surface.
2. In an insulator the charge stays where it is placed.

Teacher's Note:
1. Conductors spread charge.

Answer: (c) \( \frac49F \)

Explanation:
1. The effective distance in air is r_eff = (r − t) + t√K = r/2 + (r/2)(2) = 3r/2.
2. The force is F′ = F (r/r_eff)² = F × 4/9.

Teacher's Note:
1. Effective distance method.

Answer: (b) does not lose any charge

Explanation:
1. In induction the charging body does not touch the other body.
2. It therefore keeps all its own charge.

Teacher's Note:
1. Induction involves no loss of charge.

Answer: (a) zero

Explanation:
1. By Gauss's law a Gaussian surface inside the shell encloses no charge.
2. So the field inside is zero.

Teacher's Note:
1. Shell theorem.

Answer: (b) in the opposite direction

Explanation:
1. On the equatorial line the field is directed opposite to the dipole moment p.

Teacher's Note:
1. E_eq = −kp/r³.

Answer: (c) \( 8.85\times10^{-12}\ \text{C}^2/\text{Nm}^2 \)

Explanation:
1. ε₀ = 8.85 × 10⁻¹² C²/(N m²).

Teacher's Note:
1. ε₀ = 8.85 × 10⁻¹² C² N⁻¹ m⁻².

Answer: (d) A-4, B-1, C-2, D-3

Explanation:
1. A point charge gives q/4πε₀r² (A-4).
2. A long wire gives λ/2πε₀r (B-1).
3. A sheet gives σ/2ε₀ (C-2).
4. The field inside a shell is zero (D-3).

Teacher's Note:
1. Standard field expressions.

Answer: (d) zero

Explanation:
1. The field lines are parallel to the surface, so the angle between E and the area vector is 90°.
2. Φ = EL² cos 90° = 0.

Teacher's Note:
1. Field in the plane gives zero flux.

Answer: (c) radial, inwards

Explanation:
1. Field lines point towards a negative charge and are radial.

Teacher's Note:
1. Lines end on negative charges.

Answer: (b) \( \frac{-q}{4\pi R_1^2} \)

Explanation:
1. A charge −q is induced on the inner surface so that the field inside the metal is zero.
2. σ = −q/(4πR₁²).

Teacher's Note:
1. Induced charge is −q.

Answer: (d) \( 3.7\times10^{-3}\ \text{C} \)

Explanation:
1. Q = σ × 4πr² with r = 2.2 m.
2. Q = 60 × 10⁻⁶ × 4π × 4.84 = 3.65 × 10⁻³ ≈ 3.7 × 10⁻³ C.

Teacher's Note:
1. Q = σ·4πr².

Answer: (c) a torque as well as a translational force

Explanation:
1. A dipole at an angle to a field feels a torque.
2. In a non-uniform field the two forces also do not cancel, so there is a net translational force too.

Teacher's Note:
1. Both torque and force.

Answer: (a) II and III

Explanation:
1. CO₂ is linear and symmetric, so it is non-polar.
2. H₂O is bent and polar. Hence II and III are incorrect.

Teacher's Note:
1. Symmetry cancels the dipole.

Answer: (b) from the negative charge to the positive charge

Explanation:
1. The dipole moment, and hence the axis, points from the negative charge to the positive charge.

Teacher's Note:
1. p points from −q to +q.

Answer: (d) Coulomb's law

Explanation:
1. The force between two point charges is given by Coulomb's law.

Teacher's Note:
1. F = kq₁q₂/r².