JEE Physics Electric Charges And Fields MCQs Set 27

Practice MCQs for JEE Physics Electric Charges and Fields

Access targeted multiple-choice questions for Electric Charges and Fields designed to align with the latest JEE academic syllabus for JEE Physics. These objective practice sets help students evaluate their conceptual understanding and improve exam readiness.

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Multiple Choice Questions

Question 1: An electric dipole is placed at an angle of 30° with an electric field of intensity 2 × 10⁵ N/C. It experiences a torque of 4 N m. The charge on the dipole, if the dipole length is 2 cm, is
(a) 8 mC
(b) 4 mC
(c) 8 μC
(d) 2 mC
Show Answer & Explanation

Answer: (d) 2 mC

Explanation:
1. τ = q d E sin θ.
2. q = 4/(0.02 × 2 × 10⁵ × 0.5) = 2 × 10⁻³ C = 2 mC.

Teacher's Note:
1. τ = (qd)E sin θ.

Question 2: The electric field near a conducting surface having uniform surface charge density σ is given by
(a) \( \frac{\sigma}{\varepsilon_0}\text{ and is parallel to the surface} \)
(b) \( \frac{2\sigma}{\varepsilon_0}\text{ and is parallel to the surface} \)
(c) \( \frac{\sigma}{\varepsilon_0}\text{ and is normal to the surface} \)
(d) \( \frac{2\sigma}{\varepsilon_0}\text{ and is normal to the surface} \)
Show Answer & Explanation

Answer: (c) \( \frac{\sigma}{\varepsilon_0}\text{ and is normal to the surface} \)

Explanation:
1. Just outside a conductor E = σ/ε₀.
2. The field is normal to the surface, because a tangential component would move the charges.

Teacher's Note:
1. Coulomb's theorem.

Question 3: Which of the following statements is incorrect? I. The charge q on a body is always given by q = ne, where n is any integer, positive or negative. II. By convention, the charge on an electron is taken to be negative. III. The fact that electric charge is always an integral multiple of e is termed quantisation of charge. IV. The quantisation of charge was experimentally demonstrated by Newton in 1912.
(a) Only I
(b) Only II
(c) Only IV
(d) Only III
Show Answer & Explanation

Answer: (c) Only IV

Explanation:
1. Statements I, II and III are correct.
2. Quantisation of charge was demonstrated by Millikan, not by Newton, so IV is incorrect.

Teacher's Note:
1. Millikan's oil drop.

Question 4: A comb rubbed with dry hair attracts pieces of paper. This is because the
(a) comb polarises the piece of paper
(b) comb induces a net dipole moment opposite to the direction of the field
(c) electric field due to the comb is uniform
(d) comb induces a net dipole moment perpendicular to the direction of the field
Show Answer & Explanation

Answer: (a) comb polarises the piece of paper

Explanation:
1. The charged comb polarises the paper, bringing opposite charge nearer to it.
2. The attraction is then stronger than the repulsion.

Teacher's Note:
1. Polarisation gives attraction.

Question 5: Which of the following is correct about Gauss's theorem?
(a) It can be derived by using Coulomb's law
(b) It is valid for a conservative field obeying an inverse square law
(c) Gauss's theorem is applicable in gravitation too
(d) all of these
Show Answer & Explanation

Answer: (d) all of these

Explanation:
1. Gauss's law can be derived from Coulomb's law.
2. It holds for any inverse-square field and has a gravitational analogue.

Teacher's Note:
1. Inverse-square fields.

Question 6: The unit of electric field is not equivalent to
(a) N/C
(b) J/C
(c) V/m
(d) J/(C m)
Show Answer & Explanation

Answer: (b) J/C

Explanation:
1. N/C, V/m and J/(C m) are all units of electric field.
2. J/C is the unit of potential, not field.

Teacher's Note:
1. E = force per charge = V/m.

Question 7: The force of repulsion between two electrons at a certain distance is F. The force between two protons separated by the same distance is (m_p = 1836 m_e)
(a) 2F
(b) F
(c) 1836 F
(d) \( \frac{F}{1836} \)
Show Answer & Explanation

Answer: (b) F

Explanation:
1. The Coulomb force depends only on the charges and the distance.
2. Protons and electrons have equal magnitude of charge, so the force is F.

Teacher's Note:
1. Charge magnitudes are equal.

Question 8: Two identical metal spheres A and B are supported on insulating stands and placed in contact. What kind of charge will A and B develop when a negatively charged ebonite rod is brought near A?
(a) A will have a positive charge and B a negative charge
(b) A will have a negative charge and B a positive charge
(c) both A and B will have positive charges
(d) both A and B will have negative charges
Show Answer & Explanation

Answer: (a) A will have a positive charge and B a negative charge

Explanation:
1. The rod repels electrons to the far sphere B.
2. So A (near the rod) is left positive and B is negative.

Teacher's Note:
1. Induced charges.

Question 9: If a dipole of moment 2.57 × 10⁻¹⁷ C m is placed in an electric field of magnitude 3.0 × 10⁴ N/C such that the field lines make an angle of 30° with the dipole, what torque acts on the dipole?
(a) \( 7.7\times10^{-13}\ \text{N m} \)
(b) \( 3.855\times10^{-13}\ \text{N m} \)
(c) \( 3.855\times10^{-15}\ \text{N m} \)
(d) \( 7.7\times10^{-15}\ \text{N m} \)
Show Answer & Explanation

Answer: (b) \( 3.855\times10^{-13}\ \text{N m} \)

Explanation:
1. τ = pE sin θ = 2.57 × 10⁻¹⁷ × 3 × 10⁴ × 0.5.
2. = 3.855 × 10⁻¹³ N m.

Teacher's Note:
1. τ = pE sin θ.

Question 10: If one penetrates a uniformly charged spherical cloud, the electric field strength
(a) decreases directly as the distance from the centre
(b) increases directly as the distance from the centre
(c) remains constant
(d) none of these
Show Answer & Explanation

Answer: (b) increases directly as the distance from the centre

Explanation:
1. Inside a uniformly charged sphere, E = ρr/3ε₀.
2. So E increases directly with the distance from the centre.

Teacher's Note:
1. E ∝ r inside.

Question 11: Quantisation of charge was experimentally demonstrated by
(a) Einstein's photoelectric effect
(b) the Frank–Hertz experiment
(c) the Davisson–Germer experiment
(d) Millikan's oil drop experiment
Show Answer & Explanation

Answer: (d) Millikan's oil drop experiment

Explanation:
1. Millikan measured the charge on tiny oil drops and found it to be an integral multiple of e.

Teacher's Note:
1. Millikan's oil drop.

Question 12:
Assertion (A): Consider two identical charges Q placed a distance 2d apart along the x-axis. The equilibrium of a positive test charge placed at the point O midway between them is stable for displacements along the x-axis.
Reason (R): The force on the test charge at O is zero.
(a) Both A and R are true, and R is the correct explanation of A.
(b) Both A and R are true, but R is not the correct explanation of A.
(c) A is true, but R is false.
(d) A is false, but R is true.
Show Answer & Explanation

Answer: (b) Both A and R are true, but R is not the correct explanation of A.

Explanation:
1. For a small displacement along x the nearer charge pushes the test charge back, so the equilibrium is stable.
2. The force being zero at O is true but it does not explain stability.

Teacher's Note:
1. Stability needs a restoring force.

Question 13: If the electric field in a region is radially outward with magnitude E = Ar, the charge contained in a sphere of radius r centred at the origin is
(a) \( \frac{1}{4\pi\varepsilon_0}Ar^3 \)
(b) \( 4\pi\varepsilon_0Ar^3 \)
(c) \( \frac1{4\pi\varepsilon_0}\frac{A}{r^3} \)
(d) \( 4\pi\varepsilon_0\frac{A}{r^3} \)
Show Answer & Explanation

Answer: (b) \( 4\pi\varepsilon_0Ar^3 \)

Explanation:
1. Gauss's law: Ar × 4πr² = Q/ε₀.
2. So Q = 4πε₀Ar³.

Teacher's Note:
1. Φ = E × 4πr².

Question 14: The Gaussian surface
(a) can pass through a continuous charge distribution
(b) cannot pass through a continuous charge distribution
(c) can pass through any system of discrete charges
(d) can pass through a continuous charge distribution as well as any system of discrete charges
Show Answer & Explanation

Answer: (a) can pass through a continuous charge distribution

Explanation:
1. The field of a point charge is infinite at its position, so a surface cannot pass through a discrete charge.
2. A continuous distribution has a finite field, so a Gaussian surface may pass through it.

Teacher's Note:
1. Avoid point charges.

Question 15: A charge q is placed at the centre of the open end of a cylindrical vessel (in the plane of the opening). The flux of the electric field through the surface of the vessel is
(a) zero
(b) \( q/\varepsilon_0 \)
(c) \( q/2\varepsilon_0 \)
(d) \( 2q/\varepsilon_0 \)
Show Answer & Explanation

Answer: (c) \( q/2\varepsilon_0 \)

Explanation:
1. The charge sits in the plane of the opening, so half of the field lines pass into the vessel and half go outward.
2. The flux through the vessel is q/2ε₀.

Teacher's Note:
1. Charge on a plane: half the flux.

Question 16:
Assertion (A): The electric flux ∮E·dA is zero. The electric field is zero everywhere on the surface.
Reason (R): The charge inside the surface is zero.
(a) Both A and R are true, and R is the correct explanation of A.
(b) Both A and R are true, but R is not the correct explanation of A.
(c) A is true, but R is false.
(d) A is false, but R is true.
Show Answer & Explanation

Answer: (d) A is false, but R is true.

Explanation:
1. Zero flux means zero net charge inside, but the field on the surface need not be zero, so the assertion is false.
2. The reason is true.

Teacher's Note:
1. Zero flux does not mean zero field.

Question 17:
Assertion (A): The tyres of aircraft are slightly conducting.
Reason (R): If a conductor is connected to the ground, the extra charge induced on the conductor will flow to the ground.
(a) Both A and R are true, and R is the correct explanation of A.
(b) Both A and R are true, but R is not the correct explanation of A.
(c) A is true, but R is false.
(d) A is false, but R is true.
Show Answer & Explanation

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

Explanation:
1. Slightly conducting tyres let the static charge produced during landing flow to the ground.
2. This follows from the behaviour of a grounded conductor.

Teacher's Note:
1. Grounding removes charge.

Question 18: On charging by conduction, the mass of a body may
(a) increase
(b) decrease
(c) increase or decrease
(d) none of these
Show Answer & Explanation

Answer: (c) increase or decrease

Explanation:
1. A body that gains electrons gains mass and a body that loses electrons loses mass.
2. So the mass may increase or decrease.

Teacher's Note:
1. Electrons have mass.

Question 19: For a given surface Gauss's law is stated as ∮E·dA = 0. From this we can conclude that
(a) E is necessarily zero on the surface
(b) E is perpendicular to the surface at every point
(c) the total flux through the surface is zero
(d) the flux is only going out of the surface
Show Answer & Explanation

Answer: (c) the total flux through the surface is zero

Explanation:
1. The integral is the total flux.
2. It being zero means that the net flux through the surface is zero; the field itself need not vanish.

Teacher's Note:
1. Zero net flux.

Question 20: The total electric flux emanating from a closed surface enclosing an α-particle is (e is the electronic charge)
(a) \( \frac{2e}{\varepsilon_0} \)
(b) \( \frac{e}{\varepsilon_0} \)
(c) \( e\varepsilon_0 \)
(d) \( \frac{\varepsilon_0e}{4} \)
Show Answer & Explanation

Answer: (a) \( \frac{2e}{\varepsilon_0} \)

Explanation:
1. The α-particle carries a charge +2e.
2. Φ = 2e/ε₀.

Teacher's Note:
1. Φ = q/ε₀.

Question 21:
Assertion (A): A uniformly charged disc has a pin hole at its centre. The electric field at the centre of the disc is zero.
Reason (R): A disc can be supposed to be made up of many rings. Also the electric field at the centre of a uniformly charged ring is zero.
(a) Both A and R are true, and R is the correct explanation of A.
(b) Both A and R are true, but R is not the correct explanation of A.
(c) A is true, but R is false.
(d) A is false, but R is true.
Show Answer & Explanation

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

Explanation:
1. Each ring contributes zero field at its centre by symmetry.
2. Adding the rings gives zero at the centre.

Teacher's Note:
1. Symmetry.

Question 22:
Assertion (A): A deuteron and an α-particle are placed in an electric field. If F₁ and F₂ are the forces acting on them and a₁ and a₂ are their accelerations respectively, then a₁ = a₂.
Reason (R): The forces will be the same in an electric field.
(a) Both A and R are true, and R is the correct explanation of A.
(b) Both A and R are true, but R is not the correct explanation of A.
(c) A is true, but R is false.
(d) A is false, but R is true.
Show Answer & Explanation

Answer: (c) A is true, but R is false.

Explanation:
1. a = qE/m: the deuteron has e/2m_p and the α-particle has 2e/4m_p, which are equal.
2. The forces F = qE differ (e and 2e), so the reason is false.

Teacher's Note:
1. a = qE/m.

Question 23: When a charge q placed at the origin is displaced, the electric field at a point P at a distance OP from the origin changes
(a) at the same time when q is displaced
(b) at a time OP/c later, where c is the speed of light
(c) at a time OP cos θ/c later
(d) at a time OP sin θ/c later
Show Answer & Explanation

Answer: (b) at a time OP/c later, where c is the speed of light

Explanation:
1. Changes in the field travel at the speed of light.
2. The field at P changes after a delay OP/c.

Teacher's Note:
1. Field changes propagate at c.

Question 24: The insulating property of air breaks down when the electric field is 3 × 10⁶ V m⁻¹. The maximum charge that can be given to a sphere of diameter 5 m is approximately
(a) \( 2\times10^{-2}\ \text{C} \)
(b) \( 2\times10^{-3}\ \text{C} \)
(c) \( 2\times10^{-4}\ \text{C} \)
(d) \( 2\times10^{-5}\ \text{C} \)
Show Answer & Explanation

Answer: (b) \( 2\times10^{-3}\ \text{C} \)

Explanation:
1. E = kQ/R² gives Q = ER²/k = 3 × 10⁶ × (2.5)²/(9 × 10⁹).
2. = 2.08 × 10⁻³ ≈ 2 × 10⁻³ C.

Teacher's Note:
1. Surface field E = kQ/R².

Question 25: Equal charges +q are placed at the three corners of an equilateral triangle ABC, each at the same distance r from the centre O. The electric intensity at O is
(a) \( \frac1{4\pi\varepsilon_0}\frac qr \)
(b) \( \frac1{4\pi\varepsilon_0}\frac q{r^2} \)
(c) \( \frac1{4\pi\varepsilon_0}\frac{3q}{r^2} \)
(d) zero
Show Answer & Explanation

Answer: (d) zero

Explanation:
1. The three fields at O have equal magnitude and are 120° apart.
2. Their vector sum is zero.

Teacher's Note:
1. Symmetry.

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