Electric Charges and Fields Class 12 Assertion Reason Questions

Electric Charges and Fields — Class 12 Physics Assertion–Reason Questions
Practise 30 concept-focused Assertion–Reason questions covering electric charge, Coulomb's law, superposition, electric field, electric field lines, electric dipole, electric flux and Gauss's law.

Class 12 Physics Assertion–Reason Practice for 2026–27

This practice set is aligned with the CBSE Class 12 Physics 2026–27 syllabus for Chapter 1, Electric Charges and Fields. The current CBSE 2026–27 Physics Sample Question Paper includes four Assertion–Reason questions in Section A. This page provides a larger original practice set so students can strengthen the same reasoning skill across the chapter.

Important: These are original practice questions for learning and exam preparation. They are not official CBSE questions.

How to Attempt Assertion–Reason Questions

First judge the Assertion independently. Then judge the Reason independently. If both are true, check whether the Reason actually explains why the Assertion is true. Do not select an option merely because both statements are factually correct.

Choose the correct option:

(A) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.

(B) Both Assertion and Reason are true, but Reason is not the correct explanation of Assertion.

(C) Assertion is true, but Reason is false.

(D) Both Assertion and Reason are false.

Exam-pattern note: The A–D structure above follows the Assertion–Reason option pattern used in the CBSE Class XII Physics Sample Question Paper for 2026–27.

1. Electric Charge and Coulomb's Law

Question 1

Assertion: Electric charge is conserved in an isolated system.

Reason: Electric charge can neither be created nor destroyed in an isolated system.

Answer: A

Explanation: Both statements express the conservation of charge. In an isolated system, the total algebraic charge remains constant.

Question 2

Assertion: The electrostatic force between two point charges becomes one-fourth when their separation is doubled.

Reason: Coulomb's force varies inversely as the square of the separation between the charges.

Answer: A

Explanation: Since F ∝ 1/r², replacing r by 2r gives F′ = F/4.

Question 3

Assertion: The electrostatic force between two unlike point charges is attractive.

Reason: The force on each charge acts along the line joining the two charges and is directed towards the other charge.

Answer: A

Explanation: Opposite charges attract. The electrostatic force acts along the line joining the charges and points towards the other charge for an unlike pair.

Question 4

Assertion: Coulomb's law and Newton's law of gravitation have analogous inverse-square forms for point sources.

Reason: Both electrostatic and gravitational forces always have the same nature: they are always attractive.

Answer: C

Explanation: The Assertion is true because both laws have an inverse-square dependence. The Reason is false because electrostatic force can be attractive or repulsive, whereas gravitational force between masses is attractive.

Question 5

Assertion: The electrostatic force between two point charges is a vector quantity.

Reason: Force has both magnitude and direction.

Answer: A

Explanation: Electrostatic force has a definite magnitude and direction, so it is represented as a vector.

2. Principle of Superposition

Question 6

Assertion: The net electrostatic force on a charge due to several charges is the vector sum of the individual forces.

Reason: Electrostatic forces obey the principle of superposition.

Answer: A

Explanation: The total force is obtained by adding the individual electrostatic force vectors produced by all other charges.

Question 7

Assertion: At the midpoint between two identical positive point charges, the resultant electric field is zero.

Reason: The electric fields produced by the two charges at the midpoint have equal magnitudes and opposite directions.

Answer: A

Explanation: Equal positive charges produce equal field magnitudes at the midpoint, but the two field vectors point in opposite directions, so they cancel.

Question 8

Assertion: At the midpoint between two equal and opposite point charges, the resultant electric field is zero.

Reason: At the midpoint, the electric fields due to the two charges have equal magnitude and opposite direction.

Answer: D

Explanation: The Assertion is false because the two fields at the midpoint point in the same direction, from the positive charge towards the negative charge. The Reason is also false because it incorrectly says the fields are opposite.

Question 9

Assertion: For a continuous charge distribution, the total electric field can be obtained by integrating the field contributions of small charge elements.

Reason: A continuous charge distribution can be treated as a collection of infinitesimal charge elements and the contributions are added using superposition.

Answer: A

Explanation: The discrete-charge superposition principle extends to continuous distributions through integration, with dq representing an infinitesimal charge element.

3. Electric Field and Electric Field Lines

Question 10

Assertion: Electric field intensity at a point is defined as force per unit positive test charge placed at that point, in the limiting sense of a sufficiently small test charge.

Reason: A positive test charge is used to define the direction of the electric field.

Answer: A

Explanation: Electric field is defined by E = F/q₀ as the test charge becomes sufficiently small. Its direction is the direction of force on a positive test charge.

Question 11

Assertion: If the electric potential at a point is zero, the electric field at that point must also be zero.

Reason: If electric potential is zero at a point, its spatial gradient must also be zero at that point.

Answer: D

Explanation: The Assertion is false: zero potential at a point does not necessarily mean zero electric field. The Reason is also false because electric field depends on the spatial variation of potential, not simply on the value of potential at one point.

Question 12

Assertion: Electric field lines do not intersect one another at a point in electrostatics.

Reason: At a given point, the electric field has a unique direction.

Answer: A

Explanation: If two field lines crossed, the field at their intersection would have two directions, which is not possible for a single-valued electrostatic field.

Question 13

Assertion: Electric field lines are closer together in regions where the electric field is stronger.

Reason: The density of field-line representation is used to indicate the relative magnitude of the electric field.

Answer: A

Explanation: Field-line diagrams use greater line density to represent stronger electric fields. This is a qualitative representation, not a literal concentration of physical lines.

4. Electric Dipole

Question 14

Assertion: Electric dipole moment is directed from the negative charge to the positive charge.

Reason: By convention, the dipole moment vector points from negative charge to positive charge.

Answer: A

Explanation: For a dipole of charges ±q separated by distance d, the magnitude is p = qd and its direction is from negative to positive charge.

Question 15

Assertion: At the same distance from the centre of an electric dipole, the electric-field magnitudes on the axial and equatorial lines are equal.

Reason: The electric field of an ideal dipole has the same expression on both the axial and equatorial lines.

Answer: D

Explanation: Both statements are false. For an ideal dipole, the axial and equatorial fields have different magnitudes and directions; their standard expressions differ by a factor of 2 in magnitude at the same large distance.

Question 16

Assertion: The torque on an electric dipole in a uniform electric field is maximum when the dipole is perpendicular to the field.

Reason: The magnitude of torque is given by τ = pE sin θ.

Answer: A

Explanation: Torque is maximum when sin θ = 1, which occurs at θ = 90°.

Question 17

Assertion: The torque on an electric dipole is zero when the dipole is parallel to a uniform electric field.

Reason: For θ = 0°, sin θ = 0 in τ = pE sin θ.

Answer: A

Explanation: When the dipole is parallel to the field, θ = 0° and therefore τ = 0. The antiparallel position at 180° also has zero torque.

5. Electric Flux

Question 18

Assertion: Electric flux through a plane surface in a uniform electric field is Φ = EA cos θ.

Reason: Electric flux is the dot product of the electric field and the area vector.

Answer: A

Explanation: The dot product gives Φ = E · A = EA cos θ, where θ is the angle between the electric field and the area vector.

Question 19

Assertion: Electric flux through a plane surface is maximum when the electric field is perpendicular to the surface.

Reason: The area vector is perpendicular to the surface, so the angle between the electric field and the area vector is zero in this situation.

Answer: A

Explanation: When E is perpendicular to the surface, it is parallel to the area vector. Thus θ = 0° and Φ = EA, the maximum value for fixed E and A.

Question 20

Assertion: Electric flux through a plane surface is zero when the electric field is parallel to the surface.

Reason: When the electric field is parallel to the surface, the angle between the electric field and the area vector is zero.

Answer: C

Explanation: The Assertion is true because E is perpendicular to the area vector, so θ = 90° and Φ = 0. The Reason is false because the angle between E and the area vector is 90°, not 0°.

6. Gauss's Law and Its Applications

Question 21

Assertion: The total electric flux through a closed surface depends only on the net charge enclosed by the surface.

Reason: Gauss's law gives the net electric flux as Φ = q_enclosed/ε₀.

Answer: A

Explanation: Gauss's law directly relates the net flux through a closed surface to the algebraic sum of charge enclosed by that surface.

Question 22

Assertion: If the net charge enclosed by a closed Gaussian surface is zero, the net electric flux through the surface is zero.

Reason: Gauss's law gives the net flux as q_enclosed/ε₀.

Answer: A

Explanation: With q_enclosed = 0, Gauss's law gives Φ = 0. This does not necessarily mean that the electric field is zero at every point on the surface.

Question 23

Assertion: The electric field inside a uniformly charged thin spherical shell is zero.

Reason: A Gaussian surface completely inside the shell encloses zero charge.

Answer: B

Explanation: Both statements are true, but the Reason alone is not a complete explanation. Gauss's law gives zero net flux; spherical symmetry is also required to conclude that the electric field is zero everywhere on the interior Gaussian surface.

Question 24

Assertion: Outside a uniformly charged thin spherical shell, the electric field is the same as that of a point charge equal to the total charge concentrated at the centre.

Reason: A spherical Gaussian surface outside the shell encloses the total charge, and spherical symmetry makes the field radial with the same magnitude at every point on that Gaussian surface.

Answer: A

Explanation: Applying Gauss's law to a spherical surface of radius r gives E(4πr²) = Q/ε₀, so E = (1/4πε₀)Q/r².

Question 25

Assertion: The electric field due to an ideal infinite uniformly charged plane sheet is independent of distance from the sheet.

Reason: For an infinite plane sheet, the field magnitude is E = σ/(2ε₀).

Answer: A

Explanation: The expression contains surface charge density and permittivity but no distance term, so the field magnitude does not decrease with distance in the ideal infinite-sheet model.

Question 26

Assertion: The electric field due to an infinitely long uniformly charged straight wire decreases as 1/r.

Reason: Using a cylindrical Gaussian surface, the field magnitude is E = λ/(2πε₀r).

Answer: A

Explanation: The cylindrical Gaussian surface has area proportional to r, so Gauss's law leads to E ∝ 1/r.

Question 27

Assertion: A Gaussian surface may be chosen in any convenient closed shape, but a symmetric Gaussian surface is especially useful for calculating electric fields.

Reason: Symmetry can make the electric field magnitude constant over suitable portions of the Gaussian surface and simplify the flux integral.

Answer: A

Explanation: Gauss's law is valid for any closed surface. The practical advantage of a spherical, cylindrical or planar-symmetry-based Gaussian surface is that the flux calculation can become simple.

Question 28

Assertion: The net electric flux through a closed surface can be zero even when the electric field is non-zero at points on the surface.

Reason: Gauss's law relates the net flux to the net enclosed charge, not to whether the electric field is zero at every point on the surface.

Answer: A

Explanation: Zero net flux means the total outward and inward contributions cancel in the surface integral. It does not require E = 0 everywhere on the surface.

Question 29

Assertion: Electric flux through a closed surface is a scalar quantity.

Reason: Electric field is a vector quantity.

Answer: B

Explanation: Both statements are true, but the vector nature of E does not by itself explain why flux is scalar. Flux is obtained from the dot product E · dA and therefore has a scalar value.

Question 30

Assertion: The electrostatic force between two point charges depends on the medium between them.

Reason: The electrostatic constant depends on the permittivity of the medium.

Answer: A

Explanation: In a medium, Coulomb's law is written using the medium's permittivity ε: F = (1/4πε)|q₁q₂|/r². Therefore the force can change when the medium changes.

Answer Key

Q	Ans.	Q	Ans.	Q	Ans.
1	A	11	D	21	A
2	A	12	A	22	A
3	A	13	A	23	B
4	C	14	A	24	A
5	A	15	D	25	A
6	A	16	A	26	A
7	A	17	A	27	A
8	D	18	A	28	A
9	A	19	A	29	B
10	A	20	C	30	A

Quick Revision

• Charge: conserved and quantised; q = ne.
• Coulomb's law: F ∝ |q₁q₂|/r² for point charges in a specified medium.
• Superposition: individual electric forces and fields are added vectorially.
• Electric field: E = F/q₀ for a sufficiently small positive test charge.
• Field lines: they do not intersect; their density qualitatively represents field strength.
• Dipole: p = qd and τ = pE sin θ in a uniform electric field.
• Flux: Φ = EA cos θ for a uniform field through a plane surface.
• Gauss's law: Φ = q_enclosed/ε₀ for a closed surface.
• Infinite line charge: E = λ/(2πε₀r).
• Infinite plane sheet: E = σ/(2ε₀).
• Thin spherical shell: E = 0 inside and outside behaves like a point charge Q at the centre.

Exam Strategy

For Assertion–Reason questions, avoid treating every true statement as an explanation. A reliable method is: (1) test Assertion, (2) test Reason, (3) identify the exact physical law connecting them, and (4) check whether the Reason is sufficient to establish the Assertion.

Recommended practice: Attempt all 30 questions once without looking at the answers. Then review only the questions you got wrong and explain the correct option in your own words.

Curriculum basis: CBSE Class XII Physics 2026–27, Unit I: Electrostatics, Chapter 1: Electric Charges and Fields. The chapter includes electric charge, Coulomb's law, forces between multiple charges, superposition and continuous charge distribution, electric field, electric dipole, electric flux, and Gauss's theorem with applications to an infinite straight wire, infinite plane sheet and uniformly charged thin spherical shell.