Previous year questions: Electric Charges and Fields

Ten questions follow, drawn from official CBSE papers in the collected corpus. The label on each question gives the paper and the marks. Work the question before you read the answer, and write the direction of every vector in your own solution.

Q1 (CBSE Board 2023, 5 marks)

Define electric flux and write its SI unit. Apply Gauss's law to a uniformly charged infinite plane sheet and obtain its field. A cube of side L lies in a region whose electric field is E = (A x + B) times the unit vector along x. Compute the net charge that the cube encloses.

Answer: Electric flux through a surface is the dot product of E and the area vector, with unit N m^2 C^-1. For the sheet, take a pillbox crossing the sheet with flat faces of area A parallel to it. By symmetry the field is normal to the sheet on both sides and parallel to the curved surface, so the flux is 2 E A while the enclosed charge is sigma A. Hence E = sigma / (2 epsilon0) and the direction is away from the sheet for positive sigma. For the cube, only the faces at x = 0 and x = L contribute, giving flux -B L^2 + (A L + B) L^2 = A L^3, so the enclosed charge is epsilon0 A L^3.

Q2 (CBSE Board 2023, 2 marks)

An electric dipole of dipole moment 6 × 10^-7 C m is kept in a uniform electric field of 1 × 10^4 N C^-1 so that the dipole moment is parallel to the field. Find the potential energy of the dipole. How much work is needed to rotate it until the dipole moment is antiparallel to the field?

Answer: Potential energy is U = -p E cos theta. With theta = 0, U = -p E = -6 × 10^-7 × 1 × 10^4 = -6 × 10^-3 J. For the antiparallel position theta = 180 degrees, U = +p E = +6 × 10^-3 J. The work needed is the change in potential energy, 6 × 10^-3 - (-6 × 10^-3) = 1.2 × 10^-2 J. The external agent does this work because the field resists the rotation.

Q3 (CBSE Board 2024, 5 marks)

An electric dipole consists of charges q and -q separated by 2a and is placed along the x-axis with its centre at the origin, so the dipole moment is p times the unit vector along x. Show that at a point x on the axis with x much greater than a, the potential is p divided by 4 pi epsilon0 x squared.

Answer: The potential at x is the sum of the potentials of the two charges: k q / (x - a) plus k (-q) / (x + a). Combine over a common denominator to get k q [(x + a) - (x - a)] / (x^2 - a^2) = 2 k q a / (x^2 - a^2). Put p = 2 q a and neglect a squared compared with x squared, since x is much greater than a. The result is k p / x^2, which is p / (4 pi epsilon0 x^2), as required. The sign is positive on the side where the positive charge is nearer.

Q4 (CBSE Board 2024, 5 marks)

A triangle with equal sides carries charges q, 2q and nq at its three corners. The total potential energy of this arrangement is zero. Determine n.

Answer: Add the three pair terms for an equilateral triangle of side a. The pairs are q with 2q, 2q with nq, and nq with q. Factoring out k q^2 / a leaves 2 + 2n + n = 2 + 3n. A total potential energy of zero gives 2 + 3n = 0, so n = -2/3. The third charge must be negative and smaller in size than q.

Q5 (CBSE Board 2025, 1 mark)

An electric dipole of dipole moment 1.0 × 10^-12 C m lies along the x-axis. An electric field of magnitude 2.0 × 10^4 N C^-1 is switched on, with its direction along the positive z-axis. Find the magnitude of the torque on the dipole.

Answer: The dipole moment and the field are perpendicular, so theta = 90 degrees and the torque is p E sin 90 = p E. Substitute 1.0 × 10^-12 × 2.0 × 10^4 = 2.0 × 10^-8 N m. The direction is along the negative y-axis, from the cross product of the x-axis moment and the z-axis field.

Q6 (CBSE Board 2025, 5 marks)

Obtain an expression for the electric field of a dipole of moment p at a point on its equatorial plane. An electric field E = (10 x + 5) N C^-1 exists in a region where x is in metres, and a cube of side L is kept in that region. Find the charge enclosed by the cube.

Answer: On the equatorial plane, the two charges are each at distance sqrt(r^2 + a^2) from the point. Their fields are equal in size and their components along the axis cancel, while the components perpendicular to the axis add. The net field is 2 k q a / (r^2 + a^2)^(3/2), directed opposite to p. For r much greater than a this becomes k p / r^3. For the cube, only the two faces normal to x contribute. The flux is [(10 L + 5) - 5] L^2 = 10 L^3, so the enclosed charge is 10 epsilon0 L^3.

Q7 (CBSE Board 2026, 5 marks)

Charges q and -q sit 2a apart and make an electric dipole. Take a point on the perpendicular bisector of the pair, at distance r from its centre, and derive the field there. Then write the simpler form that holds when r is far larger than a.

Answer: Let the point be on the perpendicular bisector of the dipole. Its distance from each charge is sqrt(r^2 + a^2). Each charge gives a field k q / (r^2 + a^2). Resolve these fields along the dipole axis and perpendicular to it. The components along the axis are equal and opposite, so they cancel. The components perpendicular to the axis add, and each has a factor a / sqrt(r^2 + a^2). The net field is therefore 2 k q a / (r^2 + a^2)^(3/2) pointing opposite to the dipole moment. At a far point, replace 2 q a by p and drop a^2 beside r^2 to get k p / r^3.

Q8 (CBSE SQP 2021-22, 1 mark)

Which statement about Gauss's law is true? (i) All charges, inside or outside the Gaussian surface, contribute to the flux. (ii) The flux depends on the geometry of the Gaussian surface. (iii) Gauss's theorem can be applied to a non-uniform electric field. (iv) The field over the Gaussian surface is continuous and uniform at every point.

Answer: Statement (iii) is true. Gauss's law holds for any closed surface and for any field distribution, uniform or not. Statement (i) is false because only enclosed charges contribute to the net flux. Statement (ii) is false because the flux depends only on the enclosed charge. Statement (iv) is false because the field on the surface can vary from point to point.

Q9 (CBSE Additional PQ 2023-24, 5 marks)

Write the statement of Gauss's theorem in electrostatics. Use the theorem to obtain the field just outside a uniformly charged infinite plane sheet. A cone with base radius R and height h stands on a table, and a horizontal field of size E passes through it. Compute the flux that enters through the curved surface.

Answer: Gauss's theorem: the total flux through any closed surface is the enclosed charge divided by epsilon0. For the sheet, a pillbox of flat area A gives 2 E A = sigma A / epsilon0, so E = sigma / (2 epsilon0). For the cone, the field enters through the curved surface and leaves through the base. Since no charge is enclosed, the flux entering through the curved surface equals the flux leaving through the base. The silhouette projected on a plane perpendicular to the field is a triangle with base 2 R and height h, whose area is R h, so the entering flux has magnitude E R h. The sign is negative if entering flux is counted as negative.

Q10 (CBSE Board 2024, 5 marks)

An electric dipole consists of point charges of -1.0 pC and +1.0 pC located at (0, 0) and (3 mm, 4 mm) in the x-y plane. A uniform electric field of 1000 N C^-1 along the positive x-axis is switched on. Find the torque acting on the dipole.

Answer: The separation between the charges is the distance from (0, 0) to (3 mm, 4 mm), which is 5 mm. The dipole moment has magnitude p = q d = 1.0 × 10^-12 × 5 × 10^-3 = 5 × 10^-15 C m, and it points from the negative charge at the origin to the positive charge at (3 mm, 4 mm), so its direction is along 0.6 i + 0.8 j. The torque is the cross product of p and E. Only the y component of p contributes, so the magnitude is p E sin theta = 5 × 10^-15 × 1000 × 0.8 = 4 × 10^-12 N m. The direction is negative z, since j cross i points along negative z.