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The Vault
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The Vault
CBSE Class 12 Physics · 11 questions · 26 marks
Every rubbed comb that lifts paper bits is running the same physics your board paper will test: charges, forces, and the invisible field between them. This chapter takes you from Coulomb's inverse-square law to Gauss's law, which turns messy integrals into one-line answers for symmetric charge distributions. Get the field patterns right here and the whole electrostatics unit falls into place.
Two identical point charges separated by distance r in vacuum repel each other with force F. If the whole arrangement is immersed in a liquid of dielectric constant 4 without changing the separation, the new force becomes:
Answer
F/4 — inside a medium of dielectric constant K, Coulomb's law becomes F' = (1/4πε₀K)·q₁q₂/r², so the force is reduced by a factor K. With K = 4, the force falls to F/4. The separation is unchanged, so only the medium factor matters.
A closed cubical surface encloses a charge of +5 μC. A second charge of -8 μC is placed just outside one face of the cube. The net electric flux through the cube is:
Answer
5×10⁻⁶/ε₀ — Gauss's law counts only the charge enclosed by the surface. The -8 μC charge lies outside, so its field lines enter and leave the cube, contributing zero net flux. Hence Φ = q(enclosed)/ε₀ = 5×10⁻⁶/ε₀ N m²C⁻¹.
An electric dipole of moment p is placed in a uniform electric field E with its axis making an angle of 90° with the field. The net force and torque on it are respectively:
Answer
Zero and pE — in a uniform field the forces +qE and -qE on the two charges are equal and opposite, so the net force is zero. They form a couple of torque τ = pE sinθ; at θ = 90°, sinθ = 1, giving maximum torque τ = pE.
A hollow conducting sphere of radius 12 cm carries a charge of +6 nC spread on its surface. The electric field at a point 4 cm from the centre is:
Answer
Zero — the point at 4 cm lies inside the conducting shell (radius 12 cm). A Gaussian sphere drawn through that point encloses no charge, since all charge resides on the outer surface, so by Gauss's law E = 0 everywhere inside a charged conducting shell.
Assertion (A): The electric field just outside the surface of a charged conductor in electrostatic equilibrium is always perpendicular to the surface. Reason (R): If the field had a component along the surface, free charges on the conductor would experience a tangential force and keep moving, which contradicts electrostatic equilibrium.
Answer
Both A and R are true and R is the correct explanation of A — electrostatic equilibrium means no net motion of free charges. Any tangential field component would push the surface charges sideways indefinitely, so in equilibrium that component must vanish, leaving the field normal to the surface with magnitude σ/ε₀.
Two small spheres carrying charges of +3.0 μC and -12.0 μC are held 30 cm apart in air. Calculate the magnitude of the force between them and state its nature.
Answer
Using Coulomb's law, F = (1/4πε₀)·q₁q₂/r². Here q₁ = 3.0×10⁻⁶ C, q₂ = 12.0×10⁻⁶ C (magnitude), r = 0.30 m. F = 9×10⁹ × (3.0×10⁻⁶ × 12.0×10⁻⁶)/(0.30)² F = 9×10⁹ × 3.6×10⁻¹¹/0.09 F = 0.324/0.09 = 3.6 N. The force is 3.6 N and, since the charges are unlike, it is attractive along the line joining them.
Define electric flux and write its SI unit. A uniform electric field of 500 N/C passes through a flat square sheet of side 20 cm whose normal makes an angle of 60° with the field. Find the flux through the sheet.
Answer
Electric flux through a surface is the total number of electric field lines crossing it, defined as Φ = ∮E·dS = EA cosθ, where θ is the angle between the field and the outward normal to the area. It is a scalar quantity with SI unit N m²C⁻¹ (equivalently V m). Calculation: A = (0.20)² = 0.04 m², E = 500 N/C, θ = 60°, so cosθ = 0.5. Φ = 500 × 0.04 × 0.5 = 10 N m²C⁻¹.
State any two properties of electric field lines that distinguish them from ordinary curves drawn on paper.
Answer
1. Electric field lines never intersect one another, because at a point of intersection the field would have two different directions, which is impossible for a single-valued vector field. 2. Field lines start on positive charges and end on negative charges (or extend to infinity) and are never closed loops, since the electrostatic field is conservative. They are also always normal to the surface of a conductor and their local density represents the field strength.
Using Gauss's law, derive an expression for the electric field at a point at distance r from an infinitely long, straight, uniformly charged wire of linear charge density λ. Hence find the field 25 cm from a long wire carrying 2.5×10⁻⁸ C/m.
Answer
Derivation: Consider an infinitely long straight wire with uniform linear charge density λ (charge per unit length). By symmetry, the field at every point must be radial (pointing directly away from the wire for positive λ) and its magnitude can depend only on the perpendicular distance r. Choose a Gaussian surface that is a coaxial cylinder of radius r and length l, closed by two flat end caps. Flux through the curved surface: the field E is everywhere parallel to the outward normal, so this contribution is E × (2πrl). Flux through the two end caps: here E is perpendicular to the outward normal (θ = 90°, cosθ = 0), so each contributes zero. Total flux Φ = E × 2πrl. Charge enclosed by the cylinder = λl. Applying Gauss's law, Φ = q(enclosed)/ε₀: E × 2πrl = λl/ε₀ E = λ/(2πε₀r). So the field of a long straight charged wire falls off as 1/r, not 1/r², and is directed radially outward for positive λ and radially inward for negative λ. Numerical: λ = 2.5×10⁻⁸ C/m, r = 0.25 m. Using E = 2kλ/r with k = 9×10⁹: E = 2 × 9×10⁹ × 2.5×10⁻⁸/0.25 E = 450/0.25 = 1800 N/C, directed radially outward from the wire.
(a) Derive an expression for the electric field on the axial line of a short electric dipole at a distance r from its centre. (b) A dipole of moment 4.0×10⁻³⁰ C m is held at 30° to a uniform field of 6.0×10⁴ N/C. Find the torque acting on it.
Answer
(a) Let a dipole consist of charges -q at A and +q at B, separated by 2a, with dipole moment p = q(2a) directed from -q to +q. Take a point P on the axial line at distance r from the centre O, on the side of +q. Distance of P from +q = (r - a); distance from -q = (r + a). Field at P due to +q, directed from B to P (away from the dipole): E₊ = (1/4πε₀)·q/(r - a)² Field at P due to -q, directed from P towards A (towards the dipole): E₋ = (1/4πε₀)·q/(r + a)² Since E₊ and E₋ point in opposite directions and E₊ > E₋, the net field points away from the dipole along the axis: E = (q/4πε₀)[1/(r - a)² - 1/(r + a)²] E = (q/4πε₀) × [(r + a)² - (r - a)²]/[(r² - a²)²] (r + a)² - (r - a)² = 4ra, so E = (q/4πε₀) × 4ra/(r² - a²)² Using p = 2aq: E = (1/4πε₀) × 2pr/(r² - a²)² For a short dipole, r >> a, so a² can be neglected against r²: E(axial) = (1/4πε₀) × 2p/r³ = 2kp/r³, directed along p. (b) Torque on a dipole in a uniform field: τ = pE sinθ. τ = 4.0×10⁻³⁰ × 6.0×10⁴ × sin30° τ = 4.0×10⁻³⁰ × 6.0×10⁴ × 0.5 τ = 1.2×10⁻²⁵ N m. The torque acts so as to rotate the dipole into alignment with the field.
Read the passage and answer the questions that follow: A school laboratory has a hollow metal sphere of radius 15 cm mounted on an insulating stand. A student, Meera, transfers a charge of +9.0 nC to the sphere using a charged rod, and the charge quickly spreads over the outer surface. She then places a small uncharged metal ball inside the hollow sphere, touching nothing, and finds it stays uncharged and feels no force. (a) Explain, using Gauss's law, why the electric field inside the hollow sphere is zero. (b) Calculate the electric field at a point 30 cm from the centre of the sphere. (c) Calculate the surface charge density on the sphere. (d) If Meera doubles the charge on the sphere, how does the field at 30 cm change, and how does the field inside change?
Answer
(a) All the charge on a charged conductor resides on its outer surface. Draw a spherical Gaussian surface of radius less than 15 cm, entirely inside the metal shell. It encloses no charge, so by Gauss's law Φ = q(enclosed)/ε₀ = 0. Because of spherical symmetry the field has the same magnitude everywhere on that surface, so E × 4πr² = 0 gives E = 0 at every interior point. This is why the small ball inside feels no force. (b) Outside the shell the sphere behaves as a point charge at its centre. E = kq/r² = 9×10⁹ × 9.0×10⁻⁹/(0.30)² E = 81/0.09 = 900 N/C, directed radially outward. (c) Surface area A = 4πR² = 4 × 3.14 × (0.15)² = 4 × 3.14 × 0.0225 = 0.2826 m². σ = q/A = 9.0×10⁻⁹/0.2826 = 3.19×10⁻⁸ C/m², roughly 32 nC/m². (d) The external field is directly proportional to the enclosed charge, so doubling the charge to 18 nC doubles the field at 30 cm to 1800 N/C. The field inside remains exactly zero, because a Gaussian surface drawn inside the metal still encloses no charge no matter how much charge sits on the outer surface.
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