Physics · Electric Charges And Fields · NEET
For a conductor it is E = σ/ε₀. The value σ/2ε₀ is for a single thin charged sheet (isolated infinite plane). A conductor has charge on its surface but zero field inside, so the full field jump appears on the outside as σ/ε₀. Rule to remember: conductor surface = σ/ε₀ (bigger, factor 1), thin sheet = σ/2ε₀ (smaller, factor 1/2).
A thin sheet pushes field out on BOTH sides, so each side gets σ/2ε₀. A conductor also would give σ/2ε₀ from its own surface charge, but the rest of the charge on the conductor adds another σ/2ε₀ on the outside and cancels it on the inside. Outside: σ/2ε₀ + σ/2ε₀ = σ/ε₀. Inside: σ/2ε₀ − σ/2ε₀ = 0. That is why it is zero inside and σ/ε₀ outside.
If the field had a component ALONG the surface, the free electrons in the conductor would feel a sideways force and keep moving. In electrostatics nothing is moving, so there can be no tangential (sideways) component. Only the normal (straight-out) part survives. So at every point of the surface, E is perpendicular.
No. Just inside the surface the field is 0, but just OUTSIDE it is σ/ε₀. The surface is exactly the place where the field jumps from 0 to σ/ε₀. Where σ = 0 (no charge on that patch) the field there is zero even outside.
σ is the LOCAL surface charge density at that point — charge per unit area on that specific patch of the surface. On an irregular conductor σ is larger at sharp points and smaller on flat parts, so the field is stronger near sharp tips. Only on a sphere is σ the same everywhere.
Which of the following statements are correct? A. Inside a conductor, the electrostatic field is zero. B. Electric field at the surface of a charged conductor does not depend on its surface charge density. C. The interior of a charged conductor can have no excess charge in the static situation. D. At the surface of a charged conductor, the electrostatic field must be normal to the surface at every point. E. The electrostatic potential is zero everywhere inside a charged conductor.
Two charged spherical conductors of radii R₁ and R₂ are connected by a wire. Then the ratio of the surface charge densities of the spheres (σ₁/σ₂) is
Try the real previous-year questions from this chapter — each with the answer and a full solution.
E = σ/ε₀ n̂, where σ is the local surface charge density, ε₀ is the permittivity of free space, and n̂ is the unit vector pointing outward, normal to the surface. This is NCERT equation 2.35.
A conductor has zero field inside, so the entire field discontinuity of σ/ε₀ appears only on the outside. A thin sheet radiates on both sides, splitting the field into σ/2ε₀ on each side. Conductor = σ/ε₀, sheet = σ/2ε₀.
Yes, indirectly. The formula E = σ/ε₀ uses the LOCAL σ, and σ is larger at sharp points and edges. So sharp tips have a stronger surface field — this is why lightning rods are pointed.
Yes, in electrostatics the field is zero everywhere inside a conductor. Excess charge lives only on the surface, and the potential inside is constant (though not necessarily zero).
Any component along the surface would exert a force on the free surface charges and make them move. In the static situation nothing moves, so only the normal (perpendicular) component can exist.
