A positive charge +q is placed at the centre of a hollow metallic sphere of inner radius a and outer radius b. The electric field at a distance r from the centre is denoted by E. In this regard, which one of the following statements is correct ?
- (a)E = 0 for a < r < b
- (b)E = 0 for r < a
- (c)E = q/4πε₀r for a < r < b
- (d)E = q/4πε₀a for r < a
Correct — A, E = 0 for a < r < b. That range is the metal itself, the body of the shell between its inner and outer surfaces. In electrostatic equilibrium the free electrons in a conductor rearrange themselves until the field inside the material is exactly cancelled; if any field remained, charges would keep moving and the situation would not be static. The induced charge -q that appears on the inner surface at radius a is precisely what cancels the field of the central +q throughout the metal. Outside the shell an induced +q sits on the outer surface, so beyond r = b the field is the ordinary Coulomb field of a charge q.
- (b)E = 0 for r < a — Inside the cavity there is no shielding at all — the charge +q sits there and its field fills the hollow, falling off as the inverse square of the distance. A conductor screens the region outside itself from an enclosed charge's redistribution, not the cavity that holds the charge.
- (c)E = q/4πε₀r for a < r < b — This assigns a non-zero field to the interior of the metal, which contradicts electrostatic equilibrium. The expression is also not a Coulomb field, since the distance appears to the first power rather than squared.
- (d)E = q/4πε₀a for r < a — Inside the cavity the field must depend on where you stand, growing as you approach the central charge. Writing it in terms of the fixed inner radius a makes it constant throughout the cavity, which is wrong.
Three regions, three answers. For r less than a you are in the cavity, where only the central charge acts and E = q/(4πε₀r²) pointing outward. For r between a and b you are inside the metal, where E = 0. For r greater than b you are outside the shell, and because the total enclosed charge is still +q the field is again q/(4πε₀r²). Gauss's law gives all three in one line, since the field of a spherically symmetric distribution depends only on the charge enclosed by the Gaussian surface.
An incidental note on the printing. In the paper the expressions in options (c) and (d) are set with r and a to the first power, without the square that the Coulomb field requires; that is a typographic slip in the booklet and it does not change anything, because both options fail for the physical reason above long before the exponent matters. Answer such an item by regions, not by matching formulae: identify which region each statement refers to, and ask what the metal is doing there. This is also the principle behind the Faraday cage — a conducting enclosure keeps external fields out of the space it surrounds.
- The electrostatic field inside the material of a conductor in equilibrium is zero.
- A charge +q inside a cavity induces -q on the inner surface and +q on the outer surface of a neutral shell.
- In the cavity the field is that of the central charge alone, E = q/(4πε₀r²).
- Outside the shell the field is again q/(4πε₀r²), because the enclosed charge is still q.
- Gauss's law relates the flux through a closed surface to the charge enclosed by it, and does the whole calculation for spherical symmetry.
Only the metal itself is field-free; the cavity is not shielded from the charge it contains.
- Reading 'field inside the sphere is zero' as covering the cavity as well as the metal.
- Forgetting the induced charges on the inner and outer surfaces.
- Assuming a field expression is wrong only because of its algebra, when the physical region already rules it out.
NDA sets conductor-and-cavity items regularly, either as this region-by-region statement question or through the direction of field lines at a conducting surface.
The electric field lines from an isolated positively charged conducting sphere are
- (a) tangential to the conducting surface
- (b) at right angles to the conducting surface and towards the centre of the sphere
- (c) at any angle to the conducting surface
- (d) at right angles to the conducting surface and outwards from the centre of the sphere
Answer(d) at right angles to the conducting surface and outwards from the centre of the sphere
The companion property of a conductor in equilibrium — no field inside it, and no tangential field at its surface either.
Why are the tyres of aircrafts made of conducting rubber? 1. So that the charge accumulated on the aircraft in flight, by rubbing the air, can easily be transferred to ground on landing. 2. So that the charge accumulated due to the operation of various electronic equipments in the aircraft in flight can easily be transferred to ground on landing. Select the correct answer using the code given below.
- (a) 1 only
- (b) 2 only
- (c) Both 1 and 2
- (d) Neither 1 nor 2
Answer(a) 1 only
Conductors and static charge again, this time in an everyday setting rather than in a spherical geometry.
- practice — not a real PYQ
In electrostatic equilibrium, the electric field inside the material of a charged conductor is
- (a)Zero
- (b)Uniform and non-zero
- (c)Greatest at the centre
- (d)Directed along the surface
Answer(a) Zero — the free charges rearrange until the interior field cancels.
- practice — not a real PYQ
A charge +q at the centre of a neutral hollow conducting sphere induces on its inner surface a charge of
- (a)+q
- (b)−q
- (c)Zero
- (d)+2q
Answer(b) −q — with an equal +q appearing on the outer surface, keeping the shell neutral overall.