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Exam level3 past questions

Conductors, charge distribution, and electrostatic shielding

Understand how charges distribute on conductors, the concept of electrostatic shielding, and the minimization of potential energy in conductors.

Why this shows up in the exam

NEET expects you to know how conductors behave in electrostatic equilibrium and how shielding works.

How NEET tests this

Application · 2 QsStatement analysisNumerical

Learn the idea

In electrostatic equilibrium a conductor becomes an equipotential body, excess charge resides only on its outer surface, the interior field is zero and therefore it shields any cavity inside.

🧠 Memory hook: CAGE – Conductors Are Good Electrostatic shields, so they keep the inside field at zero and make the whole body a single potential cage.

Get this one clearly and it pays off every single time it shows up in the paper. 🎯

Formulas & facts to keep ready

  • Electric field inside a conductor in electrostatic equilibrium is zero (E=0).
  • All excess charge on a conductor resides on its outer surface.
  • Potential of an isolated spherical conductor V = kQ/R and surface charge density σ = Q/(4πR²).
  • When two conductors are connected by a conducting wire they acquire the same potential (V₁=V₂).
  • A hollow conductor shields its interior: the field in a cavity is zero irrespective of external charges.

How to approach it

  1. 1Write down the relevant relation (V = kQ/R or V = 4πkσR) for each conductor.
  2. 2Use the condition V₁ = V₂ when conductors are connected or the fact that E=0 inside a conductor.
  3. 3Replace Q by σ·4πR² if surface charge density is asked.
  4. 4Solve the resulting algebraic relation for the required quantity (σ₁/σ₂, potential, field, etc.).

Worked example — watch it click

Two charged spherical conductors of radius R₁ and R₂ are connected by a wire. Then the ratio of surface charge densities of the spheres (σ₁/σ₂) is

  • A)R₁²/R₂²
  • B)R₁/R₂
  • ✅R₂/R₁
  • D)√(R₁/R₂)

The concept behind this problem

The example asks for the ratio of surface charge densities after two spheres are joined, which requires using the equal‑potential condition and the σ‑R relationship for a sphere.

Step by step

  1. 1When two conductors are connected by a wire, they reach the same potential: V₁ = V₂.
  2. 2For a spherical conductor, V = kQ/R and surface charge density σ = Q/(4πR²).
  3. 3So V = kQ/R = k(σ·4πR²)/R = 4πkσR.
  4. 4Since V₁ = V₂: 4πkσ₁R₁ = 4πkσ₂R₂, which gives σ₁R₁ = σ₂R₂.
  5. 5Therefore σ₁/σ₂ = R₂/R₁.
  6. 6The smaller sphere has higher surface charge density.

Watch out

Students often set σ₁ = σ₂ directly, forgetting that equal potential gives σ₁R₁ = σ₂R₂, leading to the wrong ratio.

Common slip-ups that cost marks

  • •Confusing charge Q with surface charge density σ – remember σ = Q/(4πR²).
  • •Assuming charge distributes uniformly on the inner surface of a hollow conductor; it actually stays on the outer surface only.
  • •Using V = kQ/R for a conductor that is not isolated (connected by a wire) without first imposing V₁=V₂.

🌟 That's the whole idea — you've got this. Try the practice set below; every question you attempt makes it stick a little harder.

Practise it

These are real questions from past NEET papers that test this exact idea.

Question 1 of 3NEET 2022

Two hollow conducting spheres of radii R₁ and R₂ (R₁ >> R₂) have equal charges. The potential would be:

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