Q.The dimensions of an atom are of the order of an Angstrom. Thus there must be large electric fields between the protons and electrons. Why, then is the electrostatic field inside a conductor zero?

The key idea is Gauss’s Law applied to electrostatic equilibrium in a conductor.

Inside a conductor, free electrons rearrange instantly until the net electric field at every interior point is zero. If any field existed, it would push charges until they cancelled it — that is the definition of electrostatic equilibrium.

1. Consider a Gaussian surface anywhere inside the conductor. By Gauss’s Law, $\oint \vec{E} \cdot d\vec{A} = \frac{q_{\text{enc}}}{\varepsilon_0}$.
2. In equilibrium, $\vec{E} = 0$ everywhere inside, so the left side is zero. Hence $q_{\text{enc}} = 0$ — no net charge inside any interior surface. …

The question touches a subtle point that often confuses students. You are right: inside an atom, the electric field between the proton and electron is enormous — on the order of $10^{11} \, \text{V/m}$. So how can we claim the field inside a conductor is zero? The key is to distinguish between microscopic fields (inside individual atoms) and the macroscopic average field we talk about in electrostatics.

When we say "the electrostatic field inside a conductor is zero," we mean the net macroscopic field — the average field over a region containing many atoms — is zero. This is a statement about the conductor as a whole, not about the fields inside each atom.

Let’s break this down properly.

1. What does "inside a conductor" mean?

A conductor (like copper or aluminium) has a lattice of positive ions surrounded by a sea of free electrons — electrons that are not bound to any particular nucleus. These electrons can move freely throughout the material. In contrast, the electrons in an insulator are tightly bound to their atoms.

When we place a conductor in an external electric field, the free electrons respond instantly. They drift until they arrange themselves on the surface in such a way that their own field exactly cancels the external field everywhere inside the bulk of the conductor. This is the essence of electrostatic shielding.

2. The atomic fields are bound, not free

Inside each atom, the electron cloud and the nucleus produce a strong electric field — but this field is confined to the atom itself. Over distances larger than an atom (say, a few nanometres), these atomic fields average out to zero because the positive and negative charges are paired.

Think of it this way: if you take a box containing many neutral atoms, the net charge inside is zero, and the average field from all those dipoles cancels. The macroscopic field we measure is the average over many atoms, and that average is zero in a conductor at equilibrium.

3. What happens when an external field is applied?

Suppose you bring a charged rod near a metal sphere. The external field tries to push free electrons one way. But the electrons move so quickly (within $10^{-14}$ s) that they accumulate on one side of the sphere, creating an induced surface charge. This induced charge produces its own field that exactly opposes the external field inside the metal.

The result: inside the bulk (everywhere except a thin surface layer), the net field is zero. The free electrons have rearranged to shield the interior.

4. Why don't the atomic fields cause a net field? …

Concept: Electrostatic Shielding and Free Electrons in Conductors

The key idea is that conductors have mobile free electrons that can instantly rearrange themselves to cancel any internal electric field.

Method: The Free Charge Redistribution Method

Step 1: Understand the microscopic picture

• In a conductor (like copper), atoms give up their outermost electrons.
• These electrons are free to move throughout the material — they are not bound to individual atoms.
• The positive ions (nuclei + core electrons) form a fixed lattice.

Step 2: Apply an external field (or consider the atom's own field)

• If an electric field $\vec{E}$ exists inside the conductor, it exerts a force $\vec{F} = -e\vec{E}$ on each free electron.
• Electrons move instantly in the direction opposite to $\vec{E}$.

Step 3: Charge redistribution creates an opposing field

• Electrons accumulate on one surface, leaving positive charge on the opposite surface.
• This separation creates an induced electric field $\vec{E}_{\text{induced}}$ that points opposite to the original field.

Step 4: Equilibrium condition

• Electrons stop moving when:

• This happens in less than $10^{-16}$ seconds — essentially instantaneously.

Step 5: The result

• Inside the conductor, the net electrostatic field is exactly zero at equilibrium.
• The large atomic-scale fields between protons and electrons are perfectly cancelled by the rearrangement of free electrons.

Why the atomic fields don't matter …

Common Mistakes Students Make on This Question

This question tests your understanding of electrostatic equilibrium in conductors — not just the existence of electric fields inside atoms. Here are the most frequent errors:

1. Confusing Atomic Scale with Bulk Scale

Mistake: Students argue that because atoms have internal fields, the conductor must also have fields inside.

Why it's wrong: The question is about the net macroscopic field inside a conductor, not the microscopic fields between individual protons and electrons. Inside a conductor in equilibrium, charges rearrange so that the net field at any point is zero.

How to avoid: Always distinguish between:

• Microscopic fields (inside atoms, between charges)
• Macroscopic (net) field (average over many atoms)

2. Forgetting Free Electrons in Conductors

Mistake: Treating the conductor as a collection of fixed atoms, ignoring mobile electrons.

Why it's wrong: In conductors, valence electrons are free to move. When an external field is applied, these electrons redistribute until the internal field cancels the external one.

How to avoid: Remember: conductors have free charge carriers (electrons) that can move until electrostatic equilibrium is reached.

3. Misapplying Gauss's Law

Mistake: Claiming Gauss's law proves the field is zero everywhere inside, without considering charge distribution.

Why it's wrong: Gauss's law says $\oint \vec{E} \cdot d\vec{A} = \frac{Q_{\text{enc}}}{\varepsilon_0}$. For a conductor, excess charge resides on the surface, so $Q_{\text{enc}} = 0$ inside → $\vec{E} = 0$ inside. But this is a result of charge rearrangement, not a starting assumption.

How to avoid: Understand the cause-and-effect:

1. Free electrons move in response to any field
2. They accumulate on the surface
3. This creates an opposing field
4. Equilibrium: net field inside = 0

4. Ignoring the Time Scale of Rearrangement …