Electrostatics of conductors

Take a hollow metal sphere, charge it up, and push a small test charge through a tiny hole into its interior. However hard you push, however much charge sits on the outside, the test charge feels nothing once it is inside — no force, no field, total silence. This quiet interior is not an accident of one particular shape; it is a law that every conductor obeys once its free electrons have finished rearranging themselves. That rearrangement is fast (a fraction of a nanosecond for a good metal) and it always ends the same way: zero field inside, all the charge pushed out to the surface, the field just outside standing bolt upright perpendicular to the metal, and the potential the same number everywhere you probe the conductor, surface included. This chapter of "electrostatics of conductors" turns that one physical fact — free electrons hate an internal field and will move until it vanishes — into four testable properties, and then one clean formula, $E = \sigma/\varepsilon_0$ , linking the field just outside a charged surface to how densely charge is packed there. It explains why a car (or an aircraft fuselage) is one of the safer places to be in a lightning strike, why a pointed lightning rod is more effective than a flat plate, and why the metal cage of a lift shields your phone signal. None of it needs calculus beyond what you already have — just Gauss's law, applied with care to a surface drawn just inside the metal, and a steady head for what "electrostatic equilibrium" really forbids. We will state the four properties precisely, derive the surface-field formula from Gauss's law, work a board-style numeric problem and a slightly trickier shielding-and-induced-charge problem, and close with the traps that quietly cost marks — chiefly, confusing "field is zero inside the conductor" with "field is zero everywhere near it."

Key takeaways

A conductor has free (mobile) electrons; an insulator/dielectric has only bound charges that cannot travel through the bulk.
In electrostatic equilibrium, the field inside a conductor's material is exactly zero — any residual field would keep pushing free electrons, so equilibrium demands $E_{\text{inside}} = 0$ .
All net charge on a conductor sits on the outer surface only; Gauss's law on a surface drawn just inside the metal proves zero charge can hide in the bulk.
Just outside the surface, the field is perpendicular to the surface and has magnitude $E = \sigma/\varepsilon_0$ , where $\sigma$ is the local surface charge density.
The potential is uniform throughout the conductor, including its surface — since $E = 0$ inside, no work is done moving a test charge from any interior point to any other.
On an irregular shape, $\sigma$ (and hence $E$ ) is larger where the surface curves more sharply — the qualitative basis of lightning rods and corona discharge.
Key idea: free charges in a conductor redistribute until the interior field is zero, forcing all net charge onto the surface with $E = \sigma/\varepsilon_0$ just outside it and a single constant potential throughout — the physics behind electrostatic shielding.

The core insight

At electrostatic equilibrium a conductor's free charges arrange themselves so the field inside vanishes, pushing all net charge to the surface where $E = \sigma/\varepsilon_0$ acts perpendicular to it, and making the potential constant throughout the conductor including its surface.

Where it fits in the syllabus

This section sits inside the CBSE Class 12 Physics chapter "Electrostatic Potential and Capacitance" (Unit II, Electrostatics), right after potential energy in an external field and before dielectrics and polarisation. The prescribed board content is exactly the free-versus-bound-charge distinction and the standard list of conductor properties in equilibrium — this is core, examinable material, not enrichment. Two related items are explicitly kept light by the 2025-26 CBSE prescription: the Van de Graaff generator and corona discharge may be mentioned only as one-line applications, never built out as device-engineering detail, and the deeper molecular picture of how a dielectric slab polarises is left for the next node. Everything derived below — the four equilibrium properties and $E = \sigma/\varepsilon_0$ — is squarely inside the syllabus and fair game for both board and competitive exams.

The physical picture

Picture a solid or hollow chunk of metal. Deep inside the metal, at the atomic scale, some outer electrons of each atom are not bound to any single nucleus — they roam through the whole piece as a shared "sea" of free electrons, free to drift under the slightest push from an electric field. An insulator has no such sea: every electron stays clamped to its parent atom or molecule, so charge cannot flow through its bulk (though it can be slightly displaced within each atom, which is what a dielectric does — the subject of the next node, not this one). Now charge the conductor, or place it in an external field. The free electrons feel a force and start to drift. As they pile up on one side (or spread over the whole surface, for a net-charged isolated conductor), they themselves create an electric field pointing back the other way, partially cancelling whatever field was pushing them. The drift continues — it is a genuinely dynamic, if extremely fast, process — until the electrons' own field exactly cancels any field that would otherwise exist inside the metal. Only then does the motion stop; that stopped, unchanging state is electrostatic equilibrium. From that one requirement — "the free charges must stop moving" — four properties follow:

Property 1 — field inside is zero. If any field survived inside the conducting material, free electrons would keep accelerating along it, and the system would not be in equilibrium. So $E_{\text{inside}} = 0$ everywhere within the conductor's material, always, in electrostatic equilibrium.

Property 2 — all net charge lives on the surface. Draw an imaginary closed (Gaussian) surface just inside the conductor's actual surface, entirely within the metal. Since Property 1 guarantees the field is zero everywhere on this imaginary surface, Gauss's law $\oint \vec{E}\cdot d\vec{A} = q_{\text{enc}}/\varepsilon_0$ forces the enclosed charge $q_{\text{enc}}$ to be zero. This holds for any such surface drawn inside the metal, however close to the real outer surface — so no net charge can sit anywhere in the bulk; it must all be on the outer surface.

Property 3 — the field just outside is perpendicular to the surface. If the field just outside had any component running along the surface, that component would drag the free surface charges sideways, again contradicting equilibrium. So the field lines meet a charged conductor's surface at a right angle, everywhere.

Property 4 — the potential is constant throughout the conductor. Since $E_{\text{inside}} = 0$ , moving a test charge between any two points inside (or on the surface of) the conductor does zero work, and the potential difference between those two points is zero. The whole conductor — interior and surface together — sits at one single value of potential.

E = \frac{\sigma}{\varepsilon_0} \qquad \text{(field just outside a charged conductor's surface)}

where: $E$ is the magnitude of the electric field immediately outside the surface (N/C or V/m), $\sigma$ is the local surface charge density (C/m $^2$ , which may vary from point to point on an irregular conductor), and $\varepsilon_0 = 8.85\times10^{-12}\ \text{C}^2/\text{N}\,\cdot\,\text{m}^2$ is the permittivity of free space. The physical reading: wherever the free electrons have packed themselves more densely (larger $\sigma$ ), the field standing guard just outside is proportionally stronger — and on a real, irregularly shaped conductor, $\sigma$ (hence $E$ ) is largest where the surface is most sharply curved, such as a point or a thin edge, which is why lightning tends to strike the tallest, sharpest object around.

Figure

A charged metal sphere with field lines emerging perpendicular to its surface, zero field shown inside the metal, and a nearby pointed conductor where field lines are packed more closely near the sharp tip

Derivation AField just outside a charged conductor's surface

The goal is to show that the electric field immediately outside a conductor's surface must equal the local surface charge density divided by $\varepsilon_0$ , and that it must point straight out (or in) along the perpendicular to the surface — using nothing beyond Properties 1 to 3 above and a small "pillbox" Gaussian surface, which is the standard board-level route to this result and does not require anything above school algebra.

Step 1. Choose the Gaussian surface. Take a tiny cylindrical "pillbox," so small that the patch of the conductor's surface it straddles looks flat, with one circular face of area $\Delta A$ sitting just outside the conductor (in the air/vacuum) and the matching face of the same area $\Delta A$ sitting just inside the metal. The curved side wall of the cylinder is negligibly short, connecting the two faces.

Step 2. Evaluate the flux through each part. Inside the conductor, $E = 0$ (Property 1), so the inner face contributes zero flux. On the side wall, Property 3 tells us the field just outside is perpendicular to the surface — so it runs parallel to the side wall, not through it — and the wall is vanishingly short anyway, so its contribution is also zero. All the flux must therefore pass through the single outer face.

Step 3. Apply Gauss's law. The charge enclosed by the pillbox is the charge sitting on the patch of surface it straddles, $q_{\text{enc}} = \sigma\,\Delta A$ , where $\sigma$ is the local surface charge density. Gauss's law states $\oint \vec{E}\cdot d\vec{A} = q_{\text{enc}}/\varepsilon_0$ . With flux only through the outer face, this becomes $E\,\Delta A = \dfrac{\sigma\,\Delta A}{\varepsilon_0}$ .

Step 4. Cancel $\Delta A$ and read off the result. $E = \sigma/\varepsilon_0$ , directed perpendicular to the surface, outward for positive $\sigma$ and inward (toward the surface) for negative $\sigma$ . Because the pillbox was placed at an arbitrary point on the surface, this holds locally at every point — if $\sigma$ varies from point to point on an irregular conductor, so does $E$ just outside it, in direct proportion. Check: the formula has the right units, $\text{C/m}^2$ divided by $\text{C}^2/(\text{N}\,\cdot\,\text{m}^2)$ gives $\text{N/C}$ , a field; and it correctly predicts that a larger local $\sigma$ (as at a sharp point) gives a locally larger $E$ , matching the qualitative lightning-rod picture above.

Worked examples

Board level - field just outside a charged conductor's surface

A charged conductor has a surface charge density of $\sigma = 4.5\times10^{-6}\ \text{C/m}^2$ at a certain point on its surface. Find the magnitude of the electric field just outside the conductor at that point. Take $\varepsilon_0 = 8.85\times10^{-12}\ \text{C}^2/\text{N}\,\cdot\,\text{m}^2$ .

Step 1. Identify the formula that connects field just outside a conductor to surface charge density: $E = \sigma/\varepsilon_0$ .

Step 2. Substitute the given values, $\sigma = 4.5\times10^{-6}\ \text{C/m}^2$ and $\varepsilon_0 = 8.85\times10^{-12}\ \text{C}^2/\text{N}\,\cdot\,\text{m}^2$ .

Step 3. Divide: $E = \dfrac{4.5\times10^{-6}}{8.85\times10^{-12}} \approx 5.08\times10^{5}\ \text{N/C}$ .

Answer: the field just outside the conductor's surface at that point is about $5.08\times10^5\ \text{N/C}$ , directed perpendicular to the surface (outward, since $\sigma$ is positive).

Competitive level - shielding and induced charge on a hollow sphere in an external field

A hollow, initially uncharged metallic sphere of radius $R = 0.10\ \text{m}$ is placed in a uniform external electric field of magnitude $E_0 = 3.0\times10^{4}\ \text{N/C}$ . (a) What is the field inside the hollow cavity once equilibrium is reached, and why? (b) Describe, qualitatively, the induced surface charge distribution, and (c) explain why the total induced charge on the sphere must be zero even though $\sigma$ is not zero everywhere on it.

Step 1. The external field pushes the sphere's free electrons toward the face nearer the field source that repels them (or attracts them, depending on sign), so negative induced charge accumulates on one hemisphere and an equal positive induced charge (electron-depleted region) appears on the opposite hemisphere, while the sphere as a whole remains electrically neutral.

Step 2. This redistribution continues, growing the induced surface charge, until the field these induced charges produce exactly cancels the external field $E_0$ inside the conducting material itself — which must happen, because Property 1 forbids any residual field inside the metal at equilibrium.

Step 3. Since the conducting shell's material carries zero field throughout, and the hollow cavity is enclosed entirely by that shielded material with no free charge of its own inside the cavity, the field inside the empty cavity is also zero — this is electrostatic shielding: the interior of a closed (or nearly closed) conductor is protected from any external field, however strong $E_0$ is.

Step 4. The total induced charge is zero because the sphere started neutral and electrons only moved within it — no charge entered or left the sphere. So integrating $\sigma$ over the whole surface must give exactly zero net charge, even though $\sigma$ is locally negative on one hemisphere and locally positive on the other, each with the same magnitude by the symmetry of a sphere in a uniform field.

Answer: the field inside the hollow cavity is exactly zero, regardless of the strength $E_0 = 3.0\times10^4\ \text{N/C}$ of the external field — the sphere shields its interior completely — while the surface carries an induced negative charge on one hemisphere and an equal-magnitude induced positive charge on the other, summing to zero net charge overall. Timing note: this style of question rewards a clean four-line logical chain (redistribution occurs, it continues until $E_{\text{inside}}=0$ , the shielded shell protects the cavity, total induced charge is zero by neutrality) rather than any calculation — aim to write it in under three minutes, since there are no numbers to crunch, only the physics of Property 1 applied correctly.

Common traps

Believing a charged conductor's charge is spread through its whole volume — in equilibrium, all net charge sits only on the outer surface (Property 2), never in the bulk.
Assuming the field is zero everywhere near a conductor — it is zero only inside the conducting material; just outside the surface, $E = \sigma/\varepsilon_0$ is generally nonzero.
Treating surface charge density as uniform on any conductor's shape — it is uniform only for an isolated symmetric conductor like a sphere; on an irregular shape it is larger where the surface curves more sharply.
Confusing "potential is constant on the conductor" with "field is zero everywhere outside it" — potential is constant only throughout the conductor itself (interior plus surface), not in the surrounding space, where both field and potential generally vary with position.
Forgetting that shielding works for the field, not automatically for a changing magnetic field or for a conductor with a genuine hole large compared to the cavity — the syllabus-level statement is specifically about a closed (or nearly closed) conducting shell and a static electric field.

Common trapZero field inside the metal does not mean zero field everywhere outside it

The single most common slip is exporting "the field inside a conductor is zero" to the space around the conductor. The field is zero strictly inside the conducting material — the moment you step just outside the surface, the field is generally nonzero, equal to $\sigma/\varepsilon_0$ and pointing perpendicular to the surface. Likewise, "potential is constant on the conductor" is a statement about the conductor's own interior and surface, not about points in the surrounding vacuum or air, where the potential still varies smoothly with distance from the charged conductor. Keep these two zones — "inside the metal" and "the space around it" — mentally separate, because exam questions are frequently built to catch exactly this confusion.

You should be able to…

State and justify, from the requirement of electrostatic equilibrium, why the field inside a conductor's material must be zero.
Explain why all net charge on a conductor resides on its outer surface, using Gauss's law on a surface drawn just inside the metal.
Derive $E = \sigma/\varepsilon_0$ for the field just outside a charged conductor's surface using a small pillbox Gaussian surface.
Explain why the potential is constant throughout a conductor, including its surface.
Apply $E = \sigma/\varepsilon_0$ to numeric problems involving surface charge density.
Explain electrostatic shielding and predict the field inside a hollow conducting cavity placed in an external field.

Ask the tutor: want to see how these same four properties set up the parallel-plate capacitor formula $C = \varepsilon_0 A/d$ in the very next node, or how electrostatic shielding explains why the inside of a car is a relatively safe place during a lightning strike?

Why does the field have to be zero inside a conductor — what actually stops the charges from moving forever?

Think of it as a feedback loop, not a rule imposed from outside. If any field survived inside the metal, the free electrons there would keep feeling a force and keep accelerating along it — but that motion itself shifts more charge to the surface, which changes the field the moving electrons create, which changes the net field inside. The process only stops changing once the electrons' own rearranged charge exactly cancels whatever field was trying to push them, at which point there is genuinely no leftover force and nothing left to move. "Zero field inside" is therefore not an assumption we plug in — it is the necessary end state of any process that must eventually stop, and it happens extremely fast (a tiny fraction of a second) for ordinary metals.

If all the charge sits on the surface, does it matter whether the conductor is solid or hollow?

Not for the electrostatic properties in this chapter. Once you charge a conductor and let it reach equilibrium, the charge migrates to the outer surface regardless of whether the interior is solid metal or empty space — Property 2 followed purely from $E_{\text{inside}} = 0$ and Gauss's law on a surface drawn inside the material, and that argument works identically whether "inside" is metal or a hollow cavity bounded by metal. A solid sphere and a hollow spherical shell of the same outer radius, carrying the same net charge, produce exactly the same field and potential everywhere outside and on the surface, and zero field in both the metal and (for the hollow case) the cavity.

Is a Van de Graaff generator or corona discharge something I need to know in detail for boards?

No — for CBSE 2025-26, both are explicitly kept to a one-line mention as an application of "surface charge density is higher at sharp points," not core content you would be asked to derive or describe in device detail. What you should know is the underlying physics: a Van de Graaff generator exploits the fact that charge sprayed onto a large, smoothly curved conducting sphere spreads over its surface and (because the sphere's own field is being cancelled by that same charge) can accept an enormous total charge without breaking down the surrounding air, letting it build up very high potentials. Corona discharge is what happens at a sharp point instead, where locally huge $\sigma$ produces a locally huge field strong enough to ionise the surrounding air. Know the one-line idea; do not build an essay-length answer around either device.

Why does the potential stay the same everywhere inside and on a conductor, even far from where the charge actually sits?

Potential difference between two points is defined as the work done per unit charge in moving a test charge between them against the electric field, $V_A - V_B = -\int_B^A \vec{E}\cdot d\vec{l}$ . Inside the conductor's material, $\vec{E} = 0$ at every point, so that integral is zero along any path you choose between any two interior points — and since the surface is the boundary of that same zero-field region, points on the surface share the same value too. It does not matter how far apart the two points are inside the conductor, or how the conductor is shaped: as long as the whole path stays inside the material (or on the surface), zero field means zero work means zero potential difference, so the entire conductor is one single equipotential region.