Difficulty: Intermediate | Prerequisites: Gauss' law fundamentals, the infinite-plane result E = σ/(2ε₀), vector superposition.
Once you know the field from a single infinite charged plane (σ/(2ε₀), constant, independent of distance), you can analyse surprisingly complex systems just by adding fields from multiple planes together. This is the superposition principle applied to planar geometry, and it forms the basis of how capacitors work. This set of notes also introduces conductors in electrostatics, where the key rule is that the electric field inside a conductor is always zero. If you are not yet confident with the single-plane Gauss' law derivation, review that before continuing.
The electric field from an infinite charged slab is the same as from an infinite sheet: E = σ/(2ε₀), regardless of the slab's thickness or exact position. You can split a slab into multiple thinner slabs and superpose their fields. In a conductor, charge migrates to the surfaces and the interior field drops to zero, which lets you calculate the induced surface charge densities using Gauss' law.
Surface charge density (σ)
Charge per unit area, measured in C/m². For an infinite sheet or slab, σ is the total charge sitting on (or within) a unit area of the surface or slab. Think of it as how densely charge is spread across a flat region.
Superposition (for electric fields)
The principle that the total electric field at any point is the vector sum of the fields from each source, calculated independently. In simple terms, you work out the field from each charged slab on its own, then add them all up as vectors.
Conductor (electrostatic)
A material in which charges are free to move. In electrostatic equilibrium, all excess charge sits on the surface and the electric field inside is exactly zero. Think of it as a material that rearranges its charges until there is no internal force left to push them further.
Induced charge
Charge that redistributes itself on a conductor's surface in response to an external electric field. No charge is created or destroyed; it simply moves to cancel the field inside the conductor.
Gaussian pillbox
A rectangular or cylindrical Gaussian surface that straddles a flat charged surface, with one flat face on each side. It is the standard tool for applying Gauss' law to planar geometries.
Consider a thick slab of non-conducting material with total surface charge density σ, uniformly distributed throughout its volume.
A Gaussian pillbox straddling the slab encloses charge Q_enc = σA (where A is the area of the pillbox face).
Gauss' law gives: 2|E|A = σA/ε₀, so E = σ/(2ε₀) on each side, pointing away from the slab.
This result does not depend on:
The position of the slab (shifting it left or right changes nothing).
The thickness of the slab (a thicker slab with the same total σ per unit area produces the same external field).
Replacing one slab of charge density σ with two slabs, each carrying σ/2, produces the same total field at any external point.
Each slab contributes σ/(4ε₀), and at external points these contributions add constructively: σ/(4ε₀) + σ/(4ε₀) = σ/(2ε₀).
This generalises: any number of parallel infinite slabs can be analysed by summing their individual fields as vectors, considering direction carefully.
When the slab is made of metal (a conductor) instead of a non-conducting material:
Outside the slab: The field is unchanged. The total enclosed charge σ is the same, so Gauss' law gives the same result.
Inside the slab: The field becomes exactly zero. Free charges rearrange until no net force acts on them.
Charge distribution: All charge moves to the surfaces. None remains in the interior. This is a direct consequence of E = 0 inside; if there were a field, charges would keep moving.
Consider two non-conducting slabs (charge densities σ_a and σ_b) with an uncharged metal slab between them.
Finding the field at external points (A, B, C):
The metal slab's induced surface charges (σ_L on the left face, σ_R on the right face) produce fields that cancel each other at every point outside the metal. So you can ignore the metal's induced charges when computing the field at external points.
At points A and B (both to the left of the rightmost slab, or between the left slab and the metal): E_A = E_B = E_a + E_b. Both slab fields point in the same direction here.
At point C (to the right of both charged slabs): E_a points one way and E_b points the other. The net field is the difference of their magnitudes.
Worked example with σ_a = -4 C/m² and σ_b = +1.5 C/m²:
At points A and B: E = -(|σ_a| + σ_b)/(2ε₀) x̂ = -(5.5 C/m²)/(2ε₀) x̂
At point C: E = -(|σ_a| - σ_b)/(2ε₀) x̂ = -(2.5 C/m²)/(2ε₀) x̂
σ_L (left surface) is positive, σ_R (right surface) is negative. This follows from the direction of the external field: it pushes positive charges to the left and negative charges to the right.
The zero-field condition inside the metal means the fields from σ_L and σ_R must exactly cancel the combined field from slabs a and b at every point in the metal's interior.
Using Gauss' law:
2E_m = (σ_L + |σ_R|)/(2ε₀) = (|σ_a| + σ_b)/(2ε₀)
For the example above: σ_L = 2.75 C/m², σ_R = -2.75 C/m²
Conservation of charge: σ_L + σ_R = σ_m (the net charge on the metal slab). For an uncharged metal slab, σ_L = -σ_R.
If the metal slab is given a net charge σ_m = +3 C/m²:
The net charge distributes evenly: +1.5 C/m² is added to each surface first.
Then the induced redistribution from the external fields shifts charge exactly as before.
Result: σ_L = 2.75 + 1.5 = 4.25 C/m², σ_R = -2.75 + 1.5 = -1.25 C/m²
Check: σ_L + σ_R = 4.25 + (-1.25) = 3.0 C/m² = σ_m. ✓
Quantity | Formula | Notes |
|---|---|---|
Field from infinite slab/sheet | E = σ/(2ε₀) | Independent of distance and slab thickness |
Gauss' law (pillbox) | 2|E|A = σA/ε₀ | For a symmetric slab geometry |
Induced charge (uncharged metal) | σ_L = -σ_R | Conservation of charge, zero net charge |
Induced charge (charged metal) | σ_L + σ_R = σ_m | Net charge constraint |
Field inside conductor | E = 0 | Always, in electrostatic equilibrium |
This is directly how parallel-plate capacitors are designed and analysed. The two charged slabs represent the capacitor plates, and the uniform field between them stores energy. Understanding how a metal slab between two charged plates develops induced surface charges is the basis of electrostatic shielding, used in electronics to protect sensitive circuits from external fields.
Students often think that making a slab thicker will change the external field. It does not, provided the total charge per unit area σ stays the same.
Students forget that the induced charges on an uncharged metal slab cancel each other's fields outside the metal. You can ignore them when computing the field at external points, but you cannot ignore them when computing the field inside the metal.
A common error is assuming the field inside a conductor is zero only when the conductor is uncharged. The field inside is zero regardless of how much net charge the conductor carries, as long as it is in electrostatic equilibrium.
When a metal slab has net charge, students sometimes put all the charge on one surface. In a symmetric external field it splits evenly first, then redistributes in response to the external sources.
⚠️ You will almost certainly see a problem with multiple parallel charged slabs and be asked for the field at various points. Practise superposition with signs and directions until it is automatic.
⚠️ Expect a problem where a conductor is inserted between charged plates. You must use E = 0 inside the conductor to find induced surface charges.
⚠️ The relationship σ_L + σ_R = σ_m (net charge on the metal) is a standard check. If your answer violates this, something has gone wrong.
⚠️ Questions about what does and does not change when you swap a non-conducting slab for a conducting one are a classic conceptual exam question.
True or false: The electric field outside an infinite charged slab depends on the slab's thickness.
Fill in the blank: Inside a conductor in electrostatic equilibrium, the electric field is ______.
True or false: If an uncharged metal slab is placed between two charged non-conducting slabs, the field at points outside all three slabs is affected by the metal's presence.
Fill in the blank: When a metal slab with net charge σ_m is placed in an external field, the sum of its surface charges σ_L + σ_R must equal ______.
True or false: Replacing a single charged slab (charge density σ) with two slabs (each σ/2) changes the field at an external point.
Answers: 1. False. 2. Zero. 3. False (the metal's induced charges cancel outside). 4. σ_m. 5. False.
Q: A non-conducting slab of infinite area carries a uniform charge density σ = +6 C/m². What is the electric field at a point 10 m to the right of the slab?
A: E = σ/(2ε₀) = 6/(2 × 8.85 × 10⁻¹²) ≈ 3.39 × 10¹¹ N/C, pointing to the right (away from the slab). The distance of 10 m is irrelevant for an infinite plane.
Q: An uncharged metal slab is placed between two infinite non-conducting slabs with σ_a = -4 C/m² and σ_b = +1.5 C/m². What are the induced surface charge densities σ_L and σ_R on the metal?
A: σ_L = (|σ_a| + σ_b)/2 = (4 + 1.5)/2 = 2.75 C/m². σ_R = -2.75 C/m². The metal's left surface is positive because the field from the two external slabs pushes positive charge leftward.
Q: If the metal slab above is now given a net charge of +3 C/m², what are the new σ_L and σ_R?
A: The +3 C/m² splits evenly (+1.5 to each surface), then the induced shift from the external field adds the same ±2.75 as before. σ_L = 1.5 + 2.75 = 4.25 C/m², σ_R = 1.5 - 2.75 = -1.25 C/m².
Q: Why can you ignore the metal slab's induced charges when calculating the field at a point outside all three slabs?
A: Because an uncharged conductor's two surface charges are equal in magnitude and opposite in sign. Their fields cancel at every external point. Even when the metal has net charge, the induced component still cancels externally; only the evenly split net charge contributes, and that behaves like a single sheet of charge σ_m.
This connects directly to capacitors: a parallel-plate capacitor is two infinite charged sheets, and the field between them (E = σ/ε₀ when both plates are accounted for) determines the voltage and stored energy. The conductor rules here also lay the groundwork for Faraday cages and electrostatic shielding. The superposition technique you practise with slabs is the same one you will use when combining fields from point charges, lines, and spheres in more complex problems.
Superposition, infinite plane, infinite slab, charged sheet, surface charge density, σ, Gaussian pillbox, conductor, electrostatic equilibrium, induced charge, metal slab, parallel plates, capacitor, E = σ/(2ε₀), field inside conductor, charge redistribution, P212, PHYS 212, University Physics, electromagnetism