Source: Discussion Question 5D (Exam Review)
Tags: electric field, conducting slabs, induced charge, surface charge density, superposition, Gauss's Law, conductors in electrostatics, P212, PHYS 212, UIUC
Difficulty: Intermediate to Advanced Prerequisites: Electric field of an infinite sheet of charge (σ/2ε₀), superposition principle, conductors in electrostatic equilibrium (E = 0 inside), Gauss's Law.
This topic ties together several ideas from earlier weeks: the electric field produced by infinite sheets of charge, the superposition principle, and the behaviour of conductors in electrostatic equilibrium. The setup, multiple parallel conducting slabs with known or induced charges, is a classic exam problem format. Mastering it means you can handle any "find the field at point X" or "find the induced surface charge" question that appears on the midterm.
Each infinite charged sheet produces a uniform field σ/(2ε₀) on both sides. To find the total field at any point, add up the contributions from every charged surface using superposition, paying attention to direction. Inside a conductor the net field must be zero, and that constraint tells you what the induced surface charges must be.
Infinite sheet of charge
An idealised flat surface carrying uniform surface charge density σ. It produces a uniform electric field of magnitude σ/(2ε₀) pointing away from the sheet (for positive σ) or toward it (for negative σ), on both sides. Think of it as the simplest possible source of a uniform electric field.
Superposition
The total electric field at any point is the vector sum of the fields produced by each source individually. In simple terms, you find each source's contribution separately and then add them all up.
Induced charge
Charge that appears on the surfaces of a conductor in response to external electric fields. The conductor was electrically neutral, so the induced charges on its two faces are equal and opposite. Think of it as the conductor rearranging its mobile charges until its interior field is zero.
Electrostatic equilibrium
The condition in which all charges in a conductor have settled and there is no current. The electric field inside the conductor is zero, and any excess charge sits on the surface.
Three infinite metal slabs arranged in parallel:
Left plate (very thin): surface charge density σ_a = +4 µC/m²
Middle plate (very thin): surface charge density σ_b = −3 µC/m²
Right slab (thickness d = 1 cm): carries no net charge, but has induced surface charges σ₁ (left face) and σ₂ (right face), with σ₁ + σ₂ = 0
Spacing: 3 cm between left and middle plates; 6 cm between middle plate and left face of thick slab. Points A and B are located between specific plates.
Each charged surface (σ_a, σ_b, σ₁, σ₂) acts as an infinite sheet producing field magnitude σ/(2ε₀) on each side.
To find the field at any point:
List every charged surface
For each surface, determine whether its field points left or right at the point of interest (positive charge pushes the field away from the surface; negative charge pulls it toward the surface)
Add up all contributions with correct signs (pick a positive direction, typically +x to the right)
The thick slab has no net charge, so σ₁ = −σ₂. Before you know the actual values of σ₁ and σ₂, you can still say:
At any point outside the slab, the fields from σ₁ and σ₂ cancel (because they are equal and opposite and both contribute to the same side in opposite directions)
Therefore the thick slab, by itself, produces zero electric field at point B (or anywhere outside it)
This is a general result: an uncharged conductor produces no external electric field on its own.
Point B sits between the middle plate and the thick slab. Contributions:
σ_a (to the left of B): field points rightward (+) with magnitude σ_a/(2ε₀)
σ_b (to the left of B): σ_b is negative, so field points leftward toward σ_b, which is also leftward (−), magnitude |σ_b|/(2ε₀)
Wait, let me be more careful. σ_b = −3 µC/m². The field from a negative sheet points toward the sheet on both sides. At point B (to the right of σ_b), the field from σ_b points leftward (toward σ_b), i.e., in the −x direction.
From σ_a at B: σ_a is positive, B is to the right of σ_a, so field points right: +σ_a/(2ε₀)
From σ_b at B: σ_b is negative, B is to the right of σ_b, so field points left (toward σ_b): +σ_b/(2ε₀) — but σ_b is negative, so this is a negative contribution
From σ₁ and σ₂: as shown above, they cancel at B
E_B = (σ_a + σ_b)/(2ε₀) = (4 − 3) × 10⁻⁶ / (2 × 8.854 × 10⁻¹²)
E_B = 1 × 10⁻⁶ / (1.771 × 10⁻¹¹) ≈ 5.65 × 10⁴ N/C, directed to the right
More precisely: E_B = (σ_a + σ_b)/(2ε₀), pointing in the +x direction (since the net is positive).
Point A sits between σ_a and σ_b. At point A:
From σ_a (A is to the right): field points right: +σ_a/(2ε₀)
From σ_b (A is to the left of σ_b): field from a negative sheet points toward it, so rightward: −σ_b/(2ε₀) = +|σ_b|/(2ε₀)
From σ₁ and σ₂: cancel (uncharged slab)
E_A = σ_a/(2ε₀) + |σ_b|/(2ε₀) = (σ_a − σ_b)/(2ε₀) = (4 − (−3)) × 10⁻⁶ / (2ε₀) = 7 × 10⁻⁶ / (1.771 × 10⁻¹¹)
E_A ≈ 3.95 × 10⁵ N/C, directed to the right
The electric field inside the thick conducting slab must be zero. The total field inside comes from all four surfaces: σ_a, σ_b, σ₁, and σ₂.
Setting the total field inside the slab to zero determines σ₁. The external sheets (σ_a and σ_b) produce a net rightward field inside the slab (same reasoning as at point B). The induced charges must produce a field that exactly cancels this. For the induced charges to push the field leftward inside the slab, the left face must be negative (σ₁ < 0) and the right face must be positive (σ₂ > 0).
So σ₁ is negative. The conductor's mobile electrons are pulled toward the external positive charges and pushed away from the external negative charges, resulting in a negative left face and a positive right face.
Electric field from one infinite sheet:
E = σ/(2ε₀), directed away from the sheet if σ > 0, toward it if σ < 0
Superposition for multiple sheets (choosing +x to the right):
At any point, sum each sheet's contribution with the correct sign based on whether the point is to the left or right of that sheet and the sign of σ.
Conductor constraint:
E_inside = 0, which means the sum of all contributions (external + induced) at any interior point is zero.
Uncharged conductor in an external field:
σ₁ + σ₂ = 0 (no net charge), and the induced charges adjust so that the internal field vanishes.
This type of analysis describes how shielding works. A grounded conducting enclosure (a Faraday cage) develops induced surface charges that cancel external fields in its interior. The same physics explains why coaxial cables have an outer conductor: induced charges on the inner surface of the shield cancel the field from the central wire, preventing electromagnetic interference from leaking out.
Students often forget that each surface of a conductor is a separate charged sheet. A thin plate with σ_a has charge on both faces in general, but for a very thin plate carrying a specified σ, the problem is typically treating it as a single sheet. Read the problem carefully.
Mixing up field directions. For a negative sheet, the field points toward the sheet on both sides, not away from it. Drawing a quick diagram with arrows before doing algebra prevents sign errors.
Assuming the uncharged slab has no effect. While it produces no external field on its own, its induced charges do affect the field between the slab and adjacent plates. The total field in those regions is determined by superposition of all surfaces including σ₁ and σ₂.
Forgetting the factor of 2. Each sheet produces σ/(2ε₀), not σ/ε₀. The σ/ε₀ result is for the field between two plates of a parallel-plate capacitor, which is the sum of two sheets' fields.
⚠️ This is labelled "Extra Question for Exam Review," which is a strong signal that problems of this type will appear on the midterm. Practice until the superposition-of-sheets method is automatic.
⚠️ The "E = 0 inside a conductor" constraint is the single most important tool for finding induced charges. If you know the net external field inside the slab must be zero, you can solve for σ₁ directly.
⚠️ Sign errors are the most common mistake. Always define a positive direction, draw the field arrows for each sheet, and track signs carefully through the algebra.
⚠️ Be comfortable with the units: σ in µC/m², ε₀ in C²/(N·m²), and E in N/C (or equivalently V/m).
True or False: An infinite sheet with surface charge density σ = −5 µC/m² produces a field that points toward the sheet on both sides.
Fill in the blank: The electric field inside a conductor in electrostatic equilibrium is ______.
True or False: An uncharged conducting slab placed between two charged plates has no effect on the electric field anywhere.
Fill in the blank: If a conductor carries no net charge, the induced surface charge densities on its two faces satisfy σ₁ + σ₂ = ______.
True or False: The field from a single infinite sheet of charge depends on the distance from the sheet.
Answers: 1. True. 2. Zero. 3. False (it affects the field between itself and adjacent plates through its induced charges, though it produces no field outside itself when considered alone). 4. Zero. 5. False (it is uniform, independent of distance).
Q: Two infinite parallel sheets carry σ₁ = +6 µC/m² and σ₂ = −2 µC/m². What is the electric field in the region between them?
A: Between the sheets, both fields point in the same direction (from the positive toward the negative sheet). E = σ₁/(2ε₀) + |σ₂|/(2ε₀) = (6 + 2) × 10⁻⁶ / (2 × 8.854 × 10⁻¹²) ≈ 4.52 × 10⁵ N/C, directed from the positive sheet toward the negative sheet.
Q: For the same two sheets, what is the field to the right of both (beyond the −2 µC/m² sheet)?
A: To the right, σ₁'s field points right and σ₂'s field points left (toward the negative sheet). E = (σ₁ + σ₂)/(2ε₀) = (6 − 2) × 10⁻⁶ / (2 × 8.854 × 10⁻¹²) ≈ 2.26 × 10⁵ N/C, directed to the right.
Q: An uncharged conducting slab is placed in a uniform external field E₀. What are the induced surface charge densities?
A: The induced charges must produce a field inside the slab that cancels E₀. Two opposing sheets each producing σ/(2ε₀) give a combined internal field of σ/ε₀. Setting σ/ε₀ = E₀ gives σ = ε₀E₀. The face toward the field source is negative (−ε₀E₀), the opposite face is positive (+ε₀E₀).
Q: Why is σ₁ negative on the left face of the thick slab in Discussion 5D?
A: The external charges (σ_a = +4, σ_b = −3) produce a net rightward field inside the slab. To cancel this and maintain E = 0 inside, the induced charges must create a leftward field inside. A negative left face and positive right face accomplish this, with field pointing from + to − (right to left) inside the slab.
This connects to Gauss's Law (the tool used to derive the field of an infinite sheet in the first place) and to the parallel-plate capacitor (which is just two oppositely charged sheets, giving E = σ/ε₀ between them by superposition of σ/(2ε₀) from each sheet). It also leads into the concept of electrostatic shielding and Faraday cages, which you may encounter later in the course or in lab.
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