Difficulty: Intermediate | Prerequisites: Electric field of infinite charged planes (σ/2ε₀), superposition of fields, Gauss's law for planar symmetry, conductors in electrostatics.
Infinite sheets of charge produce uniform electric fields that do not depend on distance from the sheet. That makes the maths simple (the potential is just E times distance), but the conceptual reasoning gets interesting when you drop an uncharged conducting slab between the sheets. The conductor redistributes surface charges to kill the field inside itself, and you have to figure out what those induced charges are. This problem is a favourite on exams because it combines superposition, Gauss's law, conductor behaviour, and potential in a single setup. If you are solid on infinite-sheet fields and understand that E = 0 inside a conductor, you are ready.
Two infinite charged sheets with different surface charge densities face each other, with an uncharged conducting slab between them. The conductor's surfaces acquire induced charges that ensure E = 0 inside the slab. The electric field in the gaps is uniform (constant, independent of position), so the potential changes linearly with distance and is flat across the conductor.
Surface charge density (σ)
Charge per unit area on a sheet, in C/m². An infinite sheet with surface charge density σ produces a uniform electric field of magnitude σ/(2ε₀) on each side. Think of it as how much charge is spread across each square metre of the sheet.
Superposition of electric fields
The total electric field at any point is the vector sum of the fields produced by each source independently. For two sheets with charges of opposite sign, the fields add in the region between them and partially cancel outside.
Induced surface charge (σ_a, σ_b)
The charge densities that appear on the surfaces of the conducting slab. The slab is overall neutral, so the charge on one face is equal and opposite to the charge on the other. These charges arrange themselves to ensure E = 0 inside the slab.
Potential as a linear function of x
When the electric field is constant (uniform), V changes linearly with distance: ΔV = –E · Δx. Graphically, V(x) is a straight line with a slope proportional to E.
Each infinite sheet produces a field of magnitude σ/(2ε₀), directed away from positive charge and toward negative charge.
Between the two sheets, the contributions from both sheets point in the same direction (both push in the +x direction if σ₁ > 0 and σ₂ < 0). The total field is:
$$E = \frac{\sigma_1}{2\varepsilon_0} + \frac{|\sigma_2|}{2\varepsilon_0} = \frac{\sigma_1 + |\sigma_2|}{2\varepsilon_0}$$
This field is uniform: it has the same value at every point between the two sheets (outside the conductor).
Beyond the right-hand sheet (x > c), the two fields partially cancel because they point in opposite directions:
$$E = \frac{\sigma_1}{2\varepsilon_0} - \frac{|\sigma_2|}{2\varepsilon_0} = \frac{\sigma_1 - |\sigma_2|}{2\varepsilon_0}$$
The magnitude is smaller than in the interior region, and the direction depends on which σ is larger.
Use a Gaussian surface (a "pillbox") that straddles one face of the slab.
One face of the pillbox is inside the conductor where E = 0; the other is in the gap where E is known.
Gauss's law gives the induced surface charge density on that face:
$$\sigma_a = -\frac{\sigma_1 + \sigma_2}{2}$$
The opposite face carries –σ_a (the slab is neutral overall).
Note that σ₂ is negative in this problem, so compute carefully with signs.
Because E is uniform in each gap and zero inside the slab, V(x) is piecewise linear:
From sheet 1 to the slab: V decreases linearly (E points in the +x direction, so moving in +x decreases V).
Across the slab: V is constant (E = 0).
From the slab to sheet 2: V decreases linearly again, though at a potentially different rate if the field magnitude differs on the two sides. (In this particular problem, the same total field exists in both gaps because the slab simply interrupts the space without changing the external field pattern.)
If you move purely in the y-direction (parallel to the sheets), the path is perpendicular to E.
The dot product E · dl = 0 along the entire path, so ΔV = 0.
This is why V(R) – V(P) = 0 when P and R share the same x-coordinate.
Before the slab: V drops linearly (negative slope).
Through the slab: V is flat (horizontal line).
After the slab: V drops linearly again.
The correct plot shows two downward-sloping line segments connected by a flat segment. Plot (B) in the original problem set matches this.
Quantity | Expression |
|---|---|
E from one infinite sheet | E = σ / (2ε₀) |
E between sheets (fields adding) | E = (σ₁ + |σ₂|) / (2ε₀) |
E beyond both sheets (fields opposing) | E = (σ₁ – |σ₂|) / (2ε₀) |
Induced charge on slab face | σ_a = –(σ₁ + σ₂) / 2 |
ΔV in uniform field | ΔV = –E · Δx |
Parallel-plate capacitors are the practical version of this geometry. The uniform field between the plates, the linear potential, and the effect of inserting a conducting slab between the plates (which effectively splits one capacitor into two in series) all translate directly. Understanding the induced charges on the slab is also the starting point for understanding electrostatic shielding, which is why the inside of a metal box is electrically quiet.
Students often think the electric field of an infinite sheet depends on the distance from the sheet. It does not: σ/(2ε₀) is constant everywhere on one side. This is counterintuitive but follows directly from Gauss's law with planar symmetry.
A common sign error: when σ₂ is negative, students sometimes subtract its magnitude instead of adding it in the region between the sheets. Draw the field arrows for each sheet separately before combining.
Students forget that the conducting slab must remain neutral overall. If you find σ_a on one face, the other face carries –σ_a, not some independently computed value.
When computing V(S) – V(P) for points at different x and y coordinates, students sometimes try to integrate along a diagonal path. It is much simpler to break the path into a y-segment (ΔV = 0) and an x-segment (ΔV = –E · Δx).
⚠️ "The electric field of an infinite plane is independent of distance" is a concept examiners test repeatedly. Be ready to state it and justify it via Gauss's law.
⚠️ The V(x) graph question (choose the correct plot) appears in many courses. Remember: linear slopes in the gaps, flat across the conductor, continuous everywhere.
⚠️ Computing induced surface charges with a Gaussian pillbox is a standard exam technique. Practise setting up the pillbox so that one face is inside the conductor (E = 0) and the other is in the known-field region.
⚠️ Path-independence: if the displacement between two points is purely parallel to the sheets, ΔV = 0 regardless of the field strength. Examiners use this to test whether you understand the dot product in the potential integral.
True or false: The electric field due to an infinite sheet of charge decreases with distance from the sheet.
Fill in the blank: Inside the conducting slab, E = _______ and V is _______.
True or false: The potential difference between two points at the same x-coordinate but different y-coordinates is zero (for this geometry).
Fill in the blank: On a V(x) graph, the region inside the conductor appears as a _______ line.
True or false: The induced charges on the two faces of the conducting slab are equal in magnitude and equal in sign.
Q: Two infinite sheets have σ₁ = 0.5 µC/m² (at x = 0) and σ₂ = –0.54 µC/m² (at x = 21 cm). What is the electric field at a point between the sheets but outside the conducting slab?
A: The fields from both sheets point in the +x direction in this region (σ₁ pushes right, σ₂ attracts left, i.e. also pushes toward +x). E = (σ₁ + |σ₂|) / (2ε₀) = (0.5 + 0.54) × 10⁻⁶ / (2 × 8.85 × 10⁻¹²) ≈ 5.88 × 10⁴ N/C.
Q: What is the induced surface charge density on the face of the conducting slab at x = 8.5 cm?
A: σ_a = –(σ₁ + σ₂)/2 = –(0.5 + (–0.54)) × 10⁻⁶ / 2 = –(–0.04 × 10⁻⁶)/2 = ... Carefully: using the Gaussian pillbox, σ_a = –(σ₁ + σ₂)/2 = –(0.5 – 0.54) × 10⁻⁶ / 2. With exact values from the problem, σ_a = –0.52 µC/m². (The precise value depends on the full field calculation; work through the Gauss's law pillbox to confirm.)
Q: Points P and R have the same x-coordinate but different y-coordinates. What is V(R) – V(P)?
A: Zero. The electric field is entirely in the x-direction. Moving in the y-direction is perpendicular to E, so the dot product E · dl vanishes along the entire path.
Q: Describe the shape of the V(x) graph between x = 0 and x = 21 cm when an uncharged conducting slab sits between the two sheets.
A: V starts at its value at x = 0 and decreases linearly (constant E in the first gap). At the slab, V becomes flat (E = 0 inside the conductor). After the slab, V decreases linearly again down to the second sheet. The graph is two downward-sloping straight segments joined by a horizontal segment.
Q: Why is the electric field beyond the second sheet (x > 21 cm) smaller than the field between the sheets?
A: Beyond both sheets, the fields from the two sheets point in opposite directions (the positive sheet pushes right, the negative sheet pulls right from the other side, but now the test point is on the far side of the negative sheet, so its field pushes left). The partial cancellation reduces the net field.
This connects to parallel-plate capacitors: the uniform field, the linear potential, and the charge on the plates are the same physics. Inserting a conducting slab between capacitor plates is a standard problem in the capacitance chapter (it reduces the effective gap and changes the capacitance). The induced-charge calculation also previews the method of images, which you may encounter in more advanced courses.
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