Difficulty: Intermediate | Prerequisites: Coulomb's law, vector addition, basic integration (setting up and evaluating definite integrals)
Big picture: Point charges are the simplest case, but real-world charge distributions are spread out over surfaces, lines, and volumes. This topic bridges the gap between the discrete Coulomb's law problems and the continuous world. You learn two things here: how to use superposition with infinite planes (which gives clean algebraic results), and how to set up integrals for curved charge distributions like semicircles. Both techniques appear repeatedly on exams and in later topics like Gauss's law and potential.
An infinite plane of surface charge density σ produces a uniform electric field of magnitude σ/(2ε₀) on each side. For multiple planes, you superpose each plane's contribution with attention to direction. For a curved charge distribution, you write a small element of charge (dq), express the field it produces, and integrate over the whole distribution, keeping careful track of components.
Surface charge density (σ)
Charge per unit area, measured in C/m². Used for flat surfaces and planes. In simple terms, it tells you how densely packed the charge is across a sheet.
Infinite plane approximation
The assumption that a charged sheet extends infinitely in all directions. The electric field from such a sheet is uniform (same magnitude and direction everywhere on one side), with magnitude E = σ/(2ε₀). In simple terms, you are close enough to the sheet that its edges do not matter.
Linear charge density (λ)
Charge per unit length, measured in C/m. Used for wires, rods, and curves. If a semicircular arc of radius r carries total charge Q, then λ = Q/(πr), because the arc length of a semicircle is πr.
Electric field element (dE)
The tiny contribution to the electric field from a small piece of charge dq. You integrate these contributions over the whole distribution to find the total field.
A single infinite plane with surface charge density +σ produces a field of magnitude σ/(2ε₀), pointing away from the plane on both sides.
For a negative surface charge density −σ, the field points toward the plane on both sides.
When multiple parallel planes are present, the total field at any point is the vector sum of the contributions from every plane.
Because each plane's field is uniform and perpendicular to the plane, you only need to track the sign (left or right) of each contribution at the point of interest.
The planes carry surface charge densities +σ, −2σ, and −2σ from left to right. Points A and B are located between different pairs of planes.
At point A (between the first and second planes):
The +σ plane pushes the field to the right: +σ/(2ε₀)
The −2σ plane (to the right of A) pulls the field to the right: +2σ/(2ε₀)
The −2σ plane (far right) pulls the field to the right: +2σ/(2ε₀)
Total at A: (σ + 2σ + 2σ)/(2ε₀) = 5σ/(2ε₀), directed to the right.
At point B (between the second and third planes):
The +σ plane pushes field to the right: +σ/(2ε₀)
The −2σ plane (to the left of B) pulls field to the left: −2σ/(2ε₀)
The −2σ plane (to the right of B) pulls field to the right: +2σ/(2ε₀)
Total at B: (σ − 2σ + 2σ)/(2ε₀) = σ/(2ε₀), directed to the right.
The ratio E_A / E_B = [5σ/(2ε₀)] / [σ/(2ε₀)] = 5.
When charge is spread along a curve rather than at a point, you integrate:
Set up the geometry. A semicircle of radius r in the x-y plane, centred at the origin, spanning from −π/2 to +π/2. The field is to be found at the origin, or at a point like P(−r, 0).
Express dq. A small arc element subtends angle dθ, so its length is r dθ and the charge on it is dq = λ r dθ.
Find dE. The field from dq at the origin has magnitude dE = k dq / r² = kλ dθ / r.
Resolve into components. A charge element at angle θ on the semicircle contributes:
dEₓ = −(kλ/r) cos θ dθ (the minus sign comes from the geometry: the charge is to the right of the origin, so its field at the origin points to the left, in the −x direction).
dE_y = −(kλ/r) sin θ dθ
Integrate. The y-component integrates to zero by symmetry (sin θ is odd over −π/2 to +π/2). The x-component gives:
Eₓ = −∫ from −π/2 to π/2 of (kλ/r) cos θ dθ
This is the correct integral expression for the x-component of the field at the origin.
The arc length of a semicircle of radius r is πr.
If the total charge is Q = +15 μC and r = 5 cm = 0.05 m:
λ = Q / (πr) = 15 × 10⁻⁶ / (π × 0.05) ≈ 95.5 μC/m
So λ ≈ 95 μC/m.
The semicircle (positive charge on the right side of the origin) produces a net field at the origin pointing in the −x direction.
To cancel this field, you need a field at the origin pointing in the +x direction.
A point charge at P(−r, 0) (to the left of the origin) that is negative would produce a field at the origin pointing in the −x direction (toward the negative charge), which adds to the existing field rather than cancelling it.
A positive charge at P(−r, 0) produces a field at the origin pointing in the +x direction (away from the positive charge), which opposes the semicircle's field.
Therefore, a positive point charge is needed at P(−r, 0).
Field from an infinite plane:
E = σ / (2ε₀)
directed perpendicular to the plane, away from positive charge, toward negative charge.
Linear charge density from total charge on a semicircle:
λ = Q / (πr)
Field element from a line charge at distance r:
dE = kλ dθ / r (for an arc element subtending angle dθ at the centre)
x-component of the field at the centre of a semicircular arc (right semicircle, charge on right):
Eₓ = − ∫ from −π/2 to π/2 (kλ/r) cos θ dθ
The infinite plane result is the starting point for understanding parallel plate capacitors, which are everywhere: in touchscreens, camera flashes, and defibrillators. Two planes with equal and opposite charge density create a uniform field between them and zero field outside, which is exactly how a capacitor stores energy.
Integration over continuous charge distributions is how engineers model the fields around charged wires in power transmission lines and the fringing fields at the edges of real capacitor plates.
Students often get the direction wrong for a negative plane. A plane with charge density −σ has its field pointing toward the plane on both sides, not away from it. Draw the arrows before adding.
When superposing fields from multiple planes, students sometimes forget that a plane contributes to the field on both sides. Even a plane far to the right still produces a field at a point far to the left.
For the semicircular integral, students sometimes write dq = λ dθ, omitting the factor of r. The correct element is dq = λ (r dθ), since r dθ is the arc length.
Students sometimes mix up whether to use kλ/r or kλ/r² in the integrand. For a semicircular arc where every element is the same distance r from the field point at the centre, the field from dq is k dq/r² = kλ r dθ/r² = kλ dθ/r. The r² in Coulomb's law partly cancels with the r in dq.
⚠️ The ratio E_A/E_B problem for multiple infinite planes is a classic exam format. Write out each plane's contribution with its sign at each point. Forgetting one plane or getting one sign wrong changes the answer entirely.
⚠️ Setting up the integral for a continuous charge distribution is tested frequently. You are expected to identify dq, express dE, resolve into components, and write the correct integral with limits. You may not need to evaluate it, just set it up correctly.
⚠️ Symmetry arguments (the y-component integrates to zero for a symmetric distribution) are both a time-saver and a point-earner. State the symmetry clearly.
⚠️ To decide what charge cancels a field, think about which direction the existing field points and what sign of charge would produce a field in the opposite direction at the same point. Draw it.
True or false: The electric field from an infinite plane of charge depends on the distance from the plane.
Fill in the blank: The arc length of a semicircle of radius r is ________.
True or false: When integrating the electric field from a symmetric semicircular charge distribution centred on the y-axis, the y-component of the field at the centre is zero.
Fill in the blank: The field from a single infinite plane with surface charge density σ has magnitude ________.
True or false: To cancel a field pointing in the −x direction at the origin, you could place a positive point charge to the left of the origin.
Q: Three parallel infinite planes have surface charge densities +σ, −2σ, and −2σ from left to right. What is E_A/E_B, where A is between the first and second planes and B is between the second and third?
A: E_A = 5σ/(2ε₀) and E_B = σ/(2ε₀), so E_A/E_B = 5.
Q: A semicircular arc of radius r = 5 cm carries total charge Q = +15 μC. What is the linear charge density λ?
A: λ = Q/(πr) = 15 × 10⁻⁶ / (π × 0.05) ≈ 95 μC/m.
Q: What is the correct integral expression for the x-component of the electric field at the origin due to a uniformly charged right semicircle of radius r and linear charge density λ?
A: Eₓ = − ∫ from −π/2 to π/2 (kλ/r) cos θ dθ.
Q: A positively charged semicircle to the right of the origin produces a net field at the origin pointing in the −x direction. What sign of point charge placed at (−r, 0) would cancel this field?
A: A positive point charge. It produces a field at the origin pointing in the +x direction (away from the positive charge), opposing the semicircle's field.
The infinite plane result is used directly in the capacitors topic at the end of this midterm. Two oppositely charged infinite planes give E = σ/ε₀ between them and zero outside, which is the parallel plate capacitor field. The integration technique for continuous charge distributions reappears when you compute electric potential from distributed charges, and it is closely related to the integrals that appear in Gauss's law (where you integrate the field over a surface rather than summing charge contributions).
infinite plane, surface charge density, σ, sigma, electric field superposition, semicircular charge distribution, linear charge density, lambda, λ, continuous charge distribution, integration, dq, arc length, PHYS 212, Physics 212 midterm, E field from plane, multiple planes, field cancellation