Electric Fields of Charge Distributions: Study Guide (Physics C: E&M 8.4)

Course note. This guide is for the calculus-based Physics C: Electricity and Magnetism course, Topic 8.4. You will set up and evaluate integrals. The algebra-based Physics 2 course does not ask you to integrate over a charge distribution.

Constants used throughout: ε₀ = 8.85 × 10⁻¹² C²/(N·m²) and k = 1/(4πε₀) = 8.99 × 10⁹ N·m²/C².

From point charges to continuous charge

In Topic 8.3 you found the field of a few point charges by adding their fields as vectors. Real charged objects (a rod, a ring, a curved wire) hold charge spread along their length. You cannot list every charge, so you cut the object into tiny pieces. Each piece has charge dq and acts like a point charge. Its field at the point P is:

dE = k dq / r², directed along r̂ (away from dq if dq is positive)

Here r is the distance from that piece to P, and r̂ is the unit vector from the piece to P. The total field is the sum of all these small fields, which is an integral:

E = k ∫ dq / r² r̂

This is the principle of superposition turned into calculus. Because E is a vector, you never integrate the magnitudes directly unless every dE points the same way. Instead, you integrate each component: E_x = ∫ dE_x and E_y = ∫ dE_y.

To write dq, use the charge density of the object:

Object	Density	Small piece of charge
Straight rod or wire	λ (C/m), charge per unit length	dq = λ dx
Circular arc or ring of radius R	λ (C/m)	dq = λ dl = λR dθ
Surface	σ (C/m²)	dq = σ dA
Volume	ρ (C/m³)	dq = ρ dV

For a uniform rod of charge Q and length L, λ = Q/L. For a uniform semicircle of radius R, λ = Q/(πR). In this topic you integrate along lines and arcs; surfaces and volumes come back with Gauss’s law in Topic 8.6.

A method that works every time

1. Draw the object, the point P and axes. Mark one general piece dq at a general position (x or θ), not at a special point.
2. Write dq in terms of one variable: λ dx for a rod, λR dθ for an arc.
3. Write r from that piece to P in terms of the same variable.
4. Use symmetry to decide which components cancel. Keep only the ones that survive.
5. Write the surviving component, for example dE_x = (k dq/r²) cos θ, with cos θ also written in the same variable.
6. Set the limits so they cover the whole object once.
7. Integrate, then check: units, direction, and a limiting case such as “far away, it looks like a point charge”.

Symmetry does half the work

Symmetry tells you the direction of E before any calculus. Look for pairs of pieces that sit in mirror positions relative to P.

• Ring, point on its axis. A piece at the top of the ring has a partner at the bottom. Their field components across the axis are equal and opposite, so they cancel. Only the component along the axis survives.
• Rod, point on its perpendicular bisector. Pieces at +y and −y pair up. The components along the rod cancel, so E is perpendicular to the rod.
• Semicircle, point at its centre. Pieces at angle θ and π − θ pair up. E points along the line of symmetry, away from the arc’s middle for a positive charge.
• Full ring, point at its centre. Every piece has a partner directly opposite. Everything cancels: E = 0.
• Rod, point on its own line. Every dE already points the same way, along the rod’s line. Nothing cancels, but nothing needs resolving either.

If you skip this step, you may integrate a component that should be zero and get a confusing answer, or you may add magnitudes that point in different directions.

Worked example 1: a finite rod, point on its line

Question. A thin plastic rod of length L = 0.40 m carries a charge Q = +8.0 × 10⁻⁹ C spread uniformly. Point P lies on the line of the rod, d = 0.10 m beyond its right-hand end. Find E at P.

Set-up. Put the rod on the x-axis from x = −L to x = 0, with P at x = d. Take a piece of length ds at position s. Then:

• dq = λ ds, with λ = Q/L = 2.0 × 10⁻⁸ C/m
• distance from the piece to P: r = d − s
• every piece pushes a positive test charge at P in the +x direction, so all the dE vectors point the same way

Integral.

E = ∫ k λ ds / (d − s)², from s = −L to s = 0

Since d/ds [1/(d − s)] = 1/(d − s)², the antiderivative is kλ/(d − s). So:

E = kλ [1/d − 1/(d + L)] = kλL / [d(d + L)] = kQ / [d(d + L)]

Numbers. E = (8.99 × 10⁹)(8.0 × 10⁻⁹) ÷ [(0.10)(0.50)] = 71.92 ÷ 0.050 = 1438 N/C.

Answer. E ≈ 1.4 × 10³ N/C, pointing along the rod’s line, away from the rod (+x).

Checks.

• If P is far away (d much larger than L), d(d + L) ≈ d², so E ≈ kQ/d², the field of a point charge. Good.
• Treating the whole rod as a point charge at its centre (0.30 m from P) gives kQ/(0.30)² = 799 N/C, which is too small here. The near end of the rod is much closer than 0.30 m, and because of the 1/r² rule, close charge counts for more. The point-charge shortcut works only when P is far from the rod.

Worked example 2: a ring of charge, point on its axis

Question. A thin ring of radius a = 0.050 m carries Q = +6.0 × 10⁻⁹ C spread uniformly. (a) Derive E at a distance z from the centre, along the axis. (b) Evaluate E at z = 0.12 m. (c) Find where on the axis E is greatest.

(a) Derivation. Every piece dq of the ring is the same distance from P:

r = √(z² + a²)

Each dE has a component along the axis and one across it. By symmetry (Figure 1), the components across the axis cancel in pairs. The axial component of one piece is:

dE_z = (k dq / r²) cos θ, where cos θ = z/r

Here z, a and r are the same for every piece, so they come out of the integral, and ∫dq = Q:

E_z = k z Q / r³ = kQz / (z² + a²)^(3/2)

(b) Numbers. r = √(0.12² + 0.050²) = 0.13 m. Then E = (8.99 × 10⁹)(6.0 × 10⁻⁹)(0.12) ÷ (0.13)³ = 2946 N/C, so E ≈ 2.9 × 10³ N/C, pointing along the axis away from the ring.

Check the meaning: kQ/r² = 3192 N/C is the size of all the dE vectors added as if they were parallel. The factor cos θ = z/r = 0.923 removes the parts that cancel: 3192 × 0.923 = 2946 N/C.

(c) Maximum. Differentiate:

dE_z/dz = kQ [(z² + a²) − 3z²] / (z² + a²)^(5/2)

This is zero when a² − 2z² = 0, so z = a/√2 = 0.0354 m. There E_max = 2kQ/(3√3 a²) ≈ 0.385 kQ/a² = 8.3 × 10³ N/C.

Limiting cases.

• At the centre, z = 0: E = 0, as the full-ring symmetry says.
• Far away, z much larger than a: (z² + a²)^(3/2) ≈ z³, so E ≈ kQ/z², a point charge. Figure 2 shows the ring curve joining the point-charge curve.
• For z < 0 the formula gives a negative E_z: the field still points away from the ring, now in the −z direction.

The finite rod on its perpendicular bisector, and the infinite line

Put a uniform rod (charge per length λ, length L) on the y-axis from −L/2 to +L/2. P is on the x-axis at distance x. A piece dy at height y is a distance r = √(x² + y²) from P. By symmetry, the y-components cancel. The x-component of one piece is:

dE_x = (kλ dy / r²)(x / r) = kλx dy / (x² + y²)^(3/2)

Integrate from −L/2 to L/2, using the antiderivative y / [x²√(x² + y²)]:

E = kλL / [x√(x² + L²/4)] = kQ / [x√(x² + L²/4)], perpendicular to the rod

Two limits make this result worth knowing:

• Far away (x much larger than L): √(x² + L²/4) ≈ x, so E ≈ kQ/x², a point charge.
• Very long rod (L much larger than x): √(x² + L²/4) ≈ L/2, so E ≈ 2kλ/x = λ/(2πε₀x). This is the field of an infinite line of charge. It falls as 1/x, not 1/x², because there is always more charge further along the line.

The same 1/r field applies outside a long, uniformly charged cylinder, measured from its axis: from outside, the cylinder acts like a line carrying the same charge per unit length. In Topic 8.6 you will get this result again in two lines using Gauss’s law.

Arcs: a semicircle at its centre

Take a thin rod bent into a semicircle of radius R with uniform charge λ = Q/(πR). Put the centre at the origin and let the arc run from θ = 0 to θ = π (the upper half). A piece at angle θ has dq = λR dθ and is a distance R from the centre, for every θ.

Each dE at the centre points away from its piece, in the direction (−cos θ, −sin θ). Pieces at θ and π − θ cancel in x, so only y survives:

E_y = −∫ (kλR dθ / R²) sin θ = −(kλ/R) ∫₀^π sin θ dθ = −2kλ/R

So E = 2kλ/R = 2kQ/(πR²), pointing along the arc’s line of symmetry, away from the arc (for positive charge).

For part of an arc, change the limits. A quarter circle from θ = 0 to π/2 gives E_x = −kλ/R and E_y = −kλ/R, so its field is √2 kλ/R at 45°. A symmetric arc spanning angles π/2 ± θ₀ gives 2kλ sin θ₀ / R. With θ₀ = π/2 this is the semicircle again, which is a good check.

Quick numbers. A semicircle of radius 0.10 m carrying +5.0 × 10⁻⁹ C has λ = 1.59 × 10⁻⁸ C/m and E = 2kλ/R = 2.9 × 10³ N/C at its centre. A point charge Q at distance R would give kQ/R² = 4.5 × 10³ N/C. The arc’s field is smaller by the factor 2/π ≈ 0.64 because the pieces pull in different directions.

The results you are expected to derive

Charge distribution	Point P	Field (size)	Direction (positive charge)
Finite rod, length L, charge Q	On its line, distance d from the near end	kQ / [d(d + L)]	Along the line, away from the rod
Finite rod, length L, charge Q	On its perpendicular bisector, distance x	kQ / [x√(x² + L²/4)]	Perpendicular to the rod, away from it
Infinite line or long cylinder, λ	Distance r from the axis	λ/(2πε₀r) = 2kλ/r	Radially away from the axis
Ring, radius a, charge Q	On its axis, distance z from centre	kQz / (z² + a²)^(3/2)	Along the axis, away from the ring
Semicircle, radius R, λ	At its centre	2kλ/R	Along the symmetry line, away from the arc

The course expects you to integrate only for these shapes and locations. For any other shape (for example, a point off the axis of a ring) the integral is not expected.

Common misconceptions

• Adding magnitudes instead of components. ∫k dq/r² is the answer only if every dE points the same way, as for a point on the rod’s own line. Otherwise resolve first.
• “Treat the object as a point charge at its centre.” This works only far away. Close to a rod, the near end dominates (Worked example 1).
• Leaving two variables in the integral. If you integrate over x, then r and cos θ must also be written in x.
• Forgetting that the distance r changes along a rod but is constant for a ring or an arc measured from its centre.
• Integrating a component that symmetry has already cancelled. You will get zero, but you waste time and may make sign errors.
• “The field is biggest at the centre of a ring.” It is zero there. On the axis it peaks at z = a/√2.
• Using 1/r² for an infinite line. A long line gives E ∝ 1/r.
• Wrong λ for an arc. The arc length of a semicircle is πR, so λ = Q/(πR), not Q/(2πR).

Where this leads

Next, Topic 8.5, Electric Flux, measures how much field passes through a surface. That leads to Gauss’s law (Topic 8.6), which gives the line and cylinder results above much more quickly. In Unit 9 you will use the same “cut it into dq” method to find electric potential, which is easier because potential is a scalar. If the point-charge field still feels shaky, review Topic 8.3, Electric Fields.

Practise now with the practice questions, then use the revision notes and the topic checklist.