AP Physics C: Electricity and Magnetism · Unit 8 · Topic 2 of 4 · 6 min lesson

Continuous Charge Distributions

A short lesson with diagrams, a worked example, a problem to try, and a quick check. Then quiz yourself on 4 practice questions.

Quiz this topic · 4 questions

The big idea

For a spread-out charge, chop it into tiny pieces dq, each acting like a point charge: dE = k dq/r². Write dq in terms of position (dq = λ dx for a line), use symmetry to cancel components, and integrate what’s left.

1. Symmetry first

Every piece dq has a partner dq′ on the opposite side of the ring. Their sideways fields cancel.

dq = λ dx, σ dA, ρ dVline, surface, volume densities

dE = k dq / r²each piece

Ring, on axisE = kQx / (x² + R²)^3/2

Rod, on its axis, d past the endE = kQ / [d(d + L)]

Total chargeQ = ∫λ dx

2. Checking your answer

Limits: far away, every distribution should look like a point charge, E → kQ/r². The rod formula does: for d ≫ L, kQ/[d(d + L)] → kQ/d².

At the center of a ring, every piece is canceled by its opposite, so E = 0.

Non-uniform charge: if λ = αx, the total charge is ∫₀^L αx dx = αL²/2, and the field integral picks up the extra x too.

Units check: α has units of C/m² there, so αL² is a charge.

3. Example, then try it

Example

A rod lies from x = 0 to L = 2.0 m with charge density λ = αx², where α = 3.0 µC/m³. Find its total charge.

Q = ∫₀^L αx² dx = αL³/3 = (3.0 µC/m³)(8.0 m³) ÷ 3 = 8.0 µC.

Try it · work it on paper first

A rod of length L = 0.20 m carries Q = 4.0 nC spread uniformly. Find the field on its axis, d = 0.10 m beyond one end, by integrating.

Set up: put the rod from x = 0 to L and the point at x = L + d. A piece dq = λ dx at x is (L + d − x) away, so dE = kλ dx / (L + d − x)².

Integrate: E = kλ[1/(L + d − x)]₀^L = kλ[1/d − 1/(L + d)] = kλL / [d(L + d)] = kQ / [d(d + L)].

Numbers: E = (9.0 × 10⁹)(4.0 × 10⁻⁹) ÷ [(0.10)(0.30)] = 36 ÷ 0.030 = 1200 N/C, pointing away from the rod.

Check: a point charge at the rod’s center (0.20 m away) would give 36 ÷ 0.040 = 900 N/C. The near end makes the rod’s field stronger, as expected.

Answer: E = kQ/[d(d + L)] = 1200 N/C.

4. Common mistakes

Integrating magnitudes without checking direction.Only integrate components that don’t cancel.

Using the wrong distance in dE.r is from the piece dq to the field point, and it changes with x.

Forgetting that λ might vary.Use dq = λ(x) dx.

Check yourself

A rod of length L carries charge Q. Far from the rod along its axis (d ≫ L), what does its field approach?

kQ/[d(d + L)] → kQ/d² when d ≫ L: from far away, the rod looks like a point charge.

Ready to practice?

4 questions on Continuous Charge Distributions, each with a hint and an explanation for every wrong answer.

Quiz this topic All of Unit 8