Write dq for a uniformly charged rod of linear density λ when integrating over x. [1 point]

College Board AP 2022 (style)-style practice: Section II (FRQ, derivation). A thin rod of length L carries a uniform linear charge density λ. It lies along the x-axis from x = 0 to x = L. (a) Set up an integral for the electric field at a point P on the x-axis a distance d to the right of the far end (x = L). (b) Evaluate the integral. (c) Show that for d L the result reduces to a point charge.

A 6-point FRQ on integrating a 1D distribution. (a) Set-up (2 points): a slice at position x has charge dq = λ\,dx, a distance r = (L + d) - x from P. Its field is dE = k\,dqr^2 = kλ\,dx((L+d)-x)^2, pointing along +x (away from positive charge). (b) Evaluate (3 points): E = kλ_0^L dx((L+d)-x)^2. Let u = (L+d) - x, du = -dx; limits u: L+d → d. E = kλ[1u]_L+d^d = kλ(1d - 1L+d) = kλ Ld(L+d). (c) Limit (1 point): with Q = λ L, for d L, L + d ≈ d, so E ≈ kQd^2, a point charge. Markers reward the correct dq and r, the substitution, and the far-field check.

United StatesPhysics C: Electricity and MagnetismSyllabus dot point

How do you integrate over a continuous charge distribution to find the electric field it produces?

Topic 8.4 Electric Fields of Charge Distributions: set up and evaluate integrals to find the electric field of continuous charge distributions such as rods, rings and arcs.

A calculus-based answer to AP Physics C E&M Topic 8.4, covering linear, surface and volume charge densities, setting up dE integrals, exploiting symmetry, and deriving the field of rods, rings and arcs.

Generated by Claude Opus 4.812 min answerUpdated 2026-06-04

Reviewed by: AI editorial process; not yet individually human-reviewed

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Quick answer

A continuous charge distribution is described by a density: linear $\lambda = \dfrac{dq}{dx}$ (C/m), surface $\sigma = \dfrac{dq}{dA}$ (C per m squared), or volume $\rho = \dfrac{dq}{dV}$ (C per m cubed). To find its field, treat each element $dq$ as a point charge producing $d\vec{E} = \dfrac{k\,dq}{r^2}\hat{r}$ , then integrate over the whole distribution: $\vec{E} = \displaystyle\int d\vec{E}$ . Use symmetry to identify which component survives (the perpendicular components often cancel) and write the surviving component as $dE\cos\theta$ or $dE\sin\theta$ . Standard results follow this way: the on-axis field of a ring is $E = \dfrac{kQx}{(x^2+R^2)^{3/2}}$ , and a finite rod gives a closed-form integral. Always check a far-field limit reduces to a point charge.

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What this topic is asking
Charge densities
The integration recipe
Symmetry kills components
Try this

What this topic is asking

The College Board (Topic 8.4) wants you to find the electric field of a continuous charge distribution by integrating the point-charge field over the distribution. This is the most calculus-heavy electrostatics topic: you write $dq$ , find the field $dE$ it produces, exploit symmetry to keep only the surviving component, and integrate.

Charge densities

The first step in any problem is to write $dq$ in terms of the density and a coordinate, so that the integral runs over that coordinate.

The integration recipe

The field is built from the point-charge result applied to each infinitesimal element:

\vec{E} = \int d\vec{E} = \int \frac{k\,dq}{r^2}\,\hat{r}

The recipe, every time:

Choose an element $dq$ and write it with the density ( $dq = \lambda\,dx$ , etc.).
Find $r$ , the distance from $dq$ to the field point, in terms of the integration variable.
Pick the surviving component using symmetry. Write $dE_\parallel = dE\cos\theta$ (or $\sin\theta$ ), expressing $\cos\theta$ through the geometry.
Integrate over the whole distribution, with limits matching the coordinate.

Symmetry kills components

The single most important move is to spot which component cancels. On the axis of a symmetric body (a ring, a uniformly charged disk, the bisector of a rod), perpendicular components from opposite elements cancel in pairs, so only the on-axis component survives. Writing the surviving component as $dE\cos\theta$ and substituting $\cos\theta$ in terms of the geometry turns a vector integral into a single scalar integral.

Field on the axis of a charged ring

A ring of radius $R$ carries total charge $Q$ spread uniformly. Derive the field a distance $x$ along the axis from the center.

step 1 Set up the element

Take a small arc carrying charge $dq$ . Every element is the same distance $r = \sqrt{x^2 + R^2}$ from the axial field point P.

step 2 Write the field of the element

$dE = \dfrac{k\,dq}{r^2} = \dfrac{k\,dq}{x^2 + R^2}$ , directed from $dq$ toward P along the line joining them.

step 3 Keep only the axial component

By symmetry the components perpendicular to the axis cancel in pairs around the ring. The axial component is $dE_x = dE\cos\theta$ , where $\cos\theta = \dfrac{x}{r} = \dfrac{x}{\sqrt{x^2+R^2}}$ . So $dE_x = \dfrac{k\,dq}{x^2+R^2}\cdot\dfrac{x}{\sqrt{x^2+R^2}} = \dfrac{kx\,dq}{(x^2+R^2)^{3/2}}$ .

step 4 Integrate around the ring

Everything except $dq$ is constant, so $E = \dfrac{kx}{(x^2+R^2)^{3/2}}\displaystyle\int dq = \dfrac{kQx}{(x^2+R^2)^{3/2}}$ , directed along the axis.

Try this

Q1. Write $dq$ for a uniformly charged rod of linear density $\lambda$ when integrating over $x$ . [1 point]

Cue. $dq = \lambda\,dx$ .

Q2. State where, along the axis of a charged ring of radius $R$ , the field reaches its maximum. [1 point]

Cue. At $x = \dfrac{R}{\sqrt{2}}$ (found by setting $\dfrac{dE}{dx} = 0$ ).

Exam-style practice questions

Practice questions written in the style of College Board exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

AP 2023 (style)1 marksSection I (multiple choice). The electric field on the axis of a uniformly charged ring of radius

R

and total charge

Q

, at distance

x

from the center, is

E = \dfrac{kQx}{(x^2+R^2)^{3/2}}

. At the center of the ring (

x = 0

), the field is (A)

\dfrac{kQ}{R^2}

(B) maximum (C) zero (D) infinite. Justify your reasoning.

Show worked answer →

A 1-point MCQ on the ring field. The answer is (C).

Setting $x = 0$ in $E = \dfrac{kQx}{(x^2+R^2)^{3/2}}$ gives $E = 0$ . Physically, each element of the ring is matched by an element diametrically opposite whose field cancels it at the center, so the net field is zero there. The trap is (A): that is the point-charge formula, which does not apply on a ring.

AP 2022 (style)6 marksSection II (FRQ, derivation). A thin rod of length

L

carries a uniform linear charge density

\lambda

. It lies along the x-axis from

x = 0

x = L

. (a) Set up an integral for the electric field at a point P on the x-axis a distance

d

to the right of the far end (

x = L

). (b) Evaluate the integral. (c) Show that for

d \gg L

the result reduces to a point charge.

Show worked answer →

A 6-point FRQ on integrating a 1D distribution.

(a) Set-up (2 points): a slice at position $x$ has charge $dq = \lambda\,dx$ , a distance $r = (L + d) - x$ from P. Its field is $dE = \dfrac{k\,dq}{r^2} = \dfrac{k\lambda\,dx}{((L+d)-x)^2}$ , pointing along $+x$ (away from positive charge).
(b) Evaluate (3 points): $E = k\lambda\displaystyle\int_0^L \dfrac{dx}{((L+d)-x)^2}$ . Let $u = (L+d) - x$ , $du = -dx$ ; limits $u: L+d \to d$ . $E = k\lambda\Big[\dfrac{1}{u}\Big]_{L+d}^{d} = k\lambda\left(\dfrac{1}{d} - \dfrac{1}{L+d}\right) = \dfrac{k\lambda L}{d(L+d)}$ .
(c) Limit (1 point): with $Q = \lambda L$ , for $d \gg L$ , $L + d \approx d$ , so $E \approx \dfrac{kQ}{d^2}$ , a point charge.

Markers reward the correct $dq$ and $r$ , the substitution, and the far-field check.

Related dot points

Sources & how we know this

AP Physics C: Electricity and Magnetism Course and Exam Description — College Board (2024)