Electric fields of charge distributions

Superposition and Integration Strategy

Superposition means the total electric field at any point equals the vector sum of fields from every charge element.

For a continuous charge distribution, the sum becomes an integral. A small charge element dq produces a small field:

$$d\vec{E}=\frac{1}{4\pi {\epsilon }_{0}}\frac{dq}{{r}^{2}} \hat{r}$$

In this expression:

• The symbol dq represents an infinitesimal piece of the total charge.
• The vector r̂ points from dq toward the field point.
• The distance r is measured from dq to the field point.

To find the total field, integrate over the entire distribution:

$$\vec{E}=\int d\vec{E}=\frac{1}{4\pi {\epsilon }_{0}}\int \frac{dq}{{r}^{2}} \hat{r}$$

Every problem follows the same five-step process:

1. Choose a coordinate system that matches the geometry.
2. Express dq in terms of a spatial variable using charge density.
3. Write dE components using geometry (resolve into x, y, or parallel/perpendicular).
4. Identify symmetry cancellations — components that integrate to zero.
5. Integrate the surviving component(s) over the distribution.

Charge density connects dq to geometry:

For uniform distributions, λ, σ, or ρ is constant and comes outside the integral.

MisconceptionStudents often forget that r in the integrand changes with position along the distribution. Only when every element is the same distance from the field point can r come outside the integral.

Exam TipAlways express r in terms of your integration variable before integrating.

Examiner InsightFRQs typically say “derive an expression for E.” Start by writing the fundamental expression dE = (1/4πε₀)(dq/r²) and then state your superposition integral.

Exam TipBegin every derivation by naming the principle — superposition and Coulomb’s law.

Symmetry Simplifications

Symmetry considerations reduce vector integrals to scalar ones.

Before computing, inspect the charge distribution for geometric symmetry. If the distribution has mirror symmetry about a line or plane passing through the field point, the field component perpendicular to that symmetry axis cancels. Only the component along the symmetry axis survives.

Stating the cancellation explicitly earns credit on FRQs. Write: “By symmetry, the [perpendicular/transverse] components cancel, so only the [axial/radial] component survives.”

Infinite Line of Charge

The electric field from an infinitely long uniformly charged wire points radially outward (or inward for negative charge) and depends only on the perpendicular distance from the wire.

Key Equations

Electric field of an infinite line charge:

$$E=\frac{\lambda }{2\pi {\epsilon }_{0} r}$$

Variables:

• λ = linear charge density (C/m)
• r = perpendicular distance from the wire (m)
• ε₀ = permittivity of free space (C²/(N·m²))

SI unit: N/C (or equivalently V/m)

ProportionalityE is directly proportional to λ and inversely proportional to r. Doubling the distance halves the field.

Reference sheet status: Derived equation — know how to obtain from Coulomb’s law integration or Gauss’s law.

Derivation: Electric Field of an Infinite Line Charge (Integration Method)

Starting point: Coulomb’s law for a charge element and the principle of superposition.

$$dE=\frac{1}{4\pi {\epsilon }_{0}}\frac{dq}{{s}^{2}}$$

where s is the distance from dq to the field point.

Step 1 — Set up coordinates

Place the wire along the y-axis. The field point P is at perpendicular distance r from the wire on the x-axis. A small element dy at position y along the wire carries charge dq = λ dy.

Step 2 — Express geometry in terms of the integration variable y

The distance from the element to P is:

$$s=\sqrt{{r}^{2}+{y}^{2}}$$

The angle θ between the line from dq to P and the x-axis satisfies:

$$cos\theta =\frac{r}{\sqrt{{r}^{2}+{y}^{2}}}$$

Step 3 — Apply symmetry

By symmetry, the y-components of the field cancel in pairs (elements at +y and −y). Only the x-component (perpendicular to the wire) survives:

$$d{E}_{x}=\frac{1}{4\pi {\epsilon }_{0}}\frac{\lambda dy}{{r}^{2}+{y}^{2}}\cdot \frac{r}{\sqrt{{r}^{2}+{y}^{2}}}$$

Step 4 — Set up the integral

$$E=\frac{\lambda r}{4\pi {\epsilon }_{0}}\int_{-\infty }^{+\infty }\frac{dy}{({r}^{2}+{y}^{2}{)}^{3/2}}$$

The limits are −∞ to +∞ because the wire is infinite.

Step 5 — Evaluate the integral

Use the standard result:

$$\int_{-\infty }^{+\infty }\frac{dy}{({r}^{2}+{y}^{2}{)}^{3/2}}=\frac{2}{{r}^{2}}$$

Substituting:

$$E=\frac{\lambda r}{4\pi {\epsilon }_{0}}\cdot \frac{2}{{r}^{2}}$$

Step 6 — Simplify

$$E=\frac{\lambda }{2\pi {\epsilon }_{0} r}$$

Result

$$\boxed{E=\frac{\lambda }{2\pi {\epsilon }_{0} r}}$$

The field decreases as 1/r, not 1/r², because the source is one-dimensional rather than a point.

Worked Example: Field Near a Long Charged Wire

Scenario

A long straight wire carries uniform charge density λ = 5.00 × 10⁻⁹ C/m. Find the electric field magnitude at a perpendicular distance of 0.200 m from the wire.

Equation used

$$E=\frac{\lambda }{2\pi {\epsilon }_{0} r}$$

Given

$$\lambda =5.00\times {10}^{-9}\text{ C/m}$$

$$r=0.200\text{ m}$$

$${\epsilon }_{0}=8.85\times {10}^{-12}{\text{ C}}^{2}/({\text{N}\cdot\text{m}}^{2})$$

Working

$$E=\frac{5.00\times {10}^{-9}}{2\pi (8.85\times {10}^{-12})(0.200)}$$

$$E=\frac{5.00\times {10}^{-9}}{1.112\times {10}^{-11}}$$

$$E=450\text{ N/C}$$

Answer

$$E\approx 450\text{ N/C}$$

Interpretation

The field points radially away from the wire. Doubling the distance to 0.400 m would halve the field to about 225 N/C.

Finite Line of Charge

A finite wire of length L requires integration with finite limits, and the result depends on the position of the field point relative to the wire.

BoundaryThe AP exam tests the finite wire at a point collinear with the wire and at a point along the perpendicular bisector. Other field-point locations are not required.

Know these cases for the AP exam.

Case 1 — Field point on the perpendicular bisector

Place a wire of total charge Q and length L along the y-axis, centered at the origin. The field point P is on the x-axis at distance d from the center.

Derivation: E-Field on Perpendicular Bisector of a Finite Wire

Starting point: Coulomb’s law for a charge element and superposition.

$$dE=\frac{1}{4\pi {\epsilon }_{0}}\frac{dq}{{s}^{2}}$$

Step 1 — Express dq and geometry

The uniform linear charge density is λ = Q/L. An element at position y carries dq = λ dy. The distance from this element to P is:

$$s=\sqrt{{d}^{2}+{y}^{2}}$$

Step 2 — Apply symmetry

P lies on the perpendicular bisector, so elements at +y and −y produce field components parallel to the wire that cancel. Only the component along the x-axis (perpendicular to the wire) survives:

$$d{E}_{x}=\frac{\lambda dy}{4\pi {\epsilon }_{0}({d}^{2}+{y}^{2})}\cdot \frac{d}{\sqrt{{d}^{2}+{y}^{2}}}$$

Step 3 — Set up the integral

Integrate from −L/2 to +L/2:

$$E=\frac{\lambda d}{4\pi {\epsilon }_{0}}\int_{-L/2}^{+L/2}\frac{dy}{({d}^{2}+{y}^{2}{)}^{3/2}}$$

Step 4 — Evaluate the integral

Using the standard result:

$$\int_{-L/2}^{+L/2}\frac{dy}{({d}^{2}+{y}^{2}{)}^{3/2}}=\frac{2}{{d}^{2}}\cdot \frac{L/2}{\sqrt{{d}^{2}+(L/2{)}^{2}}}$$

This simplifies to:

$$\int_{-L/2}^{+L/2}\frac{dy}{({d}^{2}+{y}^{2}{)}^{3/2}}=\frac{L}{{d}^{2}\sqrt{{d}^{2}+{L}^{2}/4}}$$

Step 5 — Substitute and simplify

$$E=\frac{\lambda d}{4\pi {\epsilon }_{0}}\cdot \frac{L}{{d}^{2}\sqrt{{d}^{2}+{L}^{2}/4}}$$

$$E=\frac{\lambda L}{4\pi {\epsilon }_{0} d\sqrt{{d}^{2}+{L}^{2}/4}}$$

Since Q = λL:

$$\boxed{E=\frac{Q}{4\pi {\epsilon }_{0} d\sqrt{{d}^{2}+{L}^{2}/4}}}$$

Result

The field points perpendicular to the wire, away from it (for positive Q).

Limiting behavior: When d ≫ L, the term L²/4 is negligible, and E → Q/(4πε₀d²) — the point-charge result. When L → ∞ (with λ held constant), the expression reduces to λ/(2πε₀d).

Case 2 — Field point collinear with the wire (on the axis extended)

Place a wire of length L along the x-axis from x = a to x = a + L. The field point P is at the origin.

Derivation: E-Field at a Point Collinear with a Finite Wire

Starting point: Coulomb’s law for a charge element and superposition.

$$dE=\frac{1}{4\pi {\epsilon }_{0}}\frac{dq}{{x}^{2}}$$

Step 1 — Express dq

A small element at position x carries dq = λ dx. The distance from this element to the origin is simply x.

Step 2 — Apply symmetry

All elements lie along the same line as P, so every dE contribution points along the x-axis. No cancellation occurs — all components add directly.

Step 3 — Set up the integral

$$E=\frac{\lambda }{4\pi {\epsilon }_{0}}\int_{a}^{a+L}\frac{dx}{{x}^{2}}$$

The limits run from x = a (near end of wire) to x = a + L (far end).

Step 4 — Evaluate the integral

$$\int_{a}^{a+L}\frac{dx}{{x}^{2}}={\left[-\frac{1}{x}\right]}_{a}^{a+L}=-\frac{1}{a+L}+\frac{1}{a}=\frac{L}{a(a+L)}$$

Step 5 — Substitute

$$\boxed{E=\frac{\lambda L}{4\pi {\epsilon }_{0} a(a+L)}=\frac{Q}{4\pi {\epsilon }_{0} a(a+L)}}$$

Result

The field points along the axis, directed away from the wire (for positive charge), toward the wire for negative charge.

Limiting behavior: When a ≫ L, the product a(a + L) ≈ a², and E → Q/(4πε₀a²) — the point-charge result.

Thin Ring of Charge Along Its Axis

The electric field on the axis of a thin ring of charge exploits the full rotational symmetry of the ring.

Key Equations

Electric field on the axis of a uniformly charged ring:

$$E=\frac{Qx}{4\pi {\epsilon }_{0}({x}^{2}+{R}^{2}{)}^{3/2}}$$

Variables:

• Q = total charge on the ring (C)
• x = distance from the center of the ring along the axis (m)
• R = radius of the ring (m)
• ε₀ = permittivity of free space (C²/(N·m²))

SI unit: N/C

ProportionalityAt large distances (x ≫ R), E → Q/(4πε₀x²), the point-charge result. At the center (x = 0), E = 0 by symmetry.

Reference sheet status: Derived equation — know how to obtain from Coulomb’s law integration.

Derivation: Electric Field on the Axis of a Charged Ring

Starting point: Coulomb’s law for a charge element and superposition.

$$dE=\frac{1}{4\pi {\epsilon }_{0}}\frac{dq}{{s}^{2}}$$

Step 1 — Express geometry

Every element dq on the ring is the same distance from an axial point P:

$$s=\sqrt{{x}^{2}+{R}^{2}}$$

This is constant for all elements, so s comes outside the integral.

Step 2 — Apply symmetry

Each dq on the ring has a diametrically opposite element. Their perpendicular (radial) components cancel. Only the axial component survives:

$$d{E}_{x}=dEcos\theta =\frac{dq}{4\pi {\epsilon }_{0}({x}^{2}+{R}^{2})}\cdot \frac{x}{\sqrt{{x}^{2}+{R}^{2}}}$$

Step 3 — Set up the integral

Since x, R, and s are the same for every element:

$${E}_{x}=\frac{x}{4\pi {\epsilon }_{0}({x}^{2}+{R}^{2}{)}^{3/2}}\int dq$$

Step 4 — Evaluate the integral

$$\int dq=Q$$

Step 5 — Write the result

$$\boxed{E=\frac{Qx}{4\pi {\epsilon }_{0}({x}^{2}+{R}^{2}{)}^{3/2}}}$$

The field points along the axis, away from the ring for positive Q.

The field is zero at x = 0 and approaches the point-charge result when x ≫ R. The maximum field occurs at x = R/√2.

Worked Example: Field on Ring Axis

Scenario

A thin ring of radius 0.10 m carries total charge 6.0 × 10⁻⁸ C. Find the electric field on the axis at x = 0.15 m from the center.

Equation used

$$E=\frac{Qx}{4\pi {\epsilon }_{0}({x}^{2}+{R}^{2}{)}^{3/2}}$$

Given

$$Q=6.0\times {10}^{-8}\text{ C}$$

$$x=0.15\text{ m}$$

$$R=0.10\text{ m}$$

Working

$${x}^{2}+{R}^{2}=0.0225+0.0100=0.0325$$

$$({x}^{2}+{R}^{2}{)}^{3/2}=(0.0325{)}^{3/2}=5.86\times {10}^{-3}$$

$$E=\frac{(6.0\times {10}^{-8})(0.15)}{4\pi (8.85\times {10}^{-12})(5.86\times {10}^{-3})}$$

$$E=\frac{9.0\times {10}^{-9}}{6.53\times {10}^{-13}}$$

$$E\approx 1.38\times {10}^{4}\text{ N/C}$$

Answer

$$E\approx 1.38\times {10}^{4}\text{ N/C, directed along the axis away from the ring.}$$

Interpretation

The field is substantial because the point is fairly close to the ring. Moving much farther away (x ≫ R) would yield the point-charge approximation.

Semicircular Arc at Its Center

The electric field at the center of a semicircular arc of charge requires integrating with respect to the angle subtended.

Key Equations

Electric field at the center of a uniformly charged semicircular arc:

$$E=\frac{\lambda }{2\pi {\epsilon }_{0}R}$$

Variables:

• λ = linear charge density (C/m)
• R = radius of the arc (m)
• ε₀ = permittivity of free space (C²/(N·m²))

SI unit: N/C

Reference sheet status: Derived equation — know how to obtain from Coulomb’s law integration.

Calculus form: The integral is performed over the angle θ from 0 to π.

Derivation: Electric Field at Center of a Semicircular Arc

Starting point: Coulomb’s law for a charge element and superposition.

$$dE=\frac{1}{4\pi {\epsilon }_{0}}\frac{dq}{{R}^{2}}$$

Step 1 — Express dq in terms of angle

Place the semicircle in the x-y plane, opening to the right. An element subtending angle dθ has arc length dℓ = R dθ, so:

$$dq=\lambda R d\theta $$

Every element is at the same distance R from the center.

Step 2 — Apply symmetry

Orient the semicircle symmetrically about the x-axis, running from θ = −π/2 to θ = +π/2 (measuring from the positive x-axis). Elements at +θ and −θ produce y-components that cancel. Only the x-component survives:

$$d{E}_{x}=\frac{\lambda R d\theta }{4\pi {\epsilon }_{0} {R}^{2}}cos\theta =\frac{\lambda cos\theta d\theta }{4\pi {\epsilon }_{0} R}$$

Step 3 — Set up the integral

$$E=\frac{\lambda }{4\pi {\epsilon }_{0} R}\int_{-\pi /2}^{+\pi /2}cos\theta d\theta $$

Step 4 — Evaluate the integral

$$\int_{-\pi /2}^{+\pi /2}cos\theta d\theta =[sin\theta {]}_{-\pi /2}^{+\pi /2}=1-(-1)=2$$

Step 5 — Substitute

$$E=\frac{\lambda }{4\pi {\epsilon }_{0} R}\cdot 2$$

$$\boxed{E=\frac{\lambda }{2\pi {\epsilon }_{0} R}}$$

Result

The field at the center points along the axis of symmetry, directed away from the arc for positive charge (toward the open side).

MisconceptionStudents sometimes integrate from 0 to π instead of −π/2 to +π/2 and lose track of which component survives. The orientation determines which trigonometric function (sin or cos) gives the surviving component.

Exam TipSketch your angle convention first and identify which component survives by symmetry before writing the integral.

Worked Example: Semicircular Arc Field

Scenario

A thin semicircular arc of radius 0.050 m carries uniform linear charge density λ = 3.0 × 10⁻⁹ C/m. Find the electric field at the center of curvature.

Equation used

$$E=\frac{\lambda }{2\pi {\epsilon }_{0} R}$$

Given

$$\lambda =3.0\times {10}^{-9}\text{ C/m}$$

$$R=0.050\text{ m}$$

Working

$$E=\frac{3.0\times {10}^{-9}}{2\pi (8.85\times {10}^{-12})(0.050)}$$

$$E=\frac{3.0\times {10}^{-9}}{2.78\times {10}^{-12}}$$

$$E\approx 1080\text{ N/C}$$

Answer

$$E\approx 1.08\times {10}^{3}\text{ N/C, directed along the symmetry axis toward the open side of the arc.}$$

Interpretation

The field depends on λ/R. A tighter arc (smaller R) with the same charge density produces a stronger field.

Uniformly Charged Cylinder (Infinite)

The electric field of an infinitely long uniformly charged cylinder is most efficiently found using Gauss’s law, but understanding the integration approach connects to the superposition method.

For an insulating cylinder of radius a with uniform volume charge density ρ:

Key Equations

Outside the cylinder (r > a):

$$E=\frac{\rho {a}^{2}}{2{\epsilon }_{0} r}=\frac{\lambda }{2\pi {\epsilon }_{0} r}$$

Inside the cylinder (r < a):

$$E=\frac{\rho r}{2{\epsilon }_{0}}$$

Variables:

• ρ = volume charge density (C/m³)
• a = radius of the cylinder (m)
• r = distance from the central axis (m)
• λ = charge per unit length (C/m), related to ρ by λ = ρπa²
• ε₀ = permittivity of free space (C²/(N·m²))

SI unit: N/C

ProportionalityOutside, E ∝ 1/r (same as an infinite line). Inside, E ∝ r (field increases linearly from zero at the center).

Reference sheet status: Derived equation — obtain from Gauss’s law.

Derivation: E-Field of an Infinite Charged Cylinder Using Gauss’s Law

Starting point: Gauss’s law.

$$\oint \vec{E}\cdot d\vec{A}=\frac{{Q}_{\text{enc}}}{{\epsilon }_{0}}$$

Case: r > a (outside the cylinder)

Step 1 — Choose the Gaussian surface

A coaxial cylindrical Gaussian surface of radius r and length L, with r > a. This matches the cylindrical symmetry of the charge distribution.

Step 2 — Apply the symmetry argument

On the curved surface, E⃗ is perpendicular to the surface and constant in magnitude, so E⃗·dA⃗ = E dA. On the flat end caps, E⃗ is parallel to the surface, so E⃗·dA⃗ = 0.

Step 3 — Simplify the integral

$$\oint \vec{E}\cdot d\vec{A}=E(2\pi rL)$$

Step 4 — Calculate the enclosed charge

The Gaussian surface encloses the full cross-section of the cylinder over length L:

$${Q}_{\text{enc}}=\rho \pi {a}^{2}L$$

Step 5 — Solve for the field

$$E(2\pi rL)=\frac{\rho \pi {a}^{2}L}{{\epsilon }_{0}}$$

$$\boxed{E=\frac{\rho {a}^{2}}{2{\epsilon }_{0} r} (r>a)}$$

Case: r < a (inside the cylinder)

Step 1 — Choose the Gaussian surface

A coaxial cylindrical Gaussian surface of radius r and length L, with r < a.

Step 2 — Apply the symmetry argument

Same as above: E⃗ is perpendicular and constant on the curved surface; zero flux through end caps.

Step 3 — Simplify the integral

$$\oint \vec{E}\cdot d\vec{A}=E(2\pi rL)$$

Step 4 — Calculate the enclosed charge

Only the charge within radius r is enclosed:

$${Q}_{\text{enc}}=\rho \pi {r}^{2}L$$

Step 5 — Solve for the field

$$E(2\pi rL)=\frac{\rho \pi {r}^{2}L}{{\epsilon }_{0}}$$

$$\boxed{E=\frac{\rho r}{2{\epsilon }_{0}} (r<a)}$$

At the surface (r = a), both expressions give E = ρa/(2ε₀), confirming continuity.