AP Physics C: E&M Unit 1 — Electrostatics, Gauss’s Law, and Electric Potential (Teach-From-Scratch Notes)

Electric charge and Coulomb’s law

What electric charge is

Electric charge is a fundamental property of matter that causes objects to experience electric forces. In electrostatics (charges at rest), charge comes in two types—positive and negative—and the key behaviors are:

  • Like charges repel.
  • Opposite charges attract.

Charge is conserved: in an isolated system, the net charge cannot change. You can move charge from one object to another (for example, by rubbing or contact), but you do not “use up” charge.

In AP Physics C: E&M, you typically treat charge as quantized in units of the elementary charge e, but most macroscopic problems use total charge Q as a continuous quantity.

Why charge matters

Charge is the “source” for electric fields and electric potentials. Almost everything you do later in this unit—computing \vec{E}, using Gauss’s law, finding V—starts from knowing how charge produces forces and fields.

Coulomb’s law (force between point charges)

For two point charges q_1 and q_2 separated by distance r, the magnitude of the electrostatic force is

F = k\frac{|q_1 q_2|}{r^2}

where

k = \frac{1}{4\pi\epsilon_0}

and \epsilon_0 is the permittivity of free space.

A common numerical value you may use is

k \approx 8.99\times 10^9\ \text{N m}^2/\text{C}^2

Direction matters: the force acts along the line connecting the charges. A clean vector form (force on charge 2 due to charge 1) is

\vec{F}_{2\leftarrow 1} = k\frac{q_1 q_2}{r^2}\hat{r}_{1\to 2}

where \hat{r}_{1\to 2} is the unit vector pointing from charge 1 to charge 2. Notice the sign is handled by q_1 q_2: if the product is positive, the force on 2 points away from 1 (repulsion); if negative, it points toward 1 (attraction).

Superposition (how multiple charges combine)

Electrostatic force obeys superposition: the net force on a charge is the vector sum of the individual forces from all other charges.

This matters because most realistic configurations involve more than two charges. You do not invent a new force law—you apply Coulomb’s law repeatedly and add vectors.

Worked example: net force from two charges on a third

Suppose q is at the origin. A charge +Q sits on the positive x-axis at x=a and an identical charge +Q sits on the positive y-axis at y=a. Find the net force on q.

1) Force from the charge at x=a has magnitude

F_x = k\frac{|qQ|}{a^2}

and points along the x-axis (direction depends on sign of q).

2) Force from the charge at y=a has the same magnitude

F_y = k\frac{|qQ|}{a^2}

and points along the y-axis.

3) These forces are perpendicular, so the net magnitude is

F_{\text{net}} = \sqrt{F_x^2 + F_y^2} = \sqrt{2}\,k\frac{|qQ|}{a^2}

Direction: 45 degrees between the negative axes if q is negative (attraction toward both charges), or between the positive axes if q is positive (repulsion from both).

What commonly goes wrong

A frequent mistake is to add magnitudes instead of vectors, especially when geometry is not symmetric. Another is to forget that forces come in action-reaction pairs (Newton’s third law): the force on 1 due to 2 is equal in magnitude and opposite in direction to the force on 2 due to 1.

Exam Focus
  • Typical question patterns:
    • Compute net force on a charge from multiple point charges using symmetry and vector components.
    • Determine how force changes when distances are scaled (inverse-square reasoning).
    • Infer direction of force from signs and geometry.
  • Common mistakes:
    • Treating force as scalar instead of vector (missing components).
    • Using r instead of r^2, or confusing separation distance with coordinate values.
    • Mixing up the unit vector direction (from source to field point vs. reverse).

Electric field: concept, definition, and field lines

What an electric field is

The electric field is a vector field that assigns, to every point in space, the force per unit positive test charge that would be experienced at that point. Formally,

\vec{E} = \frac{\vec{F}}{q_{\text{test}}}

with the understanding that q_{\text{test}} is small enough not to disturb the source charges.

Why the field concept matters

Forces directly between charges are fine for a few particles, but fields scale better: once you know \vec{E}(\vec{r}), you can find the force on any charge placed there using

\vec{F} = q\vec{E}

The field idea also connects naturally to Gauss’s law and to electric potential, which are often easier ways to compute or reason about electric effects than summing forces.

Electric field of a point charge

For a point charge q, the electric field at distance r points radially and has magnitude

E = k\frac{|q|}{r^2}

A compact vector form is

\vec{E} = k\frac{q}{r^2}\hat{r}

where \hat{r} points from the charge to the field point.

Superposition for fields

Fields also obey superposition:

\vec{E}_{\text{net}} = \sum_i \vec{E}_i

This is extremely important: you nearly always compute net field by vector addition of contributions.

Field lines (how to visualize without doing calculus)

Electric field lines are a visualization tool:

  • The tangent to a field line gives the direction of \vec{E}.
  • The density of lines indicates relative magnitude of \vec{E}.
  • Field lines start on positive charge and end on negative charge (or at infinity).
  • Field lines never cross (because that would imply two directions for \vec{E} at one point).

Field lines are not “real strings,” and you should not count them as a quantitative method unless the problem explicitly sets a line-counting convention.

Worked example: electric field on the perpendicular bisector

Two equal positive charges +Q are located at x=+a and x=-a. Find \vec{E} on the y-axis at point P=(0,y).

1) By symmetry, the horizontal components cancel (equal magnitude, opposite directions).

2) The vertical components add.

Distance from each charge to P is

r = \sqrt{a^2 + y^2}

Magnitude of each field is

E_1 = k\frac{Q}{r^2}

The vertical component from one charge is

E_{1y} = E_1\frac{y}{r} = k\frac{Q}{r^2}\frac{y}{r} = k\frac{Qy}{r^3}

So the total is

E_y = 2k\frac{Qy}{(a^2+y^2)^{3/2}}

Direction is upward (positive y).

What commonly goes wrong

Students often confuse the direction of \vec{E} with the direction of force on an arbitrary charge. Remember: \vec{E} points the way a **positive** test charge would accelerate. If the actual charge is negative, \vec{F} = q\vec{E} points opposite \vec{E}.

Exam Focus
  • Typical question patterns:
    • Use symmetry to cancel components (midpoints, perpendicular bisectors, axes).
    • Compute \vec{E} from one or more point charges and then find force on a given charge.
    • Interpret field-line diagrams conceptually (direction, relative strength).
  • Common mistakes:
    • Forgetting to include vector directions or cancelations.
    • Using the wrong distance r in inverse-square expressions.
    • Confusing \vec{E} direction with force direction on a negative charge.

Continuous charge distributions and setting up integrals

Why you need calculus here

AP Physics C assumes you can model charge not only as point particles but also as spread continuously along a line, over a surface, or throughout a volume. When charge is continuous, you replace a sum of many small contributions with an integral.

The key move is always the same:

1) Break the charge distribution into tiny pieces dq.
2) Write the tiny field contribution d\vec{E} (or potential contribution dV) from dq.
3) Add them up using an integral, carefully handling geometry and direction.

Charge density definitions

To describe “how much charge per size,” you use charge densities:

  • Linear charge density \lambda for charge along a curve:

\lambda = \frac{dq}{dl}

  • Surface charge density \sigma for charge spread over a surface:

\sigma = \frac{dq}{dA}

  • Volume charge density \rho for charge throughout a volume:

\rho = \frac{dq}{dV}

Then

dq = \lambda\,dl

dq = \sigma\,dA

dq = \rho\,dV

Electric field from a continuous distribution (general idea)

For a small charge element dq at distance r from your field point, the magnitude of the contribution is

dE = k\frac{dq}{r^2}

Direction is from the element toward the field point (for positive dq), so in vectors you often write

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

Then you integrate. The hard part is usually not calculus—it is geometry: expressing r and the direction components in terms of the integration variable.

Strategy: use symmetry before integrating

Before writing an integral, ask: “Which components cancel by symmetry?” Often you only need to integrate one component.

Common symmetries:

  • A uniformly charged ring: transverse components cancel, only axial component remains.
  • An infinite line: radial symmetry implies the field points radially and depends only on distance from the line.
  • A uniformly charged disk: on-axis symmetry reduces to a 1D integral.

Worked example: field on axis of a uniformly charged ring

A ring of radius R carries total charge Q uniformly. Find the electric field on the axis a distance z from the center.

1) By symmetry, horizontal components cancel; only the axial component survives.

2) Every element dq is the same distance from the point:

r = \sqrt{R^2 + z^2}

3) Contribution magnitude:

dE = k\frac{dq}{r^2}

4) The axial component is dE\cos\theta where

\cos\theta = \frac{z}{r}

So

dE_z = k\frac{dq}{r^2}\frac{z}{r} = k\frac{z}{r^3}dq

5) Integrate around the ring. Since r and z are constant for all elements,

E_z = k\frac{z}{r^3}\int dq = k\frac{z}{(R^2+z^2)^{3/2}}Q

Direction is along the axis, away from the ring if Q>0.

What commonly goes wrong

A classic error is to integrate vectors without resolving components, leading to incorrect cancelations. Another is to treat r as constant when it depends on position along the distribution.

Exam Focus
  • Typical question patterns:
    • Set up (and sometimes evaluate) an integral for \vec{E} from a line, ring, or disk.
    • Use symmetry to argue that certain components cancel.
    • Translate between Q and \lambda, \sigma, \rho.
  • Common mistakes:
    • Using dq incorrectly (for example, writing dq=\lambda dx when the length element is not dx).
    • Forgetting geometry factors like \cos\theta when extracting a component.
    • Assuming cancelation where symmetry does not actually apply.

Electric flux: linking fields to “how much passes through” a surface

What electric flux is

Electric flux measures how much electric field “passes through” a surface. For a small patch of area dA with area vector d\vec{A} (magnitude dA, direction normal to the surface), the differential flux is

d\Phi_E = \vec{E}\cdot d\vec{A}

For a whole surface,

\Phi_E = \int \vec{E}\cdot d\vec{A}

If \vec{E} is uniform over a flat area and makes angle \theta with the outward normal, then

\Phi_E = EA\cos\theta

Why flux matters

Flux is the bridge between electric fields and charge via Gauss’s law. The dot product is crucial: only the component of the field perpendicular to the surface contributes. Tangential field “slides along” the surface and produces no flux.

Flux is not “field times area” automatically—orientation matters.

Closed surfaces and sign conventions

For a closed surface (a “Gaussian surface”), the area vector points outward by convention. Flux can be positive (net field lines leaving), negative (net entering), or zero.

A useful intuition: net flux through a closed surface depends on how much charge is inside, not on the detailed shape (that is Gauss’s law, coming next).

Worked example: flux through a sphere around a point charge

If a point charge q is at the center of a sphere of radius R, the field has constant magnitude on the surface and is radial:

E = k\frac{|q|}{R^2}

Everywhere, \vec{E} is parallel to d\vec{A}, so \vec{E}\cdot d\vec{A}=E\,dA. Then

\Phi_E = \int E\,dA = E\int dA = E(4\pi R^2)

Substitute E:

\Phi_E = \left(k\frac{q}{R^2}\right)(4\pi R^2) = 4\pi k q

Using k=1/(4\pi\epsilon_0) gives

\Phi_E = \frac{q}{\epsilon_0}

This result does not depend on R—a preview of Gauss’s law.

What commonly goes wrong

Many students mix up “flux through a closed surface” with “field at a point.” Flux is a single number for the entire surface; field is a vector defined everywhere.

Exam Focus
  • Typical question patterns:
    • Compute flux through a plane or curved surface with symmetry (uniform field cases).
    • Determine the sign of flux based on field direction relative to outward normal.
    • Use flux reasoning to prepare for Gauss’s law (qualitative arguments).
  • Common mistakes:
    • Using EA when the field is not perpendicular (forgetting \cos\theta).
    • Forgetting that closed-surface normals are outward.
    • Treating flux as a vector instead of a scalar.

Gauss’s law and how to use symmetry to find electric fields

The law and what it claims

Gauss’s law states that the net electric flux through any closed surface equals the enclosed charge divided by \epsilon_0:

\oint \vec{E}\cdot d\vec{A} = \frac{Q_{\text{enc}}}{\epsilon_0}

This is always true in electrostatics (and more generally in electromagnetism). The power of Gauss’s law is that, with the right symmetry, it lets you find \vec{E} without doing a difficult integral of Coulomb’s law.

Why symmetry is the make-or-break step

Gauss’s law gives you an integral of \vec{E} over a surface. To solve for E, you need cases where:

  • The direction of \vec{E} is known everywhere on the surface.
  • The magnitude E is constant on parts of the surface.
  • The dot product simplifies to E\,dA or 0.

These conditions happen for highly symmetric charge distributions:

  • Spherical symmetry (point charge, uniformly charged sphere).
  • Cylindrical symmetry (infinite line charge, infinite cylinder).
  • Planar symmetry (infinite sheet of charge).

If the charge distribution is not symmetric enough, Gauss’s law is still true but not directly useful for finding \vec{E}.

Choosing a Gaussian surface

The Gaussian surface is imaginary. You pick it to exploit symmetry:

  • Sphere for spherical symmetry.
  • Cylinder for line/cylindrical symmetry.
  • Pillbox for planar symmetry.

The surface does not have to match a physical boundary, but it often helps if it does (for conductors, for example).

Case 1: field of an infinite line of charge

Let an infinite line have uniform linear charge density \lambda. By symmetry:

  • \vec{E} points radially outward from the line.
  • The magnitude depends only on distance r from the line.

Choose a cylindrical Gaussian surface of radius r and length L coaxial with the line.

Flux through the curved surface:

\Phi_E = E(2\pi r L)

Flux through the end caps is zero because \vec{E} is parallel to the caps.

Enclosed charge:

Q_{\text{enc}} = \lambda L

Gauss’s law gives

E(2\pi r L) = \frac{\lambda L}{\epsilon_0}

So

E = \frac{\lambda}{2\pi\epsilon_0 r}

Direction is radially outward for \lambda>0.

Case 2: field of an infinite sheet of charge

For an infinite sheet with uniform surface charge density \sigma:

  • Field is perpendicular to the sheet.
  • Magnitude is constant (does not depend on distance), by symmetry.

Use a pillbox Gaussian surface of cross-sectional area A that straddles the sheet. Flux goes only through the two flat faces:

\Phi_E = EA + EA = 2EA

Enclosed charge:

Q_{\text{enc}} = \sigma A

Gauss’s law:

2EA = \frac{\sigma A}{\epsilon_0}

So

E = \frac{\sigma}{2\epsilon_0}

Direction: away from the sheet if \sigma>0, toward it if \sigma