AP Physics 2 · Unit 10 · Topic 10.3

Mapping Electric Fields from Source Charges

Turn a collection of source charges into a vector map of space. Define the field independently of any particular object placed there, calculate fields by superposition, and use field lines and symmetry as visual checks on magnitude and direction.

Learning Goals

Distinguish an electric field from the electric force on a specific charge.
Calculate the magnitude and direction of the field from one or more point charges.
Add electric-field vectors with components and use symmetry to predict cancellation.
Locate zero-field points for simple one-dimensional source arrangements.
Draw and interpret electric-field vector maps and field-line diagrams without treating lines as physical objects.

\(\vec E\)electric field\(\mathrm{N/C}\)

\(Q\)source chargecreates the field

\(q\)charge placed in fieldexperiences \(\vec F=q\vec E\)

\(P\)observation pointwhere the field is evaluated

1. A Field Assigns a Vector to Every Point

An electric field is a vector quantity that depends on position. Source charges modify the surrounding space, so every observation point \(P\) has a field vector with a magnitude and direction—even when no test charge is physically placed there.

A field map samples many points. A single arrow answers “what is \(\vec E\) here?”; a collection of arrows reveals how the field varies across a region.

2. Definition from Force per Charge

The field direction is defined as the direction of force on a small positive test charge:

\(\vec E=\lim_{q_0\to0}\frac{\vec F_{\mathrm e}}{q_0}\)

The ideal test charge is small enough not to rearrange the source charges. The field belongs to the sources and the location; it does not depend on which test charge is later used to measure it.

3. From Coulomb Force to the Field of a Point Charge

Place a positive test charge \(q_0\) a distance \(r\) from source charge \(Q\). Coulomb's law gives the force magnitude \(F=k_e|Qq_0|/r^2\). Divide by \(q_0\):

\(E=\frac{F}{q_0}=k_e\frac{|Q|}{r^2}\)

In vector form, with \(\hat r\) pointing from the source toward the observation point,

\(\vec E(P)=k_e\frac{Q}{r^2}\hat r\)

A positive \(Q\) produces a field away from itself. A negative \(Q\) reverses the vector, so the field points toward the source.

4. Field and Force Are Not the Same Vector

Once the field is known, the force on any charge placed at the point is

\(\vec F_{\mathrm e}=q\vec E\)

If \(q>0\), force points with the field.
If \(q<0\), force points opposite the field.
Changing \(q\) changes the force but not the preexisting field.

5. Units and Scaling

Electric field has units \(\mathrm{N/C}\). Because \(E\propto |Q|/r^2\), doubling the source charge doubles the field, while doubling distance makes the field one fourth as large.

The distance is measured from the source charge to the observation point—not between a source and some unrelated test charge shown elsewhere.

Worked Example 1 · One Source

Field from a Positive Point Charge

Find the field \(0.20\,\mathrm m\) east of a \(+3.0\,\mathrm{nC}\) point charge.

\(E=(8.99\times10^9)\frac{3.0\times10^{-9}}{(0.20)^2}\approx6.7\times10^2\,\mathrm{N/C}\)

The source is positive, so the field points away from it: approximately \(6.7\times10^2\,\mathrm{N/C}\) east.

Worked Example 2 · Force from a Known Field

A Negative Charge Reverses the Direction

A \(-2.0\,\mu\mathrm C\) particle is placed in a \(1.5\times10^3\,\mathrm{N/C}\) field directed east.

\(F=|q|E=(2.0\times10^{-6})(1.5\times10^3)=3.0\times10^{-3}\,\mathrm N\)

Because the particle is negative, its force is \(3.0\,\mathrm{mN}\) west. The field itself still points east.

A vector map samples the field throughout a regionEvery arrow is calculated at its own position. Near the sources the magnitude changes rapidly; between opposite charges the vectors reinforce from positive toward negative.

6. Superposition: One Field from Many Sources

Each source contributes its own field independently. The net field at observation point \(P\) is their vector sum:

\(\vec E_{\text{net}}(P)=\sum_i\vec E_i(P)=k_e\sum_i\frac{Q_i}{r_i^2}\hat r_i\)

Do not combine source charges first unless they occupy effectively the same location or a far-field approximation has been justified. Different sources usually have different distances and directions relative to \(P\).

7. Direction Before Algebra

Mark the observation point \(P\).
For each positive source, draw \(\vec E_i\) at \(P\) away from that source.
For each negative source, draw \(\vec E_i\) at \(P\) toward that source.
Only then assign component signs and calculate magnitudes.

8. Component Method

Resolve each field vector into perpendicular components:

\(E_x=\sum_i E_i\cos\theta_i,\qquad E_y=\sum_i E_i\sin\theta_i\)

\(E_{\text{net}}=\sqrt{E_x^2+E_y^2},\qquad \theta=\operatorname{atan2}(E_y,E_x)\)

The \(\operatorname{atan2}\) form reminds you to identify the correct quadrant instead of relying blindly on inverse tangent.

9. Symmetry Can Remove Components

Before calculating, inspect whether sources are mirror images around an axis. At a point on the perpendicular bisector of two equal like charges, the horizontal components cancel and the perpendicular components add. At the midpoint itself, every component cancels and \(\vec E_{\text{net}}=\vec0\).

Symmetry is a reason, not a guess: pair each source with an equal source at the mirrored location and compare their field components at the selected point.

Worked Example 3 · Like Charges

Midpoint Cancellation

Equal charges \(+Q\) are placed at \(x=-a\) and \(x=+a\). At the origin, each produces the same field magnitude. The left source's field points right; the right source's field points left.

\(\vec E_{\text{net}}=\frac{k_eQ}{a^2}\hat i-\frac{k_eQ}{a^2}\hat i=\vec0\)

Zero net field does not mean no sources exist. It means their vector contributions cancel at that one point.

Worked Example 4 · Opposite Charges

Midpoint Reinforcement

A \(+4.0\,\mathrm{nC}\) charge and a \(-4.0\,\mathrm{nC}\) charge are \(0.20\,\mathrm m\) apart. At the midpoint, each source is \(0.10\,\mathrm m\) away.

\(E_{\text{each}}=(8.99\times10^9)\frac{4.0\times10^{-9}}{(0.10)^2}\approx3.60\times10^3\,\mathrm{N/C}\)

Both vectors point from the positive source toward the negative source, so \(E_{\text{net}}\approx7.19\times10^3\,\mathrm{N/C}\) toward the negative charge.

Worked Example 5 · Perpendicular Contributions

Add Components, Not Magnitudes

At point \(P\), one source produces \(\vec E_1=(3.0\times10^3\,\mathrm{N/C})\hat i\) and another produces \(\vec E_2=(4.0\times10^3\,\mathrm{N/C})\hat j\).

\(E_{\text{net}}=\sqrt{(3.0\times10^3)^2+(4.0\times10^3)^2}=5.0\times10^3\,\mathrm{N/C}\)

\(\theta=\tan^{-1}(4/3)\approx53^\circ\text{ above }+x\)

The scalar sum \(7.0\times10^3\,\mathrm{N/C}\) would be wrong because the contributions are perpendicular.

10. Finding a Zero-Field Point

First decide where cancellation is geometrically possible. At a zero-field point, contributing vectors must oppose each other and have equal magnitudes. For two like charges, this can occur between them. For two opposite charges, the fields point the same way between them, so cancellation cannot occur there.

11. Unequal Sources Shift the Balance Point

Because \(E\propto |Q|/r^2\), the zero-field point lies closer to the source with smaller magnitude. Greater distance is needed to weaken the larger source enough for the two fields to match.

After solving, substitute the position back into a direction sketch; algebra alone can produce a root in a region where the fields actually reinforce.

Worked Example 6 · Unequal Like Charges

Locate the Balance Point

A \(+9q\) source is at \(x=0\) and a \(+q\) source is at \(x=4.0\,\mathrm m\). Between them, the fields oppose. Let the zero-field point be \(x\) meters from \(+9q\):

\(k_e\frac{9q}{x^2}=k_e\frac{q}{(4-x)^2}\)

\(\frac{3}{x}=\frac{1}{4-x}\quad\Rightarrow\quad x=3.0\,\mathrm m\)

The point is \(3.0\,\mathrm m\) from \(+9q\), equivalently \(1.0\,\mathrm m\) from the smaller \(+q\), exactly as the qualitative check predicts.

Field lines encode direction and relative strengthThe field vector is tangent to the curve at every point. Lines are a mapping convention, not particle paths or physical threads in space.

12. Rules for Electric-Field Lines

Lines begin on positive charge and end on negative charge, or extend to or from infinity.
The arrow direction matches the force direction on a positive test charge.
Closer line spacing indicates a stronger field in a qualitative diagram.
The number of lines attached to a source is drawn proportional to the source-charge magnitude.
Lines never cross, because the field at one point cannot have two directions.
Electrostatic field lines do not form closed loops.

13. Vector Maps Versus Field-Line Maps

Representation	What it shows	What to avoid
Vector map	A sampled arrow gives local magnitude and direction at each grid point.	Comparing arrows drawn with inconsistent scales.
Field lines	A continuous curve is tangent to the local field; spacing indicates relative strength.	Treating a line as the actual trajectory of a charge.

14. Why Field Lines Are Not Particle Paths

A released charge accelerates according to \(\vec F=q\vec E\), but its velocity may already point in another direction and its inertia matters. A negative charge accelerates opposite the field. Therefore a field line is generally not the same as a particle's trajectory.

15. Recognizing Common Source Patterns

Single positive source: radial arrows outward; field weakens as \(1/r^2\).
Single negative source: radial arrows inward.
Equal like charges: a zero-field point at the midpoint and lines that bend away from the central region.
Equal opposite charges, or a dipole: lines connect positive to negative; the field is strong between them.
Approximately uniform field: nearly parallel, equally spaced lines, such as well inside the gap between large oppositely charged plates.

16. Near Field and Far Field

Close to one source, its contribution may dominate. Far from a cluster whose net charge is nonzero, the field increasingly resembles that of a single point charge equal to the cluster's total charge. For a neutral dipole, the leading positive and negative contributions mostly cancel far away, so its field decreases faster than \(1/r^2\).

17. Conductors as a Map Check

For a conductor in electrostatic equilibrium, \(\vec E=\vec0\) inside the conducting material. Just outside the surface, the field is perpendicular to the surface; a tangential component would drive mobile charge along it. This provides a useful check when sketching fields near conductors.

Worked Example 7 · Reading a Map

Predict a Force Without Recalculating the Sources

A field map shows \(\vec E=(2.0\hat i-1.0\hat j)\times10^4\,\mathrm{N/C}\) at point \(P\). A proton-like positive charge \(+e\) is placed there.

\(\vec F=e\vec E=(3.20\hat i-1.60\hat j)\times10^{-15}\,\mathrm N\)

A particle with charge \(-e\) at the same point would experience the same force magnitude in the exactly opposite direction. Neither particle changes the field in the small-test-charge model.

18. Reliable Mapping Workflow

Choose coordinates and mark every source and observation point.
Draw each source's field direction at the observation point.
Calculate each magnitude with its own distance.
Resolve into components and add algebraically.
Check units, symmetry, limiting distance, and source signs.
Repeat at representative points before sketching smooth field lines.

19. Common Traps

Using the sign of a negative test charge to reverse the field itself.
Including the test charge in \(E=k_e|Q|/r^2\).
Adding field magnitudes instead of vectors.
Measuring \(r\) from the wrong source.
Combining separated source charges before checking geometry.
Claiming field lines cross or form electrostatic loops.
Assuming a zero-field point must be halfway between unequal charges.
Confusing zero field at one point with zero electric potential.

Mastery Check

1. A positive point charge's field is \(800\,\mathrm{N/C}\) at distance \(r\). What is the field magnitude at \(2r\)?

Show reasoning and answer

For a fixed point source, \(E\propto1/r^2\). Doubling distance gives \(E_f=800/4=200\,\mathrm{N/C}\).

2. A negative charge is placed in an electric field directed north. Which way is the electric force?

Show reasoning and answer

Because \(\vec F=q\vec E\) and \(q<0\), the force points opposite the field: south.

3. Why is the net field at the midpoint between equal positive charges zero?

Show reasoning and answer

The sources have equal magnitude and equal distance from the midpoint, so their field magnitudes match. Each field points away from its positive source, giving opposite directions and exact vector cancellation.

4. Can the field be zero between equal and opposite point charges?

Show reasoning and answer

No. Between them, the field from the positive source points toward the negative source, and the field toward the negative source points the same way. The contributions reinforce.

5. What would crossing field lines incorrectly imply?

Show reasoning and answer

At the crossing, two tangents would specify two field directions at the same point. A vector field has one unique vector at each point, so electrostatic field lines cannot cross.

6. Two sources produce \(300\,\mathrm{N/C}\) east and \(400\,\mathrm{N/C}\) north at a point. Find the net field.

Show reasoning and answer

The vectors are perpendicular: \(E=\sqrt{300^2+400^2}=500\,\mathrm{N/C}\). The direction is \(\tan^{-1}(400/300)\approx53^\circ\) north of east.

Investigation: Build and Test a Field Map

Use the open PhET Charges and Fields simulation. Place one positive and one negative source at recorded grid coordinates. Before turning on field sensors, predict directions at the midpoint, on the perpendicular bisector, and near each source. Then sample magnitude and direction at equal grid spacing. Change only one source magnitude or position and explain which map features change and which field-line rules remain valid.

For a quantitative extension, export or record at least eight measurements, compare \(E\) with \(1/r^2\) for a single source, and discuss grid resolution and placement uncertainty.

Official curriculum reference: College Board AP Physics 2 course page. The explanation and worked example are independently written for this study site.