AP Physics 2 · Unit 10 · Topic 10.1

Charge Interactions and Coulomb-Force Models

Predict how stationary charged objects attract or repel, quantify each pairwise interaction with an inverse-square model, and combine multiple electric forces as vectors.

Learning Goals

Describe charge sign, conservation, and quantization.
Apply Coulomb's law to point charges and justified spherical approximations.
Determine force direction independently from force magnitude.
Use inverse-square scaling without repeating a full numerical calculation.
Apply superposition with one-dimensional signs or two-dimensional vector components.
Evaluate experimental data for consistency with \(F\propto q_1q_2/r^2\).

1. Charge as a Physical Property

\(q\)electric chargeSI unit: coulomb \(\mathrm C\)

\(e\)elementary charge\(1.602\times10^{-19}\,\mathrm C\)

\(k_e\)Coulomb constant\(8.99\times10^9\,\mathrm{N\,m^2/C^2}\)

\(\epsilon_0\)vacuum permittivity\(k_e=1/(4\pi\epsilon_0)\)

Two charge signs exist: positive and negative.
Like signs repel; unlike signs attract.
Charge is conserved in an isolated system: it can move between objects but is not created or destroyed in ordinary charging.
For an isolated object, net charge is quantized approximately as \(q=ne\), where \(n\) is an integer.

2. Neutral Does Not Mean No Charge

A neutral object usually contains enormous amounts of positive and negative charge whose totals cancel. Charging commonly transfers electrons; protons generally remain bound inside nuclei.

A charged object can attract a neutral object by polarization. The closer induced opposite charge feels a stronger force than the farther like charge because the electric force depends on distance.

3. The Point-Charge Model

Coulomb's simplest form applies to point charges. An extended object may be approximated as a point when its size is much smaller than the separation or when spherical symmetry makes its external interaction equivalent to charge concentrated at its center.

Measure \(r\) center to center. The model becomes unreliable when irregular objects are close enough that their charge distributions rearrange significantly.

4. Coulomb's Law

Experiments show that the force magnitude between two stationary point charges is proportional to each charge magnitude and inversely proportional to the square of their separation:

\(F_e=k_e\frac{|q_1q_2|}{r^2}=\frac{1}{4\pi\epsilon_0}\frac{|q_1q_2|}{r^2}\)

The absolute value produces a nonnegative magnitude. Determine attraction or repulsion separately from the signs. In a material medium, the effective permittivity may differ from \(\epsilon_0\); unless stated otherwise, AP problems typically use the vacuum or air approximation.

\(\frac{\mathrm{N\,m^2}}{\mathrm{C^2}}\frac{\mathrm{C^2}}{\mathrm{m^2}}=\mathrm N\)

Inverse-square distance dependenceThe curve is steep at small \(r\) and flattens at large \(r\). It approaches zero but does not reach zero at any finite separation.

5. Direction Comes from the Interaction

The force lies along the line joining the charges.

Same signs: each force arrow points away from the other charge.
Opposite signs: each force arrow points toward the other charge.

A safe method is to calculate \(F_e\) with charge magnitudes, then draw the direction physically. This prevents using the negative sign of a charge twice.

6. Newton's Third-Law Pair

The two charges exert equal-magnitude, opposite-direction forces:

\(\vec F_{1\leftarrow2}=-\vec F_{2\leftarrow1}\)

A charge with larger magnitude does not feel a larger interaction force. The accelerations can differ because \(a=F/m\) and the objects may have different masses.

7. Vector Form and Subscript Meaning

Let \(\hat r_{1\to2}\) point from source charge \(q_1\) toward charge \(q_2\). The force on 2 due to 1 is

\(\vec F_{2\leftarrow1}=k_e\frac{q_1q_2}{r_{12}^2}\hat r_{1\to2}\)

If \(q_1q_2>0\), the force points along \(\hat r_{1\to2}\); if \(q_1q_2<0\), it points opposite. Always define the unit vector before using a signed vector formula.

8. Scaling without Recalculation

\(\frac{F_f}{F_i}=\frac{|q_{1f}q_{2f}|}{|q_{1i}q_{2i}|}\left(\frac{r_i}{r_f}\right)^2\)

Change	Force factor
Double one charge magnitude	\(\times2\)
Double both charge magnitudes	\(\times4\)
Double separation	\(\times1/4\)
Triple separation	\(\times1/9\)
Halve separation	\(\times4\)

Changing a charge's sign reverses attraction versus repulsion but does not change the magnitude if \(|q|\) stays fixed.

9. Superposition Principle

Each source charge exerts its pairwise force as though the other source charges were absent. The net force on a chosen test charge \(Q\) is the vector sum:

\(\vec F_{\mathrm{net}}=\sum_i\vec F_i=k_eQ\sum_i\frac{q_i}{r_i^2}\hat r_{i\to Q}\)

Superposition does not mean adding all charge magnitudes first and using one average distance. Calculate each source-test pair separately unless symmetry or identical distances justify simplification.

Force-vector superpositionPairwise force arrows share a common tail at the test charge. The resultant is determined by component addition, not by adding arrow lengths.

10. One-Dimensional Workflow

Choose \(+x\), usually to the right.
Select the charge whose net force is requested.
Draw one arrow for every source charge.
Calculate each magnitude using its own separation.
Attach a positive or negative coordinate sign to each force.
Add algebraically.

11. Two-Dimensional Workflow

Resolve each pairwise force into components:

\(F_{ix}=F_i\cos\theta_i,\qquad F_{iy}=F_i\sin\theta_i\)

\(F_x=\sum_iF_{ix},\quad F_y=\sum_iF_{iy}\)

\(F_{\mathrm{net}}=\sqrt{F_x^2+F_y^2},\qquad \theta=\operatorname{atan2}(F_y,F_x)\)

Use the diagram to choose signs. A calculator's inverse tangent alone may place the direction in the wrong quadrant.

12. Symmetry Before Arithmetic

Geometry can eliminate components before any calculation. Equal source charges placed symmetrically around a test charge produce equal force magnitudes. Components perpendicular to the symmetry axis may cancel while components along it add.

A zero net force does not mean that no electric forces act. It can mean several nonzero vectors cancel exactly.

13. Where Can Net Force Be Zero?

For two like-sign source charges, the zero-force point for a test charge lies between them and closer to the smaller-magnitude source. There, the forces oppose.

For opposite-sign sources, the forces point in the same direction between the charges, so cancellation cannot occur there. A zero point, if one exists, lies outside the pair on the side of the smaller-magnitude charge.

14. Force versus Electric Field

Electric force describes an interaction involving a particular test charge. Electric field describes what the source charges establish in space:

\(\vec E=\frac{\vec F}{q_{\mathrm{test}}},\qquad \vec F=q_{\mathrm{test}}\vec E\)

The field direction is defined by the force on a positive test charge. A negative test charge feels force opposite the field.

15. Comparison with Gravitation

Feature	Electric force	Gravitational force
Magnitude model	\(k_e\|q_1q_2\|/r^2\)	\(Gm_1m_2/r^2\)
Direction	Attractive or repulsive	Attractive for ordinary positive mass
Shielding/cancellation	Positive and negative sources can cancel	No known negative mass analogue in introductory mechanics
Relative microscopic strength	Often enormously stronger	Dominates astronomy because bulk matter is nearly neutral

16. Worked Example — Magnitude and Direction

Problem: Charges \(q_1=+2.0\,\mu\mathrm C\) and \(q_2=-3.0\,\mu\mathrm C\) are \(0.40\,\mathrm m\) apart.

\(F=k_e\frac{|q_1q_2|}{r^2}=(8.99\times10^9)\frac{(2.0\times10^{-6})(3.0\times10^{-6})}{(0.40)^2}\approx0.337\,\mathrm N\)

Answer: each charge feels \(0.337\,\mathrm N\) toward the other charge. The forces are attractive, equal in magnitude, and opposite in direction.

17. Worked Example — Collinear Superposition

Problem: A test charge \(Q=+2.0\,\mu\mathrm C\) is at \(x=0\). A source \(q_1=+3.0\,\mu\mathrm C\) is at \(x=-0.30\,\mathrm m\), and \(q_2=-4.0\,\mu\mathrm C\) is at \(x=+0.40\,\mathrm m\).

The positive left source repels \(Q\) rightward. The negative right source attracts \(Q\) rightward.

\(F_1=(8.99\times10^9)\frac{(2.0\times10^{-6})(3.0\times10^{-6})}{(0.30)^2}\approx0.599\,\mathrm N\)

\(F_2=(8.99\times10^9)\frac{(2.0\times10^{-6})(4.0\times10^{-6})}{(0.40)^2}\approx0.450\,\mathrm N\)

\(\vec F_{\mathrm{net}}\approx(1.05\,\mathrm N)\hat x\)

Answer: \(1.05\,\mathrm N\) to the right. The forces add because their directions match, despite the source charges having opposite signs.

18. Worked Example — Zero-Force Location

Problem: Charges \(+4Q\) at \(x=0\) and \(+Q\) at \(x=d\) are fixed. Find the point between them where a small test charge has zero net force.

\(k_e\frac{4Q|q_t|}{x^2}=k_e\frac{Q|q_t|}{(d-x)^2}\)

\(\frac{2}{x}=\frac{1}{d-x}\quad\Rightarrow\quad x=\frac{2d}{3}\)

Answer: the point is \(2d/3\) from \(+4Q\), or \(d/3\) from the smaller \(+Q\). It lies closer to the weaker source so that its smaller charge is offset by a shorter distance.

19. Worked Example — Perpendicular Components

Problem: A positive test charge feels \(0.30\,\mathrm N\) in \(+x\) from one source and \(0.40\,\mathrm N\) in \(+y\) from another.

\(F_{\mathrm{net}}=\sqrt{(0.30)^2+(0.40)^2}=0.50\,\mathrm N\)

\(\theta=\tan^{-1}(0.40/0.30)\approx53.1^\circ\)

Answer: \(0.50\,\mathrm N\), \(53.1^\circ\) above \(+x\). Adding the magnitudes directly would incorrectly give \(0.70\,\mathrm N\).

20. Testing Coulomb's Law with Data

For fixed charges, several graph choices reveal the inverse-square model:

A plot of \(F\) versus \(r\) is curved.
A plot of \(F\) versus \(1/r^2\) should be linear through the origin, with slope \(k_e|q_1q_2|\).
A log-log plot follows \(\ln F=\ln(k_e|q_1q_2|)-2\ln r\), so its slope should be approximately \(-2\).

Systematic deviations may come from charge leakage, uncertain center-to-center distance, nearby polarization, finite object size, humidity, or an unaccounted force.

21. Reliable Problem-Solving Workflow

Choose the target charge and draw only forces acting on it.
Convert \(\mu\mathrm C\) and \(\mathrm{nC}\) to coulombs.
Measure each center-to-center separation.
Use charge magnitudes for each force magnitude.
Assign directions from attraction or repulsion.
Add vectors by components and check symmetry.

22. Common Traps

Using \(1/r\) instead of \(1/r^2\).
Forgetting to square the entire distance in meters.
Using edge-to-edge rather than center-to-center separation.
Adding force magnitudes when directions differ.
Including a charge's sign in the magnitude and then reversing direction again.
Assuming the larger charge experiences the larger third-law force.
Combining source charges before checking their different locations.
Treating an extended polarized object as a point without justification.

Mastery Check

1. The separation between fixed charges triples. What happens to the force magnitude?

Show reasoning and answer

Because \(F\propto1/r^2\), \(F_f/F_i=(r_i/3r_i)^2=1/9\). The force becomes one ninth as large.

2. A \(+q\) and a \(-4q\) interact. Which feels the larger force?

Show reasoning and answer

Neither. Newton's third law gives equal magnitudes and opposite directions. Their accelerations may differ if their masses differ.

3. Two equal positive source charges are symmetrically placed left and right of a positive test charge. Find the net force.

Show reasoning and answer

Each source repels the test charge with the same magnitude, but the directions are opposite. The vector sum is zero even though two nonzero forces act.

4. If both charge magnitudes double and their separation halves, by what factor does \(F\) change?

Show reasoning and answer

The charge product contributes \(2\times2=4\). Halving distance contributes \(1/(1/2)^2=4\). The total factor is \(16\).

5. Can the net electric force be zero between two opposite-sign source charges?

Show reasoning and answer

No. Between opposite signs, the forces on any test charge point in the same direction. A zero-force point, if the magnitudes differ, must lie outside the pair on the side of the smaller source magnitude.

6. Why should \(F\) versus \(1/r^2\) be linear for fixed point charges?

Show reasoning and answer

Coulomb's law can be written \(F=(k_e|q_1q_2|)(1/r^2)\). The coefficient is constant, so it is the slope; the ideal intercept is zero.

Investigation: Identify the Distance Exponent

Use a safe electrostatic force sensor or a provided dataset to record force magnitude at several center-to-center separations while keeping the charge preparation consistent. Compare linear fits of \(F\) versus \(1/r\) and \(F\) versus \(1/r^2\), then fit a log-log graph. Use residuals and uncertainty—not just visual appearance—to decide whether the exponent is consistent with \(-2\).

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