Coulomb's Law: Electric Force Between Two Charges

Coulomb's law gives the electric force between two point charges: F = k q1 q2 / r^2, with k = 9.0 x 10^9 N m^2/C^2. Opposite charges attract, like charges repel, and the force falls off with the square of the distance: doubling r cuts the force to a quarter.

AP Physics: Unit 10 (topics 10.1 Electric Charge and Electric Force, 8.1 Electric Charge and Electric Force). In AP Physics 2, Coulomb's law is Topic 10.1 in Unit 10 (Electric Force, Field, and Potential), a unit weighted at 15 to 18 percent. In AP Physics C: Electricity and Magnetism, it is Topic 8.1 in Unit 8 (Electric Charges, Fields, and Gauss's Law), worth 15 to 25 percent.

The electric force formula

Coulomb's law gives the magnitude of the electric force between two point charges:

F=kq1q2r2F = \frac{k q_1 q_2}{r^2}

Here q1q_1 and q2q_2 are the charge magnitudes in coulombs (C), rr is the center-to-center distance in meters (m), and kk is the Coulomb constant. The AP Physics 2 equation sheet prints k=9.0×109 N⋅m2/C2k = 9.0 \times 10^9 \, \text{N}\cdot\text{m}^2/\text{C}^2, so use that value in every calculation. The force comes out in newtons.

Both charges feel the same force magnitude, whatever their sizes. That is Newton's third law at work: the two forces form an action-reaction pair, so a tiny charge pulls on a huge one exactly as hard as the huge one pulls back. Only the accelerations differ.

Want to check your homework arithmetic fast? Plug your numbers into the Coulomb's law calculator and compare.

Attraction or repulsion: let the signs decide

Like charges repel; opposite charges attract. Two positives (or two negatives) push each other directly apart along the line joining them, while a positive and a negative pull straight toward each other.

The cleanest workflow for AP problems: plug charge magnitudes into the formula to get the force size, then use the signs to decide direction. If you substitute signed values instead, a negative result signals attraction, and that shortcut works fine for a single pair. It falls apart the moment a problem has three or more charges and you need to add force vectors, because a bare minus sign no longer tells you which way anything points.

Build the habit of sketching each force as an arrow before computing, exactly as you would for a free-body diagram. Direction from the picture, magnitude from the formula.

Inverse-square scaling

The r2r^2 in the denominator makes the force fall off fast with distance. Double the separation and the force drops to one quarter. Triple it and you are down to one ninth. This kind of ratio reasoning shows up constantly in multiple-choice questions because it needs no calculator.

ChangeNew force
Double rrF/4F/4
Triple rrF/9F/9
Halve rr4F4F
Double one charge2F2F
Double both charges, double rrFF (unchanged)

To work these, write the proportionality F∝q1q2/r2F \propto q_1 q_2 / r^2 and track each factor separately. Say one charge doubles while the distance also doubles: the force picks up a factor of 2×14=122 \times \tfrac{1}{4} = \tfrac{1}{2}, so an 8.0 N force becomes 4.0 N. No constants, no unit conversions, just factors.

How Coulomb's law compares to gravity

Coulomb's law and Newton's law of gravitation have the same mathematical shape:

FE=kq1q2r2Fg=Gm1m2r2F_E = \frac{k q_1 q_2}{r^2} \qquad F_g = \frac{G m_1 m_2}{r^2}

Both are inverse-square laws, and both act along the line joining the two objects. The differences matter more than the resemblance:

  • Gravity only attracts. The electric force attracts or repels depending on the signs.
  • The constants sit twenty orders of magnitude apart: k=9.0×109k = 9.0 \times 10^9 versus G=6.67×10−11G = 6.67 \times 10^{-11} in SI units.
  • For subatomic particles, electricity dominates completely. The electric attraction between the proton and the electron in a hydrogen atom is about 103910^{39} times stronger than their gravitational attraction. Worked example 3 below runs those numbers.

Gravity ends up ruling planets and orbits only because bulk matter is almost perfectly neutral: the enormous electric forces cancel almost exactly, and the feeble but always-attractive gravity is what remains.

Where Coulomb's law sits on the AP exams

In AP Physics 2, Coulomb's law is Topic 10.1, Electric Charge and Electric Force, the opening topic of Unit 10 (Electric Force, Field, and Potential). Unit 10 carries a 15 to 18 percent exam weighting. In AP Physics C: Electricity and Magnetism, the same idea appears as Topic 8.1 in Unit 8 (Electric Charges, Fields, and Gauss's Law), worth 15 to 25 percent.

Expect three problem flavors: a direct calculation, ratio reasoning when charges or distances change, and net-force problems where two or more charges act on a third and you add the forces as vectors, the same skill as finding net force in mechanics.

Coulomb's law also feeds straight into the next topic: divide the force by a test charge and you get the electric field, covered in electric field and potential. Physics C students then extend the idea to continuous charge distributions and Gauss's law.

Common mistakes

  • Forgetting to square rr. This is the single most common slip. Write the squaring as its own line: (0.50 m)2=0.25 m2(0.50 \, \text{m})^2 = 0.25 \, \text{m}^2.
  • Botching unit conversions. Charges usually arrive in microcoulombs (1 μC=10−6 C1 \, \mu\text{C} = 10^{-6} \, \text{C}) or nanocoulombs (1 nC=10−9 C1 \, \text{nC} = 10^{-9} \, \text{C}). Convert to coulombs before substituting.
  • Plugging signed charges into the formula, getting a negative force, then losing track of what the sign means. Use magnitudes; get direction from the signs separately.
  • Assuming the bigger charge exerts the bigger force. The two forces are always equal in magnitude by Newton's third law, whatever the charges.
  • Adding force magnitudes when the forces point in different directions. Net-force problems need vector addition, components and all.

None of these are physics failures. They are bookkeeping failures, which is good news: a short checklist fixes every one of them.

Force between two charges

A charge q1=+3.0 μCq_1 = +3.0 \, \mu\text{C} and a charge q2=−4.0 μCq_2 = -4.0 \, \mu\text{C} are held 0.50 m0.50 \, \text{m} apart. Find the magnitude and direction of the electric force on each charge.

  1. Convert to SI units and take magnitudes: q1=3.0×10−6 Cq_1 = 3.0 \times 10^{-6} \, \text{C} and q2=4.0×10−6 Cq_2 = 4.0 \times 10^{-6} \, \text{C}. Save the signs for the direction call at the end.

  2. Set up Coulomb's law: F=kq1q2r2=(9.0×109)(3.0×10−6)(4.0×10−6)(0.50)2F = \dfrac{k q_1 q_2}{r^2} = \dfrac{(9.0 \times 10^9)(3.0 \times 10^{-6})(4.0 \times 10^{-6})}{(0.50)^2}.

  3. Numerator: (9.0×109)(3.0×10−6)=2.7×104(9.0 \times 10^9)(3.0 \times 10^{-6}) = 2.7 \times 10^4, then (2.7×104)(4.0×10−6)=0.108 N⋅m2(2.7 \times 10^4)(4.0 \times 10^{-6}) = 0.108 \, \text{N}\cdot\text{m}^2.

  4. Denominator: (0.50 m)2=0.25 m2(0.50 \, \text{m})^2 = 0.25 \, \text{m}^2. Divide: F=0.108/0.25=0.432 N≈0.43 NF = 0.108 / 0.25 = 0.432 \, \text{N} \approx 0.43 \, \text{N}.

  5. Direction: the signs are opposite, so the force is attractive. Each charge is pulled toward the other with the same 0.43 N, an action-reaction pair.

F≈0.43 NF \approx 0.43 \, \text{N} on each charge, directed toward the other charge (attractive).

Solving for an unknown charge

Two identical small spheres carry equal charges qq. When their centers are 0.30 m0.30 \, \text{m} apart, each sphere experiences a repulsive force of 0.036 N0.036 \, \text{N}. Find qq.

  1. With equal charges, Coulomb's law becomes F=kq2r2F = \dfrac{k q^2}{r^2}. Solve for the charge: q=Fr2/kq = \sqrt{F r^2 / k}.

  2. Substitute: q=(0.036 N)(0.30 m)29.0×109 N⋅m2/C2q = \sqrt{\dfrac{(0.036 \, \text{N})(0.30 \, \text{m})^2}{9.0 \times 10^9 \, \text{N}\cdot\text{m}^2/\text{C}^2}}.

  3. Inside the root, top first: (0.30 m)2=0.090 m2(0.30 \, \text{m})^2 = 0.090 \, \text{m}^2, so (0.036 N)(0.090 m2)=3.24×10−3 N⋅m2(0.036 \, \text{N})(0.090 \, \text{m}^2) = 3.24 \times 10^{-3} \, \text{N}\cdot\text{m}^2.

  4. Divide by kk: (3.24×10−3)/(9.0×109)=3.6×10−13 C2(3.24 \times 10^{-3}) / (9.0 \times 10^9) = 3.6 \times 10^{-13} \, \text{C}^2.

  5. Take the square root: q=6.0×10−7 C=0.60 μCq = 6.0 \times 10^{-7} \, \text{C} = 0.60 \, \mu\text{C}. The repulsion tells you the two charges share the same sign, but not which sign it is.

q=6.0×10−7 Cq = 6.0 \times 10^{-7} \, \text{C} (0.60 μC\mu\text{C}); both charges positive or both negative.

Electric force vs gravity in a hydrogen atom

In a hydrogen atom, the electron (me=9.11×10−31 kgm_e = 9.11 \times 10^{-31} \, \text{kg}) orbits about r=5.3×10−11 mr = 5.3 \times 10^{-11} \, \text{m} from the proton (mp=1.67×10−27 kgm_p = 1.67 \times 10^{-27} \, \text{kg}). Each has charge magnitude e=1.6×10−19 Ce = 1.6 \times 10^{-19} \, \text{C}. Compare the electric and gravitational forces between them.

  1. Electric force: FE=ke2r2=(9.0×109)(1.6×10−19)2(5.3×10−11)2F_E = \dfrac{k e^2}{r^2} = \dfrac{(9.0 \times 10^9)(1.6 \times 10^{-19})^2}{(5.3 \times 10^{-11})^2}.

  2. Work the pieces: (1.6×10−19)2=2.56×10−38 C2(1.6 \times 10^{-19})^2 = 2.56 \times 10^{-38} \, \text{C}^2, so the numerator is (9.0×109)(2.56×10−38)=2.30×10−28 N⋅m2(9.0 \times 10^9)(2.56 \times 10^{-38}) = 2.30 \times 10^{-28} \, \text{N}\cdot\text{m}^2. The denominator is (5.3×10−11 m)2=2.81×10−21 m2(5.3 \times 10^{-11} \, \text{m})^2 = 2.81 \times 10^{-21} \, \text{m}^2.

  3. Divide: FE=(2.30×10−28)/(2.81×10−21)=8.2×10−8 NF_E = (2.30 \times 10^{-28}) / (2.81 \times 10^{-21}) = 8.2 \times 10^{-8} \, \text{N}.

  4. Gravitational force: Fg=Gmempr2F_g = \dfrac{G m_e m_p}{r^2}. The mass product is (9.11×10−31)(1.67×10−27)=1.52×10−57 kg2(9.11 \times 10^{-31})(1.67 \times 10^{-27}) = 1.52 \times 10^{-57} \, \text{kg}^2; multiplying by G=6.67×10−11G = 6.67 \times 10^{-11} gives 1.01×10−671.01 \times 10^{-67}, and dividing by 2.81×10−21 m22.81 \times 10^{-21} \, \text{m}^2 gives Fg=3.6×10−47 NF_g = 3.6 \times 10^{-47} \, \text{N}.

  5. Compare: FE/Fg=(8.2×10−8)/(3.6×10−47)≈2.3×1039F_E / F_g = (8.2 \times 10^{-8}) / (3.6 \times 10^{-47}) \approx 2.3 \times 10^{39}. The electric force is about 39 orders of magnitude stronger.

FE≈8.2×10−8 NF_E \approx 8.2 \times 10^{-8} \, \text{N} versus Fg≈3.6×10−47 NF_g \approx 3.6 \times 10^{-47} \, \text{N}: the electric force is about 2.3×10392.3 \times 10^{39} times stronger.

Frequently asked questions

What does Coulomb's law say?

The electric force between two point charges equals k times the product of the charge magnitudes, divided by the square of the distance between them: F = k q1 q2 / r^2, with k = 9.0 x 10^9 N m^2/C^2. Opposite charges attract; like charges repel.

What value of k should I use on the AP exam?

Use k = 9.0 x 10^9 N m^2/C^2, the value printed on the AP Physics 2 equation sheet, so your numbers match the reference table you have during the exam. Many textbooks list 8.99 x 10^9, which changes answers by about 0.1 percent. In Physics C you will also see k written as 1/(4 pi epsilon_0), with epsilon_0 = 8.85 x 10^-12 C^2/(N m^2) on the equation sheet.

What happens to the electric force if the distance between two charges doubles?

It drops to one quarter of its original value. Coulomb's law is an inverse-square law, so the force scales as 1/r^2: doubling r multiplies the force by 1/4, tripling r multiplies it by 1/9, and halving r multiplies it by 4.

Do I plug negative charges into Coulomb's law?

Use magnitudes in the formula and handle direction separately: like signs repel, opposite signs attract. Substituting signed values gives a negative force for attraction, which works for a single pair but becomes confusing when several charges act at once and you need to add force vectors.

Why is the electric force so much stronger than gravity?

The constants set the scale: k is about 9.0 x 10^9 while G is only 6.67 x 10^-11. For the proton and electron in a hydrogen atom, the electric force is roughly 10^39 times the gravitational force. Gravity dominates at planetary scales only because bulk matter is almost exactly neutral, so the huge electric forces cancel.