Coulomb's law and electric fields, worked examples

Coulomb's law gives the force between two point charges. The electric field turns that into a property of space around one charge, so you can find the force on any other charge placed there. Those two ideas open College Physics 2, and every later electricity topic builds on them.

The formulas

• Coulomb's law: F = k |q1 q2| / r^2
• Field of a point charge: E = k |Q| / r^2
• Force from a field: F = qE

The constants, as OpenStax College Physics gives them:

• k = 8.99 x 10^9 N m^2/C^2
• Elementary charge e = 1.60 x 10^-19 C (a proton is +e, an electron is -e)

Use the magnitudes in the formula and get the direction from the physics: like charges repel, opposite charges attract. The field points away from a positive charge and toward a negative one. A positive test charge feels a force along the field; a negative charge feels a force opposite to it.

Charges are often given with prefixes: 1 microcoulomb (µC) = 10^-6 C and 1 nanocoulomb (nC) = 10^-9 C. Distances must be in meters.

Example 1: force between two charges

A +2.00 µC charge and a -3.00 µC charge are 5.00 cm apart. Find the force.

• Convert: q1 = 2.00 x 10^-6 C, q2 = 3.00 x 10^-6 C (magnitude), r = 0.0500 m
• F = (8.99 x 10^9)(2.00 x 10^-6)(3.00 x 10^-6) / (0.0500)^2
• Numerator: 8.99 x 10^9 x 6.00 x 10^-12 = 5.394 x 10^-2
• Denominator: 0.0500^2 = 2.50 x 10^-3
• F = 21.6 N, attractive, because the signs are opposite

Each charge feels 21.6 N toward the other, equal and opposite, as Newton's third law requires.

Example 2: the inverse square

Move the same two charges to 10.0 cm apart. The distance doubles, so the force drops by a factor of 2^2 = 4: 21.6 / 4 = 5.39 N. You can confirm with the full formula, but spotting the ratio is faster and is a common multiple-choice question. Tripling r cuts F by 9; halving r multiplies it by 4.

Example 3: electron and proton in hydrogen

In the Bohr model of hydrogen, the electron sits 0.530 x 10^-10 m from the proton (OpenStax Example 18.1).

• F = (8.99 x 10^9)(1.60 x 10^-19)^2 / (0.530 x 10^-10)^2
• F = 8.19 x 10^-8 N, attractive

That looks tiny, but compare it with gravity between the same two particles. With G = 6.67 x 10^-11 N m^2/kg^2, m_e = 9.11 x 10^-31 kg and m_p = 1.67 x 10^-27 kg, F_gravity = 3.61 x 10^-47 N. The electric force is about 2.27 x 10^39 times stronger. That is why gravity is ignored at the atomic scale and electric forces run chemistry.

Example 4: field of a point charge

Find the electric field 2.00 mm from a +5.00 nC charge (OpenStax Example 18.2).

• E = (8.99 x 10^9)(5.00 x 10^-9) / (2.00 x 10^-3)^2
• Numerator: 44.95; denominator: 4.00 x 10^-6
• E = 1.12 x 10^7 N/C, pointing directly away from the positive charge

The field does not depend on any test charge. It is the force per coulomb that a charge would feel if you put one there.

Example 5: force on a charge in a field

An electron is in a uniform field of 1.00 x 10^4 N/C pointing east.

• F = qE = (1.60 x 10^-19)(1.00 x 10^4) = 1.60 x 10^-15 N
• Direction: west, opposite the field, because the electron is negative

Its acceleration is a = F/m = 1.60 x 10^-15 / 9.11 x 10^-31 = 1.76 x 10^15 m/s^2. Tiny forces on tiny masses give enormous accelerations, which is how old CRT screens and particle accelerators move electrons.

Example 6: superposition of two charges

Fields from several charges add as vectors. Put q1 = +4.00 µC at x = 0 and q2 = +1.00 µC at x = 0.300 m.

Field at the midpoint (x = 0.150 m, 0.150 m from each):

• From q1: E1 = (8.99 x 10^9)(4.00 x 10^-6) / (0.150)^2 = 1.60 x 10^6 N/C, pointing +x (away from q1)
• From q2: E2 = (8.99 x 10^9)(1.00 x 10^-6) / (0.150)^2 = 4.00 x 10^5 N/C, pointing -x (away from q2)
• Net: 1.60 x 10^6 - 4.00 x 10^5 = 1.20 x 10^6 N/C in the +x direction

A +2.00 nC charge placed there would feel F = qE = (2.00 x 10^-9)(1.20 x 10^6) = 2.40 x 10^-3 N in the +x direction.

Where is the field zero? Between two like charges the fields point opposite ways, so there is a point where they cancel. Set the magnitudes equal at a distance x from q1:

• k(4.00 µC) / x^2 = k(1.00 µC) / (0.300 - x)^2
• Take square roots: 2.00 / x = 1.00 / (0.300 - x)
• 0.600 - 2.00x = x, so x = 0.200 m

Check: at x = 0.200 m both fields are 8.99 x 10^5 N/C, equal and opposite. The zero point sits closer to the smaller charge, which should make intuitive sense.

Field lines and conductors

You will also be asked to read field-line pictures. The rules: lines start on positive charges and end on negative ones, they never cross, their density shows field strength, and the number of lines is proportional to the charge. For a conductor in electrostatic equilibrium, the field inside is zero, any excess charge sits on the outer surface, and the field just outside is perpendicular to the surface.

Common mistakes

• Not squaring r, or squaring it before converting cm to m.
• Putting signs into the formula and then getting confused by a negative force. Use magnitudes, then decide attract or repel.
• Adding field magnitudes that point in different directions. Draw arrows first, then add as vectors.
• Forgetting the prefix. Leaving µC as a whole number makes the answer 10^12 times too big for a two-charge problem.

Where this goes next

The same inverse-square setup leads directly to electric potential, V = kQ/r, which is the next topic in the course. For the full sequence, see what's in College Physics 2. Chemistry uses the same law: ionic bond strength and lattice energy grow with the charges and shrink with distance, which is why it appears in the bonding unit of General Chemistry I. Once charges start moving through wires, series vs parallel circuits, explained takes over, and the resistor calculator checks that math. For study method, how to study for a physics exam and the testing effect: why quizzing beats review are the two to read.

Encodr's free College Physics 2 course, with cards on Coulomb's law and fields, is coming soon; the current courses are on the decks page.