Coulomb's Law Calculator (Electric Force)

Enter two charges and their separation in the calculator above to get the electric force from Coulomb's law, F = k|q1 q2|/r^2, with k = 9.0 x 10^9 N m^2/C^2. Microcoulomb inputs are converted automatically. The formula gives the magnitude; the charge signs tell you whether it attracts or repels.

Charge q1 (microC)Sign matters: negative charges are fine.Charge q2 (microC)Sign matters: negative charges are fine.Separation r (m)Center-to-center distance.

electric force magnitude |F|

0.216 N

Verdict: attractive. Opposite signs pull toward each other, each charge feeling 0.216 N.

Steps

1.convert to coulombs: q1 = 2 microC = 2e-6 C; q2 = -3 microC = -3e-6 C
2.|F| = k |q1| |q2| / r^2, with k = 9.0 x 10^9 N m^2/C^2 (the AP sheet value)
3.|F| = 9.0 x 10^9 x 2e-6 x 3e-6 / (0.5)^2
4.|F| = 0.054 / 0.25 = 0.216 N
5.the charges have opposite signs, so the force is attractive (they pull together)

AP Physics: Unit 10 (topics 10.1 Electric Charge and Electric Force, 8.1 Electric Charge and Electric Force). Matches AP Physics 2 Unit 10 topic 10.1 (Electric Charge and Electric Force), in a unit worth 15-18% of the multiple-choice section, and AP Physics C: Electricity and Magnetism Unit 8 topic 8.1, in a unit worth 15-25%.

What the Calculator Above Does

The calculator above computes the magnitude of the electric force between two point charges using Coulomb's law, $F = k\frac{|q_1 q_2|}{r^2}$ , with the Coulomb constant $k = 9.0 \times 10^9 \, \mathrm{N \cdot m^2/C^2}$ , the value printed on the AP Physics 2 equation sheet. Type in both charges, choose coulombs or microcoulombs for the units, enter the center-to-center separation $r$ in meters, and the tool returns the force in newtons. Microcoulomb inputs are converted for you, so $3.0 \, \mu\mathrm{C}$ is treated as $3.0 \times 10^{-6} \, \mathrm{C}$ before anything gets squared or multiplied. If you want the concept itself, with sign conventions, derivations, and practice problems, the Coulomb's law guide covers that in depth. This page focuses on getting the numbers right and reading the result correctly.

Magnitude vs Direction

Coulomb's law, written with absolute value bars, gives you only the size of the force. Direction comes from a separate and simpler rule: like charges repel, opposite charges attract. The force on each charge points along the line connecting the two centers, either toward the other charge (attraction) or away from it (repulsion). That is why the calculator reports a positive force magnitude no matter what signs you enter: a $-4.0 \, \mu\mathrm{C}$ charge and a $+3.0 \, \mu\mathrm{C}$ charge feel exactly the same size force as two positive charges of the same magnitudes at the same separation. On free-response questions, compute the magnitude from the formula, then state the direction in words or with a vector arrow. Do not substitute negative charge values and report a 'negative force': a force magnitude is never negative.

The Inverse Square: Double the Distance, Quarter the Force

Because $r$ is squared in the denominator, the force falls off with the square of the separation. Double the distance and the force drops to one fourth. Triple it and the force drops to one ninth. Halve it and the force quadruples. Test this in the calculator above by keeping both charges fixed and changing only $r$ .

Separation	Force (relative)
$r$	$F$
$2r$	$F/4$
$3r$	$F/9$
$r/2$	$4F$

This is the same inverse square pattern gravity follows, which is why AP questions like to pair the two force laws. A prompt such as 'each charge is doubled and the distance is doubled' tests whether you can track both effects at once: doubling each charge multiplies the force by 4, doubling the distance divides it by 4, so the force is unchanged.

Microcoulombs to Coulombs: The Conversion That Breaks Answers

A coulomb is an enormous amount of charge, so AP problems almost always state charges in microcoulombs ( $1 \, \mu\mathrm{C} = 10^{-6} \, \mathrm{C}$ ) or nanocoulombs ( $1 \, \mathrm{nC} = 10^{-9} \, \mathrm{C}$ ). The calculator above accepts microcoulomb inputs directly, but on the exam you convert by hand. Two habits prevent most errors:

Convert to base SI units before substituting. Write $5.0 \, \mu\mathrm{C}$ as $5.0 \times 10^{-6} \, \mathrm{C}$ on paper first, then plug in.
Sanity-check the size. Two microcoulomb-scale charges have a product near $10^{-12} \, \mathrm{C^2}$ , and multiplying by $k \approx 10^{10}$ puts the force in the range of newtons (or fractions of a newton) for separations near a meter. An answer like $10^{12} \, \mathrm{N}$ means you skipped the conversion.

Exponent slips in this conversion are the most common Coulomb's law mistake, so the size check is worth the ten seconds it takes.

Common Mistakes to Avoid

Most wrong answers with Coulomb's law come from a short list of slips, and the calculator above can help you catch every one of them by comparison:

Forgetting to square $r$ . Dividing by $0.50$ instead of $0.25$ halves your answer.
Leaving the distance in centimeters. Coulomb's law needs $r$ in meters; $20 \, \mathrm{cm}$ is $0.20 \, \mathrm{m}$ .
Entering microcoulombs as whole coulombs, which inflates the force by a factor of $10^{12}$ .
Substituting charge signs into the formula and reporting a negative magnitude. Use signs only to decide attract or repel.
Measuring $r$ between sphere surfaces instead of centers. For point charges and uniformly charged spheres, $r$ is the center-to-center distance.

If a homework answer disagrees with the tool, one of these five is almost always the reason.

Where Coulomb's Law Sits in the AP Courses

Coulomb's law opens the electrostatics sequence in two AP courses. In AP Physics 2 it is topic 10.1, Electric Charge and Electric Force, in Unit 10 (Electric Force, Field, and Potential), a unit worth 15-18% of the multiple-choice section. In AP Physics C: Electricity and Magnetism it is topic 8.1 in Unit 8 (Electric Charges, Fields, and Gauss's Law), worth 15-25%. Both courses build directly on it: divide the Coulomb force by a small test charge and you get the field of a point charge, the starting point of electric field and potential. Physics C then extends the same idea to continuous charge distributions and Gauss's law. A four-function, scientific, or graphing calculator is allowed on both sections of every AP physics exam, so clean substitution practice here transfers straight to test day.

Force between two microcoulomb charges

A $+3.0 \, \mu\mathrm{C}$ charge and a $-4.0 \, \mu\mathrm{C}$ charge are held $0.50 \, \mathrm{m}$ apart. Find the magnitude and direction of the electric force each charge exerts on the other.

Convert to coulombs: $q_1 = 3.0 \times 10^{-6} \, \mathrm{C}$ and $|q_2| = 4.0 \times 10^{-6} \, \mathrm{C}$ . Only the magnitudes go into the formula.
Substitute into Coulomb's law: $F = \frac{(9.0 \times 10^9)(3.0 \times 10^{-6})(4.0 \times 10^{-6})}{(0.50)^2}$ .
Multiply the numerator: $(9.0 \times 10^9)(3.0 \times 10^{-6})(4.0 \times 10^{-6}) = 0.108 \, \mathrm{N \cdot m^2}$ .
Square the separation: $(0.50 \, \mathrm{m})^2 = 0.25 \, \mathrm{m^2}$ , so $F = 0.108 / 0.25 = 0.432 \, \mathrm{N}$ .
Direction: the charges have opposite signs, so the force is attractive. Each charge is pulled toward the other with the same magnitude.

$F \approx 0.43 \, \mathrm{N}$ , attractive: each charge is pulled directly toward the other.

Checking the inverse square with the calculator

Two $+2.0 \, \mu\mathrm{C}$ charges are placed $0.10 \, \mathrm{m}$ apart, then moved to $0.20 \, \mathrm{m}$ apart. Find the force in each case and compare.

At $r = 0.10 \, \mathrm{m}$ : $F_1 = \frac{(9.0 \times 10^9)(2.0 \times 10^{-6})(2.0 \times 10^{-6})}{(0.10)^2}$ .
The numerator is $(9.0 \times 10^9)(4.0 \times 10^{-12}) = 0.036 \, \mathrm{N \cdot m^2}$ , and $(0.10)^2 = 0.010 \, \mathrm{m^2}$ , so $F_1 = 0.036 / 0.010 = 3.6 \, \mathrm{N}$ .
At $r = 0.20 \, \mathrm{m}$ : the numerator is unchanged, but $(0.20)^2 = 0.040 \, \mathrm{m^2}$ , so $F_2 = 0.036 / 0.040 = 0.90 \, \mathrm{N}$ .
Compare: $F_1 / F_2 = 3.6 / 0.90 = 4.0$ . Doubling the separation cut the force to one fourth, exactly the inverse square prediction.
Both charges are positive, so the force is repulsive in both cases.

$F_1 = 3.6 \, \mathrm{N}$ and $F_2 = 0.90 \, \mathrm{N}$ , both repulsive. Doubling $r$ divides the force by 4.

Solving for the separation

Two identical $+5.0 \, \mu\mathrm{C}$ charges repel each other with a force of $2.0 \, \mathrm{N}$ . How far apart are they?

Rearrange Coulomb's law for distance: $r = \sqrt{\frac{k |q_1 q_2|}{F}}$ .
Compute the numerator: $k|q_1 q_2| = (9.0 \times 10^9)(5.0 \times 10^{-6})(5.0 \times 10^{-6}) = (9.0 \times 10^9)(2.5 \times 10^{-11}) = 0.225 \, \mathrm{N \cdot m^2}$ .
Divide by the force: $r^2 = 0.225 / 2.0 = 0.1125 \, \mathrm{m^2}$ .
Take the square root: $r = \sqrt{0.1125} = 0.335 \, \mathrm{m}$ .
Sanity check by substituting back: $0.225 / (0.335)^2 = 0.225 / 0.112 \approx 2.0 \, \mathrm{N}$ . Consistent.

$r \approx 0.34 \, \mathrm{m}$ , about 34 centimeters apart.

Frequently asked questions

What value of the Coulomb constant does this calculator use?

It uses k = 9.0 x 10^9 N m^2/C^2, the value printed on the AP Physics 2 equation sheet. Some textbooks use 8.99 x 10^9, but the difference is under 0.2 percent and AP scoring expects answers consistent with 9.0 x 10^9.

Does the calculator tell me whether the force is attractive or repulsive?

The formula itself only produces a magnitude, so read direction from the signs of the charges: two like charges (both positive or both negative) repel, and opposite charges attract. State that direction in words on free-response work.

Why is my answer off by a factor of a trillion?

You almost certainly entered microcoulomb values as whole coulombs. Each microcoulomb is 10^-6 C, and Coulomb's law multiplies the two charges together, so skipping both conversions inflates the force by 10^6 times 10^6, a factor of 10^12.

Do both charges feel the same force even if one is much bigger?

Yes. By Newton's third law the two forces are equal in magnitude and opposite in direction. A 1 microcoulomb charge and a 100 microcoulomb charge pull (or push) on each other with exactly the same strength; only their accelerations differ if their masses differ.

Can I use this calculator for three or more charges?

Yes, one pair at a time. The force on any charge is the vector sum of the individual Coulomb forces from every other charge (superposition). Compute each pairwise magnitude with the calculator, assign directions from the signs, then add the forces as vectors.