AI Tutor
If you strip all the electrons from two 1-gram marbles and place them one meter apart, the repulsion force between them could lift the entire planet Earth. What rule governs this immense force?

Coulomb's Law

In 1785, French physicist Charles-Augustin de Coulomb used a delicate twisting instrument called a torsion balance to measure the tiny forces between charged spheres. He discovered that electrostatic force follows an inverse-square law, weakening rapidly as distance grows.
The magnitude of the electrostatic force FF between two point charges q1q_1 and q2q_2 separated by a straight-line distance rr is given by F=ke∣q1q2∣r2F = k_e \frac{|q_1 q_2|}{r^2} where keβ‰ˆ8.99Γ—109Β Nβ‹…m2/C2k_e \approx 8.99 \times 10^9 \text{ N}\cdot\text{m}^2/\text{C}^2 is Coulomb's constant.
Like charges (positive-positive or negative-negative) repel, pushing directly away from one another along the line connecting them. Opposite charges (positive-negative) attract, pulling directly toward each other.
Notice that according to Newton's Third Law, both charges feel forces of the exact same magnitude, even if one charge is a million times larger than the other.
What happens when three or more charges push and pull on the same target at the same time?

1D Vector Superposition

The principle of superposition states that the total electrostatic force on any single charge is the vector sum of the individual forces exerted by every other charge independently. That means we calculate each pairwise interaction on its own, then add the resulting vectors.
Let's solve the net force on q2=βˆ’2.0Γ—10βˆ’6Β Cq_2 = -2.0 \times 10^{-6}\text{ C} located at x=0.04Β mx = 0.04\text{ m}, between q1=+4.0Γ—10βˆ’6Β Cq_1 = +4.0 \times 10^{-6}\text{ C} at x=0Β mx = 0\text{ m} and q3=+6.0Γ—10βˆ’6Β Cq_3 = +6.0 \times 10^{-6}\text{ C} at x=0.08Β mx = 0.08\text{ m}.
First, find the magnitude of the force from q1q_1 on q2q_2 across the 0.04Β m0.04\text{ m} gap: F12=(8.99Γ—109)(4.0Γ—10βˆ’6)(2.0Γ—10βˆ’6)(0.04)2=44.95Β NF_{12} = (8.99 \times 10^9) \frac{(4.0 \times 10^{-6})(2.0 \times 10^{-6})}{(0.04)^2} = 44.95\text{ N} Because opposite charges attract, q1q_1 pulls q2q_2 to the left (βˆ’x-x direction), so Fβƒ—12=βˆ’44.95Β NΒ i^\vec{F}_{12} = -44.95\text{ N }\hat{i}.