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Charge, Coulomb's law, and fields

Charges push and pull across empty space, and the electric field is the bookkeeping that turns those pushes into a property of space itself.

Charge

Charge comes in two signs, positive and negative, measured in coulombs. Like charges repel and opposite charges attract, and the smallest free unit is the electron charge e = 1.6e-19 C. Charge is conserved: it can move around but never appears or vanishes on its own.

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

Superposition and field

When several charges act at once, the forces add as vectors, each pair independent of the others. The electric field strips out the test charge: define E as force per unit positive charge, so a point charge q makes a field pointing away from q if it is positive and toward it if negative. The constant is k = 8.99e9 N m^2/C^2.

E = \frac{F}{q} = k\frac{q}{r^2}

Worked example

Two positive charges sit 0.30 m apart: q1 = 2.0 microcoulombs and q2 = 5.0 microcoulombs. Coulomb's law gives F = (8.99e9)(2.0e-6)(5.0e-6)/(0.30)^2 = 0.0899/0.09, which is about 1.0 N of repulsion. The field from q1 alone at that spot is E = (8.99e9)(2.0e-6)/(0.30)^2, about 2.0e5 N/C pointing away from q1.

F = \frac{(8.99\times 10^9)(2.0\times 10^{-6})(5.0\times 10^{-6})}{(0.30)^2} \approx 1.0\ \text{N}

Flashcard

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

Coulomb's law: force between two point charges

The force magnitude falls off as the square of the separation and grows with the product of the charges. It is attractive for opposite signs and repulsive for like signs, with k = 8.99e9 N m^2/C^2.

Check your understanding

A point charge of +4.0 nanocoulombs sits at the origin. What is the electric field magnitude 2.0 m away?

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