Electric charge conservation and Coulomb interaction

Electric charge is a scalar quantity that can be positive or negative; in an isolated system the algebraic sum of charges is conserved: $\sum q_i = \mathrm{const}$. Therefore, when bodies are rubbed or conductors touch, charge is neither created nor destroyed but only redistributed. Charge is discrete: $q = Ne$, where $e \approx 1.60\cdot10^{-19}\ \mathrm{C}$, but in macroscopic problems it is convenient to treat charge as continuous. Stationary point charges interact with the Coulomb force, whose magnitude is $F = k\frac{|q_1 q_2|}{r^2}, \qquad k \approx 8.99\cdot10^9\ \mathrm{N\cdot m^2/C^2}.$ The force is directed along the line joining the charges: like charges repel, unlike charges attract. By Newton's third law the forces acting on the two charges are equal in magnitude and opposite in direction, and for several charges one uses the superposition principle together with a free-body diagram — the same tools as in mechanics.

Two point charges $q_1 = +2.0\ \mu\mathrm{C}$ and $q_2 = -3.0\ \mu\mathrm{C}$ are in vacuum at a distance $r = 0.30\ \mathrm{m}$. Find the magnitude and direction of the Coulomb force acting on charge $q_1$.

Two identical metal spheres carry charges $q_1 = +5.0\ \mu\mathrm{C}$ and $q_2 = -1.0\ \mu\mathrm{C}$. The spheres were brought into contact and then separated to a distance $r = 0.20\ \mathrm{m}$. Find the charge of each sphere after contact and the magnitude of their interaction force.

Three point charges lie on the $x$ axis: $q_1 = +2.0\ \mu\mathrm{C}$ at $x = 0$, $q_2 = -2.0\ \mu\mathrm{C}$ at $x = 0.10\ \mathrm{m}$, and $q_3 = +4.0\ \mu\mathrm{C}$ at $x = 0.30\ \mathrm{m}$. Find the resultant force acting on charge $q_2$ and state its direction.

Safe plan: computer simulation only. Open the PhET "Coulomb's Law" simulation (phet.colorado.edu) and complete three steps.

1. Place two point charges and record the force $F$ at distances $r = 0.5, 1.0, 2.0$ (arbitrary units). Verify that doubling $r$ reduces the force by a factor of 4, that is, $F \propto 1/r^2$.
2. Change the sign of one charge and confirm that the force reverses direction (repulsion $\leftrightarrow$ attraction) while its magnitude stays the same.
3. Set charges $+5\ \mu\mathrm{C}$ and $-1\ \mu\mathrm{C}$ and first calculate the charge each identical sphere will get after contact. Then set those charges in the simulation and check your calculation using the force.

Fill in a table "$r$ — $F$ — $F r^2$" and state your conclusion. Do not use real high-voltage sources: carry out all observations in the simulation.

Show that the Coulomb force is directed along the line joining the point charges, and that the forces acting on the two charges are equal in magnitude and opposite in direction (Newton's third law).

Sketch: two point charges $+q_1$ and $-q_2$ on one straight line; the segment of length $r$ is marked between them; two equal-length arrows — $\vec F_{12}$ applied to $q_1$ and pointing toward $q_2$, $\vec F_{21}$ applied to $q_2$ and pointing toward $q_1$; the formula $F = k|q_1 q_2|/r^2$ is written beside the sketch.

A phone battery with a capacity of $3.0\ \text{A}\cdot\text{h}$ is charged by a current of $2.0\ \text{A}$ for $1.5\ \text{h}$ from a $10\ \text{W}$ charger. The charge that passes through the wire is $Q = I t = 2.0 \cdot 5400 = 10\,800\ \text{C}$, about $6.8\cdot10^{22}$ electrons.

This is a direct application of charge conservation: a charger does not create electrons — as many electrons leave one battery terminal as arrive at the other, so the total charge of the closed circuit stays zero. The formula $Q = It$ accounts for the transfer of charge that already exists, not for its creation.

After $5\ \text{s}$ of rubbing against hair the balloon acquires a charge of $-2.0\cdot10^{-7}\ \text{C}$. Bring it up to a rubbed plastic rod carrying $+2.0\cdot10^{-7}\ \text{C}$ at a distance of $2.0\ \text{cm} = 0.020\ \text{m}$: $F = k\frac{|q_1 q_2|}{r^2} = 8.99\cdot10^{9}\cdot\frac{4.0\cdot10^{-14}}{4.0\cdot10^{-4}} \approx 0.90\ \text{N}.$ That is about the weight of a $90\ \text{g}$ bar of chocolate.

This uses Coulomb's law: the force is proportional to the product of the magnitudes of the charges and inversely proportional to the square of the distance between them, with $k = 8.99\cdot10^{9}\ \text{N}\cdot\text{m}^2/\text{C}^2$. The signs of the charges fix only the direction of the force — for $q_1 q_2 < 0$ it is attraction — and the distance $r$ is measured between the centres of the charged bodies.

The base of a thundercloud carries a charge of $-15\ \text{C}$, while the patch of ground beneath it ends up with $+15\ \text{C}$. Treating the cloud and the ground as point charges separated by $r = 800\ \text{m}$ gives $F = k\frac{|q_1 q_2|}{r^2} = 8.99\cdot10^{9}\cdot\frac{225}{6.4\cdot10^{5}} \approx 3.2\cdot10^{6}\ \text{N}.$

The scenario tests two ideas of the topic at once. The magnitude of the attraction between the cloud and the ground is computed with Coulomb's law. And the appearing $+15\ \text{C}$ in the ground is a consequence of charge conservation: those coulombs are not created but merely redistributed by induction, and the total charge of the cloud–ground system stays unchanged right up to the discharge, when the charge simply flows along the lightning channel.