Electric Force and Electric Field

Lessons

Senior 4 · Physics

Electric Force and Electric Field

5 lessons

Lesson 1 of 5

Electric Charge and Coulomb's Law

25 min

A comb that bends water

Run a dry plastic comb through dry hair a few times, then hold it close to a thin, steady stream of water from a tap. The stream bends towards the comb, although nothing touches it. On a dry day a pullover crackles as you take it off, and dust clings to a screen you have just wiped. All of these are electric forces between charges at rest, the subject of electrostatics. They are enormously strong: the electric attraction that holds an electron in a hydrogen atom is about $10^{39}$ times its gravitational attraction. In this lesson you state the rule that gives the size of the force between charges, Coulomb's law, and use it to add the forces from several charges.

Electric charge

Matter is built from protons (charge $+ e$ ), electrons (charge $- e$ ) and neutrons (no charge), where the elementary charge is $e = 1.602 \times 10^{-19}\,\mathrm{C}$ . An object is charged when it has more electrons than protons (negative) or fewer (positive). Three facts follow.

Two kinds, two rules: like charges repel and unlike charges attract (Figure 1.1).
Charge is quantised: every free charge is a whole-number multiple of $e$ : $Q = n e$ .
Charge is conserved: it is never created or destroyed, only moved. When a comb is rubbed on hair, electrons move from the hair to the comb; the comb becomes negative and the hair equally positive.

In conductors such as metals some electrons are free to move through the material; in insulators such as plastic, glass and dry air they are not, so charge stays where it is put. A charged comb attracts a neutral stream of water because it pulls the water molecules' negative ends slightly further from their positive ends (it polarises them), and the nearer, unlike charge is attracted more strongly than the further, like charge is repelled.

Coulomb's law

In 1785 Charles-Augustin de Coulomb measured the tiny forces between charged spheres with a torsion balance: a light rod hung on a thin fibre, which twisted by an amount proportional to the force. He found that the force between two small charged bodies (point charges) $q_{1}$ and $q_{2}$ , a distance $r$ apart,

is proportional to the product of the charges, $q_{1} q_{2}$ ;
is inversely proportional to the square of the distance, $r^{2}$ ;
acts along the line joining them: repulsive for like charges, attractive for unlike.

F = k\,\frac{q_1 q_2}{r^{2}}, \qquad k = \frac{1}{4\pi\varepsilon_0} = 8.99 \times 10^{9}\,\mathrm{N\,m^{2}\,C^{-2}}

Here $\varepsilon_0 = 8.85 \times 10^{-12}\,\mathrm{F\,m^{-1}}$ is the permittivity of free space. If the charges are in an insulating medium, the force is smaller by the medium's relative permittivity $\varepsilon_r$ : $F = \frac{q_1 q_2}{4\pi\varepsilon_0\varepsilon_r r^{2}}$ . For air $\varepsilon_r \approx 1.0006$ , so air can be treated as a vacuum; for water $\varepsilon_r \approx 80$ , which is why water pulls salts apart so easily.

Figure 1.1: The force between two point charges acts along the line joining them: repulsive for like charges, attractive for unlike.

Definition

Coulomb's law: the electric force between two point charges is directly proportional to the product of their charges and inversely proportional to the square of the distance between them, $F = \frac{1}{4\pi\varepsilon_0}\frac{q_1 q_2}{r^{2}}$ . The SI unit of charge is the coulomb (C), the charge carried by a current of 1 A in 1 s. Example: two charges of $+1.0\,\mu\mathrm{C}$ one metre apart in air repel with $F = 8.99 \times 10^{9} \times \frac{(1.0 \times 10^{-6})^{2}}{1.0^{2}} = 9.0 \times 10^{-3}\,\mathrm{N}$ .

Worked example 1: two charged spheres

A sphere with charge $+3.0\,\mu\mathrm{C}$ is 0.20 m from a sphere with charge $-5.0\,\mu\mathrm{C}$ . Find the force between them.

F = k\frac{q_1 q_2}{r^{2}} = 8.99 \times 10^{9} \times \frac{(3.0 \times 10^{-6})(5.0 \times 10^{-6})}{0.20^{2}} = 3.4\,\mathrm{N}

The force is attractive, because the charges are unlike. Use the sizes of the charges in the formula and decide the direction from the signs.

Worked example 2: electric against gravitational force

In a hydrogen atom the electron and proton are about $5.29 \times 10^{-11}\,\mathrm{m}$ apart. Compare the electric and gravitational forces between them ( $m_e = 9.11 \times 10^{-31}\,\mathrm{kg}$ , $m_p = 1.67 \times 10^{-27}\,\mathrm{kg}$ , $G = 6.67 \times 10^{-11}\,\mathrm{N\,m^{2}\,kg^{-2}}$ ).

F_e = \frac{8.99 \times 10^{9} \times (1.602 \times 10^{-19})^{2}}{(5.29 \times 10^{-11})^{2}} = 8.2 \times 10^{-8}\,\mathrm{N}

F_g = \frac{G m_e m_p}{r^{2}} = \frac{6.67 \times 10^{-11} \times 9.11 \times 10^{-31} \times 1.67 \times 10^{-27}}{(5.29 \times 10^{-11})^{2}} = 3.6 \times 10^{-47}\,\mathrm{N}

$\frac{F_e}{F_g} = 2.3 \times 10^{39}$ . Both forces follow an inverse-square law, so the ratio is the same at any distance. Gravity rules the motion of planets only because large bodies carry almost exactly equal amounts of positive and negative charge.

Several charges: the principle of superposition

When several charges act on one charge, each pair interacts as if the others were not there, and the resultant force is the vector sum of the separate forces:

\vec{F} = \vec{F}_1 + \vec{F}_2 + \vec{F}_3 + \dots

Find the size of each force with Coulomb's law, mark its direction from the signs, then add the forces as vectors: along a line by adding with signs, in a plane by resolving into components (Figure 1.2).

Figure 1.2: Superposition: the resultant force on C is the vector sum of the forces from A and B (drawn to scale).

Worked example 3: charges in a line, and a point of no force

Charges $q_1 = +2.0\,\mu\mathrm{C}$ and $q_2 = +4.0\,\mu\mathrm{C}$ are fixed 0.30 m apart. (a) Find the resultant force on $q_3 = -1.0\,\mu\mathrm{C}$ placed between them, 0.10 m from $q_{1}$ . (b) Where between them would any third charge feel no resultant force?

(a) $F_{13} = \frac{8.99 \times 10^{9} \times 2.0 \times 10^{-6} \times 1.0 \times 10^{-6}}{0.10^{2}} = 1.80\,\mathrm{N}$ towards $q_{1}$ ; $F_{23} = \frac{8.99 \times 10^{9} \times 4.0 \times 10^{-6} \times 1.0 \times 10^{-6}}{0.20^{2}} = 0.90\,\mathrm{N}$ towards $q_{2}$ . Resultant: $1.80 - 0.90 = 0.90\,\mathrm{N}$ towards $q_{1}$ .

(b) At distance $x$ from $q_{1}$ the two forces must be equal: $\frac{q_1}{x^{2}} = \frac{q_2}{(0.30 - x)^{2}} \Rightarrow \frac{0.30 - x}{x} = \sqrt{2} \Rightarrow x = \frac{0.30}{1 + \sqrt{2}} = 0.124\,\mathrm{m}$ from $q_{1}$ , nearer the smaller charge.

Worked example 4: forces at right angles

In Figure 1.2, C ( $+2.0\,\mu\mathrm{C}$ ) is 0.30 m from A ( $+4.0\,\mu\mathrm{C}$ ) and 0.40 m from B ( $-3.0\,\mu\mathrm{C}$ ), with CA perpendicular to CB. Find the resultant force on C.

F_A = \frac{8.99 \times 10^{9} \times 2.0 \times 10^{-6} \times 4.0 \times 10^{-6}}{0.30^{2}} = 0.80\,\mathrm{N} \text{ (away from A)}, \qquad F_B = \frac{8.99 \times 10^{9} \times 2.0 \times 10^{-6} \times 3.0 \times 10^{-6}}{0.40^{2}} = 0.34\,\mathrm{N} \text{ (towards B)}

F = \sqrt{0.80^{2} + 0.34^{2}} = 0.87\,\mathrm{N}, \qquad \tan\alpha = \frac{0.34}{0.80} \Rightarrow \alpha = 22.9^\circ

The resultant is 0.87 N, directed $22.9^\circ$ from the line AC (extended beyond C) towards B.

A physics teacher touches the dome of a Van de Graaff generator during a class in 2008. His body and hair become charged with the same sign, so the strands repel one another and stand out from his head. Photo: Drbaker171, public domain, via Wikimedia Commons.

Warning

Common mistakes. (1) Forgetting to square the distance, or to convert μC ( $10^{-6}\,\mathrm{C}$ ) and nC ( $10^{-9}\,\mathrm{C}$ ) and cm into SI units. (2) Putting the signs of the charges into Coulomb's law and then trusting the sign of the answer for the direction in two dimensions: use sizes, and draw the direction from "like repel, unlike attract". (3) Adding the sizes of forces that are not along the same line: add vectors by components. (4) Using Coulomb's law for large charged bodies that are close together: it holds for point charges and, by symmetry, for uniformly charged spheres measured from their centres.

Tip

Size, then direction, then add. For every charge acting on the one you care about: (1) size from $k\frac{|q_1 q_2|}{r^{2}}$ ; (2) direction from the signs, drawn on a sketch; (3) add the arrows as vectors. Doubling the distance quarters the force; doubling one charge doubles it.

Coulomb's Law (UCLA Physics). A lecture demonstration of the force between charged objects. Watch how the force changes as the separation changes, and compare it with the inverse-square rule.

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Practice

How many electrons must be removed from a neutral metal sphere to give it a charge of $+8.0\,\mathrm{nC}$ ?
Two point charges of $+1.0\,\mu\mathrm{C}$ are 1.0 m apart in air. Find the force between them. What is the force if the distance is halved?
Two identical small metal spheres carry $+6.0\,\mu\mathrm{C}$ and $-2.0\,\mu\mathrm{C}$ . They are touched together and then placed 0.10 m apart. Find the charge on each and the force between them.
A charge of $+1.0\,\mu\mathrm{C}$ is placed midway between charges of $+2.0\,\mu\mathrm{C}$ and $-2.0\,\mu\mathrm{C}$ that are 0.20 m apart. Find the resultant force on it.
Three charges of $+1.0\,\mu\mathrm{C}$ sit at the corners of an equilateral triangle of side 0.10 m. Find the resultant force on one of them.
A charge of $+9.0\,\mu\mathrm{C}$ is at $x = 0$ and a charge of $+1.0\,\mu\mathrm{C}$ at $x = 0.40\,\mathrm{m}$ . Where on the line between them would a third charge feel no resultant force?
Two ions with charges $+ e$ and $- e$ are 0.30 nm apart. Find the force between them in air and in water ( $\varepsilon_r = 80$ ), and use your answer to explain why table salt dissolves in water.
The electric force is about $10^{39}$ times stronger than gravity between an electron and a proton. Evaluate the statement: "Electric forces should therefore control the motion of the Moon around the Earth."

Show answersHide answers

$n = \frac{Q}{e} = \frac{8.0 \times 10^{-9}}{1.602 \times 10^{-19}} = 5.0 \times 10^{10}$ electrons.
$F = 8.99 \times 10^{9} \times \frac{(1.0 \times 10^{-6})^{2}}{1.0^{2}} = 9.0 \times 10^{-3}\,\mathrm{N}$ ; halving $r$ multiplies $F$ by 4: $3.6 \times 10^{-2}\,\mathrm{N}$ .
Total charge $+4.0\,\mu\mathrm{C}$ shared equally: $+2.0\,\mu\mathrm{C}$ each; $F = 8.99 \times 10^{9} \times \frac{(2.0 \times 10^{-6})^{2}}{0.10^{2}} = 3.6\,\mathrm{N}$ , repulsive.
Each charge is 0.10 m away; each force is $8.99 \times 10^{9} \times \frac{2.0 \times 10^{-6} \times 1.0 \times 10^{-6}}{0.10^{2}} = 1.8\,\mathrm{N}$ , both towards the negative charge: resultant 3.6 N towards the $-2.0\,\mu\mathrm{C}$ charge.
Each force is $8.99 \times 10^{9} \times \frac{(1.0 \times 10^{-6})^{2}}{0.10^{2}} = 0.90\,\mathrm{N}$ , at $60^\circ$ to each other: resultant $2 \times 0.90 \times \cos 30^\circ = 1.56\,\mathrm{N}$ , directed away from the centre of the triangle.
$\frac{9.0}{x^{2}} = \frac{1.0}{(0.40 - x)^{2}} \Rightarrow \frac{x}{0.40 - x} = 3 \Rightarrow x = 0.30\,\mathrm{m}$ from the $+9.0\,\mu\mathrm{C}$ charge.
In air $F = \frac{8.99 \times 10^{9} \times (1.602 \times 10^{-19})^{2}}{(3.0 \times 10^{-10})^{2}} = 2.6 \times 10^{-9}\,\mathrm{N}$ ; in water 80 times smaller, $3.2 \times 10^{-11}\,\mathrm{N}$ . The weakened attraction lets the jostling water molecules pull the sodium and chloride ions apart.
The statement is wrong. The Earth and the Moon each carry almost equal amounts of positive and negative charge, so their net charges, and the electric force between them, are tiny. Gravity, which only attracts and adds up over every particle, dominates on large scales; the enormous electric force matters inside atoms, molecules and materials.

Quick check

Key points

Charge is of two kinds, is quantised ( $Q = n e$ , $e = 1.602 \times 10^{-19}\,\mathrm{C}$ ) and is conserved; like charges repel, unlike attract.
Coulomb's law: $F = \frac{1}{4\pi\varepsilon_0}\frac{q_1 q_2}{r^{2}}$ , $k = 8.99 \times 10^{9}\,\mathrm{N\,m^{2}\,C^{-2}}$ ; in a medium divide by $\varepsilon_r$ .
The electric force between an electron and a proton is about $10^{39}$ times the gravitational force.
Superposition: the resultant force is the vector sum of the forces from each charge separately.

Sources and credits

Image: Coulomb's torsion balance, plate from C.-A. de Coulomb's memoir in the Mémoires de l'Académie royale des sciences (1785), Wikimedia Commons, public domain.
Photo: Muller Van de Graaff generator demonstration by Drbaker171, Wikimedia Commons, public domain.
Video: Coulomb's Law, UCLA Physics (uclaphysicsvideo), embedded from YouTube.

Written by Ganzaa: the lesson text, derivations, diagrams, worked examples and practice exercises.

1A comb that bends water

2Electric charge

3Coulomb's law

4Several charges: the principle of superposition

A comb that bends water

Electric charge

Coulomb's law

Several charges: the principle of superposition