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1.1 - Electric Charge and Electric Force

Fundamental Properties of Electric Charge

Electric charge is a fundamental property of all matter. For most materials, the net charge is neutral, because there are equal amounts of positive and negative charge.
Key characteristics: - Electric charge is a scalar, so it has no direction. - The sign (+ or −) tells you whether there’s an excess or a deficit of electrons. - Negative charge: an excess of electrons. - Positive charge: a deficit of electrons.

Elementary Charge

All charge is quantized in multiples of the elementary charge:
e=1.6×10−19 Ce = 1.6 \times 10^{-19} \text{ C}
where C stands for coulombs, the SI unit of electric charge.
The elementary charge is the size of the charge on a single proton or a single electron, and it’s the smallest amount of charge you’ll ever find on its own. You can’t split it into smaller free pieces, so every object’s charge is a whole number of elementary charges:
∣q∣=ne|q| = ne
where nn is the number of extra electrons (for a negative object) or missing electrons (for a positive object).
The three particles in an atom: - Proton: charge +e+e - Electron: charge −e-e - Neutron: no electric charge
A proton and an electron carry exactly the same amount of charge, even though a proton has about 1,800 times more mass.
Quick example: a balloon rubbed on your hair ends up with a charge of −3.2-3.2 nC. How many extra electrons does it have?
n=∣q∣e=3.2×10−9 C1.60×10−19 C=2.0×1010 electronsn = \frac{|q|}{e} = \frac{3.2 \times 10^{-9} \text{ C}}{1.60 \times 10^{-19} \text{ C}} = 2.0 \times 10^{10} \text{ electrons}
Twenty billion electrons sounds like a lot, but it’s a tiny fraction of the electrons already in the balloon. Everyday charges contain so many elementary charges that charge acts smooth and continuous, which is why Topic 1.4 can treat it as spread out over a line, a surface, or a volume.

Coulomb’s Law

Coulomb’s Law describes the electrostatic force between two charged objects.

The Math

Fe=kq1q2r2F_e = k\frac{q_1q_2}{r^2}
where: - FeF_e = electric force (N) - kk = the electrostatic constant - q1,q2q_1, q_2 = the charges (C) - rr = distance between the centers of the charges (m)
Note: the distance rr is sometimes written as dd in other references.
The point charge model. Coulomb’s Law is written for point charges: charged objects whose size is so small compared with the distance between them that you can treat all of their charge as sitting at a single point. Two charged beads a few millimeters wide and a meter apart are point charges. Two charged metal plates 1 cm apart are not, because different parts of each plate are at very different distances from the other plate. (A sphere with its charge spread evenly around its center is a special case: from outside, it acts exactly like a point charge at its center. Topic 1.6 shows why.)
On the equation sheet, Coulomb’s Law has absolute value bars, so it gives the size of the force:
∣F⃗E∣=14πϵ0∣q1q2∣r2=k∣q1q2∣r2|\vec{F}_E| = \frac{1}{4\pi\epsilon_0}\frac{|q_1q_2|}{r^2} = k\frac{|q_1q_2|}{r^2}
Find the size from the formula, then get the direction from the signs (see below). If you plug the signs in instead, like the example later on this page does, a negative answer means attractive and a positive answer means repulsive.
Scaling. The force is proportional to each charge and to 1/r21/r^2: - Double one charge: the force doubles. - Double both charges: the force goes up by a factor of 4. - Double the distance: the force drops to 1/4. - Triple the distance: the force drops to 1/9.

The Electrostatic Constant

You can write the electrostatic constant three equivalent ways:
k=14πϵ0=9×109N⋅m2/C2k = \frac{1}{4\pi\epsilon_0} = 9 \times 10^9 \mathrm{ N·m}^2\text{/C}^2
where ϵ0\epsilon_0 is the permittivity of free space (also called vacuum permittivity):
ϵ0=8.85×10−12 C2/(N⋅m2)\epsilon_0 = 8.85 \times 10^{-12} \text{ C}^2\mathrm{/(N·m}^2\text{)}

Electric Permittivity

Electric permittivity measures how much a material or medium polarizes when it sits in an electric field. Even empty space has a permittivity, ϵ0\epsilon_0, which is why ϵ0\epsilon_0 shows up in Coulomb’s Law.
Polarization is an induced rearrangement of electrons. An outside electric field pulls the electrons in each atom or molecule a little to one side and pushes the positive nuclei a little to the other side. The material ends up with its positive and negative charge slightly separated, even though its net charge is still zero. (Topic 1.2 shows how this lets a charged object attract a neutral one.)
Key points: - Free space (vacuum) has a constant permittivity, ϵ0=8.85×10−12\epsilon_0 = 8.85 \times 10^{-12} C²/(N·m²). Equations with ϵ0\epsilon_0 alone (no κ\kappa) assume the charges sit in vacuum (air is close enough). - Matter has a permittivity ϵ\epsilon that’s different from ϵ0\epsilon_0. The value depends on what the material is made of and how its atoms are arranged. - The more easily a material’s electrons can shift into new arrangements, the more the material polarizes, and the larger its permittivity.
Conductors vs. insulators. How easily charge moves inside a material also sorts materials into two groups: - Conductors (metals like copper and aluminum) are made of electrically conducting materials. Some of their electrons aren’t tied to any one atom, so these charge carriers move easily through the whole object. - Insulators (rubber, glass, plastic, dry wood) are made of electrically nonconducting materials. Their electrons stay bound to their own atoms or molecules and can’t move through the material easily. They can only shift slightly, which is how an insulator polarizes.
The ratio κ=ϵ/ϵ0\kappa = \epsilon/\epsilon_0 (on the equation sheet) is called the dielectric constant. You’ll use it with capacitors in Topic 3.4.

Electric Force vs. Gravitational Force

Coulomb’s Law looks strikingly similar to Newton’s Law of Universal Gravitation.
Gravitational force:
Fg=Gm1m2r2F_g = G\frac{m_1m_2}{r^2}
Electric force:
Fe=kq1q2r2F_e = k\frac{q_1q_2}{r^2}

Key Differences

Property
Gravitational Force
Electric Force
Nature
Always attractive
Attractive or repulsive
Depends on
Mass
Charge
Relative magnitude
Far weaker
Far stronger (about 103610^{36} times gravity for two protons)
Scale where it dominates
Large scales (planets, stars, galaxies)
Atomic, molecular, and everyday scales

Worked Example: How Much Stronger?

Two protons are 1.0×10−101.0 \times 10^{-10} m apart (about the width of an atom). Compare the electric and gravitational forces between them, using mp=1.67×10−27m_p = 1.67 \times 10^{-27} kg and G=6.67×10−11G = 6.67 \times 10^{-11} N·m²/kg² from the equation sheet.
FE=ke2r2=(9.0×109 N\cdotpm2/C2)(1.60×10−19 C)2(1.0×10−10 m)2=2.3×10−8 NF_E = k\frac{e^2}{r^2} = (9.0 \times 10^9\text{ N·m}^2\text{/C}^2)\frac{(1.60 \times 10^{-19} \text{ C})^2}{(1.0 \times 10^{-10} \text{ m})^2} = 2.3 \times 10^{-8} \text{ N}Fg=Gmp2r2=(6.67×10−11 N\cdotpm2/kg2)(1.67×10−27 kg)2(1.0×10−10 m)2=1.9×10−44 NF_g = G\frac{m_p^2}{r^2} = (6.67 \times 10^{-11} \text{ N·m}^2\text{/kg}^2)\frac{(1.67 \times 10^{-27} \text{ kg})^2}{(1.0 \times 10^{-10} \text{ m})^2} = 1.9 \times 10^{-44} \text{ N}FEFg=ke2Gmp2=1.2×1036\frac{F_E}{F_g} = \frac{ke^2}{Gm_p^2} = 1.2 \times 10^{36}
Both forces depend on 1/r21/r^2, so r2r^2 cancels in the ratio: the electric force is about 103610^{36} times stronger at any distance.

Why Gravity Wins at Large Scales

If the electric force is so much stronger, why do planets orbit because of gravity? Because large objects are almost perfectly electrically neutral. A planet has nearly equal numbers of protons and electrons, so the electric pulls and pushes between two planets cancel almost completely. Mass comes in only one kind and gravity is always attractive, so every bit of mass adds to the pull. At large scales, gravity is the force that doesn’t cancel.

Direction of the Electric Force

The direction of the electrostatic force comes down to the signs of the charges.
The rule: likes repel, opposites attract. - Repulsive when both charges have the same sign (++ or −−). - Attractive when the charges have opposite signs (+− or −+).
The force always acts along the line joining the two charges (the line of separation). Each charge pushes or pulls the other straight along that line. By Newton’s third law, the two forces are equal in size and opposite in direction, even if one charge is much bigger than the other.

How It Connects to Other Forces

The electrostatic force can create or influence other mechanical forces like: - Tension. - Friction. - Normal force.
For example, a charged object hanging below another charged object can create tension in the string connecting them, thanks to their electrostatic interaction.
Why contact forces are really electric. Normal force, friction, and tension all come from electric forces between atoms. When you push on a wall, the electron clouds of the atoms in your hand repel the electron clouds of the atoms in the wall. A rope holds together because of electric attractions between its molecules. There are far too many of these atom-to-atom interactions to add up one at a time, so in mechanics we treat their total as a single contact force. Contact forces aren’t new fundamental forces, just a convenient way to describe the combined electric forces between objects that touch.

Working with Multiple Charges

Electric force follows the principle of superposition. When several charged particles interact, you calculate the force from each pair and then combine them as vectors.

Three-Dimensional Problems with Vector Notation

You can write positions with unit vector notation: - i^\hat{i} for the x-direction - j^\hat{j} for the y-direction - k^\hat{k} for the z-direction
A position like 4i^−2j^+5k^4\hat{i} - 2\hat{j} + 5\hat{k} is the point (4, −2, 5).

The 3D Distance Formula

When charges sit apart in 3D space, find the distance with:
r=(x2−x1)2+(y2−y1)2+(z2−z1)2r = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2 + (z_2-z_1)^2}

Example Problem: Force Between Two Charges in 3D Space

Given: two charged particles. - Charge 1: q1=3×10−6q_1 = 3 \times 10^{-6} C at position (−2i^+3j^+7k^)(-2\hat{i} + 3\hat{j} + 7\hat{k}) m. - Charge 2: q2=−5×10−6q_2 = -5 \times 10^{-6} C at position (6i^+8j^−2k^)(6\hat{i} + 8\hat{j} - 2\hat{k}) m.
Find: the electric force between them.

Solution

Step 1: Calculate the distance
Using the 3D distance formula:
r=(6−(−2))2+(8−3)2+(−2−7)2r = \sqrt{(6-(-2))^2 + (8-3)^2 + (-2-7)^2}r=(8)2+(5)2+(−9)2r = \sqrt{(8)^2 + (5)^2 + (-9)^2}r=64+25+81r = \sqrt{64 + 25 + 81}r=170=13.0 mr = \sqrt{170} = 13.0 \text{ m}
Step 2: Apply Coulomb’s Law
Fe=kq1q2r2F_e = k\frac{q_1q_2}{r^2}Fe=(9×109)(3×10−6)(−5×10−6)170F_e = (9 \times 10^9)\frac{(3 \times 10^{-6})(-5 \times 10^{-6})}{170}Fe=(9×109)−15×10−12170F_e = (9 \times 10^9)\frac{-15 \times 10^{-12}}{170}Fe=−7.94×10−4 NF_e = -7.94 \times 10^{-4} \text{ N}
The read: the negative sign means the force is attractive, which checks out since the charges have opposite signs. The magnitude is about 7.9×10−47.9 \times 10^{-4} N. (Use r2=170r^2 = 170 m² straight from Step 1. Squaring the rounded 13.0 m gives 169 m², which throws off the last digit.)
Step 3: Find the direction
The force acts along the line joining the charges. The vector from q2q_2 to q1q_1 is (−2−6)i^+(3−8)j^+(7−(−2))k^=−8i^−5j^+9k^(-2 - 6)\hat{i} + (3 - 8)\hat{j} + (7 - (-2))\hat{k} = -8\hat{i} - 5\hat{j} + 9\hat{k} m. The force on q2q_2 is attractive, so it points along this vector, toward q1q_1. Divide by the vector’s length (170\sqrt{170} m) to get a unit vector, then multiply by the magnitude:
F⃗1 on 2=(7.94×10−4 N)−8i^−5j^+9k^170=(−4.87i^−3.05j^+5.48k^)×10−4 N\vec{F}_{1 \text{ on } 2} = (7.94 \times 10^{-4} \text{ N})\frac{-8\hat{i} - 5\hat{j} + 9\hat{k}}{\sqrt{170}} = (-4.87\hat{i} - 3.05\hat{j} + 5.48\hat{k}) \times 10^{-4} \text{ N}
The force on q1q_1 is the exact opposite: (4.87i^+3.05j^−5.48k^)×10−4(4.87\hat{i} + 3.05\hat{j} - 5.48\hat{k}) \times 10^{-4} N.

Example Problem: Net Force on a Charge in a Line

Given: three point charges on the x-axis. - q1=+2.0q_1 = +2.0 μC at x=0x = 0 - q2=−3.0q_2 = -3.0 μC at x=0.20x = 0.20 m - q3=+4.0q_3 = +4.0 μC at x=0.50x = 0.50 m
Find: (a) The net electric force on q2q_2. (b) The net force on q2q_2 if q1q_1 is doubled to +4.0+4.0 μC (nothing moves).
1.1 - Electric Charge and Electric Force

Solution

Step 1: The force from the first charge
The charges have opposite signs, so q2q_2 is pulled toward q1q_1, in the −x-x direction:
F1 on 2=k∣q1q2∣r122=(9.0×109 N\cdotpm2/C2)(2.0×10−6 C)(3.0×10−6 C)(0.20 m)2=1.35 NF_{1 \text{ on } 2} = k\frac{|q_1q_2|}{r_{12}^2} = (9.0 \times 10^9\text{ N·m}^2\text{/C}^2)\frac{(2.0 \times 10^{-6} \text{ C})(3.0 \times 10^{-6} \text{ C})}{(0.20 \text{ m})^2} = 1.35 \text{ N}
Step 2: The force from the third charge
Also opposite signs, so q2q_2 is pulled toward q3q_3, in the +x+x direction:
F3 on 2=(9.0×109 N\cdotpm2/C2)(4.0×10−6 C)(3.0×10−6 C)(0.30 m)2=1.20 NF_{3 \text{ on } 2} = (9.0 \times 10^9\text{ N·m}^2\text{/C}^2)\frac{(4.0 \times 10^{-6} \text{ C})(3.0 \times 10^{-6} \text{ C})}{(0.30 \text{ m})^2} = 1.20 \text{ N}
Step 3: Add them as vectors
Take +x+x as positive, and keep the third digit until the end:
Fnet=−1.35 N+1.20 N=−0.15 NF_{net} = -1.35 \text{ N} + 1.20 \text{ N} = -0.15 \text{ N}
Answer (a): 0.150.15 N in the −x-x direction, toward q1q_1. The two pulls nearly cancel: q1q_1 is the smaller charge, but it’s closer.
Step 4: Double the first charge
F1 on 2F_{1 \text{ on } 2} is proportional to q1q_1, so it doubles to 2.702.70 N. The force from q3q_3 doesn’t change:
Fnet=−2.70 N+1.20 N=−1.50 NF_{net} = -2.70 \text{ N} + 1.20 \text{ N} = -1.50 \text{ N}
Answer (b): 1.51.5 N toward q1q_1. You didn’t need Coulomb’s Law again, just the scaling.

Important Reminders

  1. Unit conversions. Watch the prefixes.
      • micro (μ\mu) = ×10−6\times 10^{-6}
      • nano (n) = ×10−9\times 10^{-9}
      • milli (m) = ×10−3\times 10^{-3}
  1. Sign convention. The sign of the force tells you attractive versus repulsive, not a vector direction in space.
  1. Superposition. For systems with more than two charges, calculate the forces pairwise, then add them as vectors.
  1. Distance matters. Always measure rr as the distance between the centers of the charged objects.

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