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2.1 - Electric Charge and Electric Force

Fundamental Properties of Charge

What Is Charge?

Charge is a fundamental property of all matter, a lot like mass. Some key facts: - Charge comes in two types: positive (+) and negative (−). - The elementary charge e=1.60×10−19e = 1.60 \times 10^{-19} C is the smallest indivisible amount. - An electron has a charge of −e-e. - A proton has a charge of +e+e. - A neutron has a charge of 00, so it’s neutral.
Charge is quantized, meaning every charge is a whole-number multiple of ee:
q=ne where n=±1,±2,±3,...q = ne \text{ where } n = \pm1, \pm2, \pm3, ...
Charge is also conserved, so the total charge in an isolated system never changes.

The Point Charge Model

A point charge is a model where the physical size of a charged object is tiny compared to the distances involved. This lets us treat charged objects as single points, which makes the calculations simpler.

Coulomb’s Law

The Force Between Charges

Coulomb’s law gives the electrostatic force between two point charges:
Fe=14πϵ0∣q1q2∣r2=k∣q1q2∣r2F_e = \frac{1}{4\pi\epsilon_0} \frac{|q_1 q_2|}{r^2} = k\frac{|q_1 q_2|}{r^2}
where: - FeF_e = magnitude of the electric force (N) - k=14πϵ0=8.99×109k = \frac{1}{4\pi\epsilon_0} = 8.99 \times 10^9 N·m²/C² (Coulomb’s constant) - q1,q2q_1, q_2 = charges (C) - rr = distance between the charges (m) - ϵ0=8.85×10−12\epsilon_0 = 8.85 \times 10^{-12} C²/(N·m²) (permittivity of free space)
Two relationships to notice: - Force is proportional to the product of the charges, so doubling either charge doubles the force. - Force is proportional to 1r2\frac{1}{r^2}, which is the inverse square law, just like gravity.

Direction of the Electric Force

The force points along the line connecting the two charges. The rule is: - Like charges repel (++ or −−), so the force pushes them apart. - Unlike charges attract (+− or −+), so the force pulls them together.
Be careful with the vector nature of this. The equation only gives you the magnitude, so you figure out the direction separately from the charge signs.
On the AP exam you’ll only have to calculate the net electric force from four or fewer charges, unless the setup is very symmetric (like equal charges spaced evenly around a circle, where a lot of the forces cancel).

Electric Force vs. Gravitational Force

Comparison Table

Property
Electric Force
Gravitational Force
Can be
Attractive or repulsive
Always attractive
Relative strength
Strong
Weak
Depends on
Charge
Mass
Law
Fe=kq1q2r2F_e = k\frac{q_1q_2}{r^2}
Fg=Gm1m2r2F_g = G\frac{m_1m_2}{r^2}

Relative Magnitudes

For objects that have both mass and charge, the electric force usually dominates:
FeFg=kq1q2Gm1m2\frac{F_e}{F_g} = \frac{kq_1q_2}{Gm_1m_2}
For two protons, Fe/Fg≈1036F_e/F_g \approx 10^{36}. For an electron and a proton it’s about 103910^{39} (see Practice Problem 3), and for two electrons it’s about 104210^{42}. Either way, the electric force is enormously stronger.
So why does gravity win at large scales? - Big objects like planets and stars are electrically neutral, with equal amounts of positive and negative charge. - Their electric forces cancel out. - Gravity is always attractive and never cancels. - So only gravity is left at astronomical scales.

Electric Permittivity

Permittivity of Materials

Electric permittivity (ϵ\epsilon) measures how much a material gets polarized by an electric field.
For free space (a vacuum): ϵ0=8.85×10−12\epsilon_0 = 8.85 \times 10^{-12} C²/(N·m²).
Inside a material: ϵ=κϵ0\epsilon = \kappa \epsilon_0, where κ\kappa is the dielectric constant, usually greater than 1.
Why do different materials have different permittivities? It comes down to what the material is made of and how its atoms and molecules are arranged. What matters most is how easily the electrons in the material can shift into a new arrangement when a field is applied: - Electrons that shift easily mean the material polarizes a lot, so its permittivity is high. - Electrons that are held tightly in place mean the material barely polarizes, so its permittivity is close to ϵ0\epsilon_0 (air is almost exactly ϵ0\epsilon_0).

Polarization

Electric polarization is the way charges rearrange inside a material when an external electric field is applied. The mechanism goes: - An external field is applied. - Electrons shift slightly toward the positive side. - This creates a separation of positive and negative charges within the atoms or molecules. - The material develops induced dipoles.
This affects the force. In a material with permittivity ϵ\epsilon:
Fe=14πϵq1q2r2F_e = \frac{1}{4\pi\epsilon} \frac{q_1 q_2}{r^2}
The force is smaller than in a vacuum by a factor of κ\kappa.

Conductors vs. Insulators

Conductors (metals, salt water): - Their charge carriers, the electrons, move freely. - They act as if their permittivity were extremely high. - Electrons rearrange easily.
Insulators (rubber, glass, plastic): - Their charge carriers can’t move easily. - Electrons are bound to atoms. - Polarization is limited.

Contact Forces Are Really Electric

Here’s a useful insight: everyday forces like the normal force, friction, and tension are all fundamentally electric forces between atoms. - When you touch a table, the electron clouds repel each other. - Friction is an interaction between the atoms on two surfaces. - What we call “contact” is really electric repulsion at the atomic scale.
Because the number of particle interactions is so huge, we just use simplified “contact force” models instead of tracking every one.

Practice Problem 1: Basic Coulomb’s Law

Problem: Two point charges, q1=+3.0×10−6q_1 = +3.0 \times 10^{-6} C and q2=−5.0×10−6q_2 = -5.0 \times 10^{-6} C, are 0.20 m apart. Find the magnitude of the electric force between them.
Solution:
Fe=k∣q1q2∣r2=(8.99×109)∣(3.0×10−6)(−5.0×10−6)∣(0.20)2F_e = k\frac{|q_1 q_2|}{r^2} = (8.99 \times 10^9) \frac{|(3.0 \times 10^{-6})(-5.0 \times 10^{-6})|}{(0.20)^2}Fe=(8.99×109)15.0×10−120.04F_e = (8.99 \times 10^9) \frac{15.0 \times 10^{-12}}{0.04}Fe=3.37 NF_e = 3.37 \text{ N}
Direction: the signs are opposite, so the force is attractive and the charges pull toward each other.

Practice Problem 2: Multiple Charges

Problem: Three charges sit on a line: q1=+2.0q_1 = +2.0 μC at x=0x = 0, q2=−4.0q_2 = -4.0 μC at x=3.0x = 3.0 cm, and q3=+3.0q_3 = +3.0 μC at x=6.0x = 6.0 cm. Find the net force on q2q_2.
Solution:
Force from q1q_1 on q2q_2: - Distance: r12=0.030r_{12} = 0.030 m. - Signs are +−, so it’s attractive and pulls left, toward q1q_1.
F12=k∣q1q2∣r122=(8.99×109)(2.0×10−6)(4.0×10−6)(0.030)2=80 N (left)F_{12} = k\frac{|q_1 q_2|}{r_{12}^2} = (8.99 \times 10^9)\frac{(2.0 \times 10^{-6})(4.0 \times 10^{-6})}{(0.030)^2} = 80 \text{ N (left)}
Force from q3q_3 on q2q_2: - Distance: r32=0.030r_{32} = 0.030 m. - Signs are +−, so it’s attractive and pulls right, toward q3q_3.
F32=k∣q3q2∣r322=(8.99×109)(3.0×10−6)(4.0×10−6)(0.030)2=120 N (right)F_{32} = k\frac{|q_3 q_2|}{r_{32}^2} = (8.99 \times 10^9)\frac{(3.0 \times 10^{-6})(4.0 \times 10^{-6})}{(0.030)^2} = 120 \text{ N (right)}
Net force, taking right as positive:
Fnet=−80+120=+40 NF_{net} = -80 + 120 = +40 \text{ N}
Answer: 40 N to the right.

Practice Problem 3: Comparing Forces

Problem: An electron and a proton are 5.0×10−115.0 \times 10^{-11} m apart, which is roughly the size of a hydrogen atom. Find: (a) the electric force between them (b) the gravitational force between them (c) the ratio Fe/FgF_e/F_g
Given: me=9.11×10−31m_e = 9.11 \times 10^{-31} kg, mp=1.67×10−27m_p = 1.67 \times 10^{-27} kg, G=6.67×10−11G = 6.67 \times 10^{-11} N·m²/kg².
Solution:
  1. Electric force:
      2. Fe=ke2r2=(8.99×109)(1.60×10−19)2(5.0×10−11)2=9.2×10−8 NF_e = k\frac{e^2}{r^2} = (8.99 \times 10^9)\frac{(1.60 \times 10^{-19})^2}{(5.0 \times 10^{-11})^2} = 9.2 \times 10^{-8} \text{ N}
  1. Gravitational force:
      2. Fg=Gmempr2=(6.67×10−11)(9.11×10−31)(1.67×10−27)(5.0×10−11)2=4.1×10−47 NF_g = G\frac{m_e m_p}{r^2} = (6.67 \times 10^{-11})\frac{(9.11 \times 10^{-31})(1.67 \times 10^{-27})}{(5.0 \times 10^{-11})^2} = 4.1 \times 10^{-47} \text{ N}
  1. Ratio:
      2. FeFg=9.2×10−84.1×10−47≈2.2×1039\frac{F_e}{F_g} = \frac{9.2 \times 10^{-8}}{4.1 \times 10^{-47}} \approx 2.2 \times 10^{39}
Answer: the electric force is about 103910^{39} times stronger. Gravity is completely negligible at atomic scales.

Practice Problem 4: Forces in Two Dimensions

Problem: Three point charges sit at the corners of a right triangle. q1=+2.0q_1 = +2.0 μC is at the origin, q2=−3.0q_2 = -3.0 μC is 0.30 m to its right on the x-axis, and q3=+4.0q_3 = +4.0 μC is 0.40 m above it on the y-axis. Find the magnitude and direction of the net electric force on q1q_1.
2.1 - Electric Charge and Electric Force
Solution:
Force from q2q_2 on q1q_1: - Distance: r21=0.30r_{21} = 0.30 m. - Signs are +−, so it’s attractive and pulls q1q_1 right, toward q2q_2 (+x).
F21=k∣q2q1∣r212=(8.99×109)(3.0×10−6)(2.0×10−6)(0.30)2=0.60 N (+x)F_{21} = k\frac{|q_2 q_1|}{r_{21}^2} = (8.99 \times 10^9)\frac{(3.0 \times 10^{-6})(2.0 \times 10^{-6})}{(0.30)^2} = 0.60 \text{ N } (+x)
Force from q3q_3 on q1q_1: - Distance: r31=0.40r_{31} = 0.40 m. - Signs are ++, so it’s repulsive and pushes q1q_1 down, away from q3q_3 (−y-y).
F31=k∣q3q1∣r312=(8.99×109)(4.0×10−6)(2.0×10−6)(0.40)2=0.45 N (−y)F_{31} = k\frac{|q_3 q_1|}{r_{31}^2} = (8.99 \times 10^9)\frac{(4.0 \times 10^{-6})(2.0 \times 10^{-6})}{(0.40)^2} = 0.45 \text{ N } (-y)
The two forces are perpendicular, so they’re already the x- and y-components of the net force: Fx=+0.60F_x = +0.60 N and Fy=−0.45F_y = -0.45 N. Combine them with the Pythagorean theorem:
Fnet=Fx2+Fy2=(0.60)2+(0.45)2=0.75 NF_{net} = \sqrt{F_x^2 + F_y^2} = \sqrt{(0.60)^2 + (0.45)^2} = 0.75 \text{ N}θ=tan⁡−1(0.450.60)=37∘\theta = \tan^{-1}\left(\frac{0.45}{0.60}\right) = 37^\circ
Answer: 0.75 N, pointing 37° below the +x axis (down and to the right).
The thing to notice: you can’t just add 0.60 N and 0.45 N. Forces in different directions have to be added as vectors, using components.

Common Mistakes

  • Sign errors. The Coulomb’s law formula uses absolute values to get the magnitude. Work out the direction separately from the charge signs, using attract versus repel.
  • Unit confusion. Charges often come in μC or nC. Convert to coulombs: 1 μC = 10−610^{-6} C.
  • Distance mistakes. Use the center-to-center distance, and convert cm to m.
  • Vector addition. With more than one charge, find each force on its own, then add them as vectors with their directions.
  • Inverse square confusion. Doubling the distance makes the force 1/4 as big, not 1/2.

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