Two Types:

• Positive (+): Protons
• Negative (-): Electrons

Key Properties:

1. Like charges repel, opposite charges attract
2. Charge is quantized: Comes in multiples of elementary charge $e = 1.60 \times 10^{-19} \text{ C}$
3. Charge is conserved: Total charge in isolated system is constant
4. Charge is invariant: Doesn't depend on reference frame

Units of Charge

Coulomb (C): SI unit of electric charge

Elementary charge: $e = 1.60 \times 10^{-19}$ C

• Charge of proton: $+e$
• Charge of electron: $-e$

Any charge: $q = ne$ where $n$ is an integer

Charging Methods

1. Friction (Triboelectric Effect)

• Rubbing transfers electrons
• Example: Rubbing balloon on hair

2. Conduction (Contact)

• Direct contact transfers charge
• Charge distributes between objects

3. Induction

• Charge separation without contact
• Grounding removes one type of charge

Conductors vs. Insulators

Conductors

• Allow charges to move freely
• Examples: Metals, graphite, salt water
• Excess charge distributes on surface

Insulators (Dielectrics)

• Charges cannot move freely
• Examples: Rubber, glass, plastic, wood
• Charge stays where placed

Semiconductors

• Properties between conductors and insulators
• Examples: Silicon, germanium

Coulomb's Law

Coulomb's Law gives the electrostatic force between two point charges:

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

where:

• $F$ = electrostatic force (N)
• $k = 8.99 \times 10^9$ N·m²/C² (Coulomb's constant)
• $q_1, q_2$ = charges (C)
• $r$ = distance between charges (m)

Vector Form:

$\vec{F} = k\frac{q_1 q_2}{r^2}\hat{r}$

Direction:

• Same signs → repulsive (away from each other)
• Opposite signs → attractive (toward each other)

Coulomb's Constant

$k = \frac{1}{4\pi\epsilon_0} = 8.99 \times 10^9 \text{ N·m}^2\text{/C}^2$

where $\epsilon_0 = 8.85 \times 10^{-12}$ C²/(N·m²) is the permittivity of free space.

Often approximated as: $k \approx 9.0 \times 10^9$ N·m²/C²

Comparison with Gravity

Similarities:

• Both are inverse square laws: $F \propto 1/r^2$
• Both are long-range forces
• Both act along line connecting objects

Differences:

Electric force is ~10³⁹ times stronger than gravity!

Superposition Principle

For multiple charges, total force = vector sum of individual forces:

$\vec{F}_{total} = \vec{F}_1 + \vec{F}_2 + \vec{F}_3 + ...$

Each force calculated using Coulomb's Law independently.

Steps:

1. Calculate force from each charge separately
2. Determine direction of each force
3. Resolve into components if needed
4. Add vectors (components)

Problem-Solving Strategy

1. Draw diagram showing all charges and distances
2. Identify charge signs (+ or -)
3. Apply Coulomb's Law for each pair
4. Determine force directions:
   • Like charges → repel
   • Unlike charges → attract
5. Use superposition for multiple charges
6. Add as vectors (use components if needed)

Common Mistakes

❌ Forgetting to square the distance in denominator ❌ Using wrong sign convention (use magnitude, then add direction) ❌ Not converting units (must use meters, coulombs) ❌ Adding forces as scalars instead of vectors ❌ Confusing force on A from B with force on B from A (equal magnitude, opposite direction)

Find: Force $F$

Solution:

Apply Coulomb's Law: $F = k\frac{|q_1 q_2|}{r^2}$ $F = (9.0 \times 10^9)\frac{|(3.0 \times 10^{-6})(-2.0 \times 10^{-6})|}{(0.50)^2}$ $F = (9.0 \times 10^9)\frac{6.0 \times 10^{-12}}{0.25}$ $F = (9.0 \times 10^9)(2.4 \times 10^{-11})$ $F = 0.216 \text{ N} \approx 0.22 \text{ N}$

Direction: Opposite signs → attractive force

Answer: 0.22 N, attractive

Find: Net force on $q_2$

Solution:

Force from q₁ on q₂:

Distance: $r_{12} = 0.30$ m $F_{12} = k\frac{|q_1 q_2|}{r_{12}^2} = (9.0 \times 10^9)\frac{(2.0 \times 10^{-6})(3.0 \times 10^{-6})}{(0.30)^2}$ $F_{12} = (9.0 \times 10^9)\frac{6.0 \times 10^{-12}}{0.09} = 0.60 \text{ N}$

Direction: Opposite signs → attractive → toward q₁ (left, negative) $\vec{F}_{12} = -0.60 \text{ N}$

Force from q₃ on q₂:

Distance: $r_{23} = 0.60 - 0.30 = 0.30$ m $F_{23} = k\frac{|q_2 q_3|}{r_{23}^2} = (9.0 \times 10^9)\frac{(3.0 \times 10^{-6})(1.0 \times 10^{-6})}{(0.30)^2}$ $F_{23} = (9.0 \times 10^9)\frac{3.0 \times 10^{-12}}{0.09} = 0.30 \text{ N}$

Direction: Opposite signs → attractive → toward q₃ (right, positive) $\vec{F}_{23} = +0.30 \text{ N}$

Net force (superposition): $F_{net} = F_{12} + F_{23} = -0.60 + 0.30 = -0.30 \text{ N}$

Answer: 0.30 N to the left (toward q₁)

Find: Net force on $q_2$

Solution:

Force from q₁ on q₂:

Distance: $r_{12} = 0.30$ m $F_{12} = k\frac{|q_1 q_2|}{r_{12}^2} = (9.0 \times 10^9)\frac{(2.0 \times 10^{-6})(3.0 \times 10^{-6})}{(0.30)^2}$ $F_{12} = (9.0 \times 10^9)\frac{6.0 \times 10^{-12}}{0.09} = 0.60 \text{ N}$

Direction: Opposite signs → attractive → toward q₁ (left, negative) $\vec{F}_{12} = -0.60 \text{ N}$

Force from q₃ on q₂:

Distance: $r_{23} = 0.60 - 0.30 = 0.30$ m $F_{23} = k\frac{|q_2 q_3|}{r_{23}^2} = (9.0 \times 10^9)\frac{(3.0 \times 10^{-6})(1.0 \times 10^{-6})}{(0.30)^2}$ $F_{23} = (9.0 \times 10^9)\frac{3.0 \times 10^{-12}}{0.09} = 0.30 \text{ N}$

Direction: Opposite signs → attractive → toward q₃ (right, positive) $\vec{F}_{23} = +0.30 \text{ N}$

Net force (superposition): $F_{net} = F_{12} + F_{23} = -0.60 + 0.30 = -0.30 \text{ N}$

Answer: 0.30 N to the left (toward q₁)

Find: Net force on one corner charge

Solution:

Consider force on charge at origin (0,0). Other charges at:

• (a, 0): distance = a
• (0, a): distance = a
• (a, a): distance = $a\sqrt{2}$ (diagonal)

Force from adjacent charges (two): $F_{adj} = k\frac{Q^2}{a^2} = (9.0 \times 10^9)\frac{(5.0 \times 10^{-6})^2}{(0.20)^2}$ $F_{adj} = (9.0 \times 10^9)\frac{2.5 \times 10^{-11}}{0.04} = 5.625 \text{ N}$

Each points along edge (90° angle between them).

Force from diagonal charge: $F_{diag} = k\frac{Q^2}{(a\sqrt{2})^2} = k\frac{Q^2}{2a^2} = \frac{F_{adj}}{2} = 2.813 \text{ N}$

Points along diagonal toward corner.

Components:

From right: $\vec{F}_1 = (5.625, 0)$ N From top: $\vec{F}_2 = (0, 5.625)$ N From diagonal: At 45°, points toward (along diagonal from origin) $$\vec{F}_3 = (2.813\cos 45°, 2.813\sin 45°) = (1.99, 1.99)$ N

Total: $F_x = 5.625 + 0 + 1.99 = 7.62 \text{ N}$ $F_y = 0 + 5.625 + 1.99 = 7.62 \text{ N}$

$F_{net} = \sqrt{F_x^2 + F_y^2} = \sqrt{7.62^2 + 7.62^2} = 10.8 \text{ N}$

$\theta = \tan^{-1}\frac{F_y}{F_x} = 45°$

Answer: 10.8 N at 45° from the two adjacent sides (along diagonal outward)

By symmetry, this makes sense - force points directly away from center!

Find: Net force on one corner charge

Solution:

Consider force on charge at origin (0,0). Other charges at:

• (a, 0): distance = a
• (0, a): distance = a
• (a, a): distance = $a\sqrt{2}$ (diagonal)

Force from adjacent charges (two): $F_{adj} = k\frac{Q^2}{a^2} = (9.0 \times 10^9)\frac{(5.0 \times 10^{-6})^2}{(0.20)^2}$ $F_{adj} = (9.0 \times 10^9)\frac{2.5 \times 10^{-11}}{0.04} = 5.625 \text{ N}$

Each points along edge (90° angle between them).

Force from diagonal charge: $F_{diag} = k\frac{Q^2}{(a\sqrt{2})^2} = k\frac{Q^2}{2a^2} = \frac{F_{adj}}{2} = 2.813 \text{ N}$

Points along diagonal toward corner.

Components:

From right: $\vec{F}_1 = (5.625, 0)$ N From top: $\vec{F}_2 = (0, 5.625)$ N From diagonal: At 45°, points toward (along diagonal from origin) $$\vec{F}_3 = (2.813\cos 45°, 2.813\sin 45°) = (1.99, 1.99)$ N

Total: $F_x = 5.625 + 0 + 1.99 = 7.62 \text{ N}$ $F_y = 0 + 5.625 + 1.99 = 7.62 \text{ N}$

$F_{net} = \sqrt{F_x^2 + F_y^2} = \sqrt{7.62^2 + 7.62^2} = 10.8 \text{ N}$

$\theta = \tan^{-1}\frac{F_y}{F_x} = 45°$

Answer: 10.8 N at 45° from the two adjacent sides (along diagonal outward)

By symmetry, this makes sense - force points directly away from center!