Electrical Charges and Coulomb's Law Notes

Fundamental building blocks of matter are atoms, which consist of:
- Electrons (negative charge)
- Neutral neutrons (no charge)
- Protons (positive charge)
Neutral atom - electron = Positive ion
1 electron charge = $-1.602 \times 10^{-19} C$
Neutral atom + electron = Negative ion

Electrostatics: Study of electric charge at rest.
Electric charge: A fundamental property of matter.
- Two types:
  - Positive charge: Every proton has a single positive charge.
  - Negative charge: Every electron has a single negative charge.

Electron and proton have the same amount of charge, but opposite signs.
Electronic charge (e): The magnitude of the charge of a single proton or electron, generally considered a positive value. We add the negative sign when we need to: $q = -e$ ; $q = +e$ .
Charge and Mass:
- Electron (e):
  - Charge: $-1.6021917 \times 10^{-19} C$
  - Mass: $9.1095 \times 10^{-31} kg$
- Proton (p):
  - Charge: $+1.6021917 \times 10^{-19} C$
  - Mass: $1.67261 \times 10^{-27} kg$
- Neutron (n):
  - Charge: 0
  - Mass: $1.67492 \times 10^{-27} kg$

Two kinds of charges exist in nature:
- Like charges repel each other.
- Unlike charges attract each other.
Charge is conserved.
Charge is quantized.

Electrostatic Attraction/Repulsion:
- Attraction: + -
- Repulsion: + +
- Repulsion: - -

An object becomes electrostatically charged by:
1. Friction: Transfers electrons between two objects in contact.
2. Contact: Transfer of electrons from a charged body.
3. Induction: Charge redistribution of electrons in a material.

Four fundamental forces of nature:
- Gravitational Force
- Electromagnetic Force
- Strong Nuclear Force
- Weak Nuclear Force

Electric field vector $\vec{E}$ at a point in space: Electric force $\vec{F_e}$ acting on a positive test charge $q$ placed at that point, divided by the test charge.
Magnitude of the electric force: $F = K \frac{q \mid q \mid}{r^2}$
Coulomb's constant: $k = 8.9875 \times 10^9 Nm^2/C^2$
Electric Field Formula: $\vec{E} = \frac{\vec{F}}{q}$
Electric Field Formula: $E = K \frac{\mid q \mid}{r^2}$
Force on a charge in an electric field: $\vec{F_1} = q \vec{E}$
Units of Electric Field: Newtons per coulomb (N/C)

Electric field lines for a point charge:
- Positive point charge: Lines directed radially outward.
- Negative point charge: Lines directed radially inward.
Electric field lines extend away from positive charge (originate) and towards negative charge (terminate).

$\vec{F_{12}}$ = force on 1 due to 2
$\vec{F_{21}}$ = force on 2 due to 1
Direction of the force depends on whether the charges have the same sign or opposite signs.

Coulomb's law equation for the magnitude of the electric force (Coulomb force) between two point charges: $Fe = k \frac{q1 q_2}{r^2}$ where
- $k$ is the Coulomb constant
- $q1$ and $q2$ are the charges
- $r$ is the distance between the charges
The force decreases with distance between the charges, similar to gravity.

$Fe = k \frac{q1 q_2}{r^2}$
The value of the Coulomb constant depends on the choice of units. The SI unit of charge is the coulomb (C).
Coulomb constant $ke$ in SI units: $ke = 8.9875 \times 10^9 N \cdot m^2/C^2$
Also written as:
$ke = \frac{1}{4 \pi \epsilon0}$
where the constant $\epsilon0$ is known as the permittivity of free space: $\epsilon0 = (8.854187817) \times 10^{-12} C^2/(N \cdot m^2)$
Conversions:
- 1 Coulomb = $10^6$ microCoulomb
- 1 Coulomb = $10^9$ nanoCoulomb

Average distance $r$ between the electron and proton in a hydrogen atom: $5.3 \times 10^{-11}$ m.
- (a) Magnitude of the average electrostatic force?
- (b) Magnitude of the average gravitational force?
Solution:
- Electrostatic force:
 $Fe = K \frac{q1 q_2}{r^2} = \frac{(8.99 \times 10^9 N \cdot m^2/C^2) (1.60 \times 10^{-19} C)^2}{(5.3 \times 10^{-11} m)^2} = 8.2 \times 10^{-8} N$
- Gravitational force:
 $FG = G \frac{m1 m_2}{r^2} = \frac{(6.67 \times 10^{-11} N \cdot m^2/kg^2)(9.11 \times 10^{-31} kg)(1.67 \times 10^{-27} kg)}{(5.3 \times 10^{-11} m)^2} = 3.61 \times 10^{-47} N$

The nucleus of an iron atom has a radius of about $4 \times 10^{-15}$ m and contains 26 protons. What repulsive electrostatic force acts between two protons in such a nucleus if a distance of one radius separates them?
Solution:
$Fe = K \frac{q1 q_2}{r^2} = \frac{(8.99 \times 10^9 N \cdot m^2/C^2)(1.60 \times 10^{-19} C)^2}{(4 \times 10^{-15} m)^2} = 14.4 N$

Two balloons with charges of $+3.37 \mu C$ and $-8.21 \mu C$ attract each other with a force of 0.0626 N. Determine the separation distance between the two balloons.
Solution:
- Given:
 - $q_1 = +3.37 \mu C = +3.37 \times 10^{-6} C$
 - $q_2 = -8.21 \mu C = -8.21 \times 10^{-6} C$
 - F = -0.0626 N (attractive force, so negative sign)
 - $k = 8.99 \times 10^9 N \cdot m^2/C^2$
- Find: r = ???
 $Fe = K \frac{q1 q2}{r^2}$ $r^2 = K \frac{q1 q2}{Fe}$
 $r = \sqrt{K \frac{q1 q2}{F_e}} = \sqrt{\frac{(8.99 \times 10^9 N \cdot m^2/C^2) (-8.21 \times 10^{-6} C)(+3.37 \times 10^{-6} C)}{(-0.0626 N)}} = \sqrt{3.98 m^2} = 1.99 m$