Coulomb Force & Electric Fields - Lecture 2 Notes

Recap of the Coulomb force as the electric force between two charged particles. This is an action at a distance, interpreted as a field force.
Field forces can act instantaneously without matter in between.
Gravity is another field force that doesn't require contact.
Modern physics explains magnetic field force as an exchange of photons, which are particles propagating the electric force.

All field forces have a particle mediating them (e.g., medicine ball analogy).
Particles in physics are classified as bosons or fermions.
Bosons: unlimited number can occupy the same quantum state.
Fermions: cannot have an infinite number in the same quantum state. This explains electron shells around an atom's nucleus, where electrons occupy different orbitals because they are fermions.
Photons are bosons; in a laser, photons have the same wavelength and direction, making it a coherent light source.
The nature of the magnetic force is propagated by the exchange of photons between charged particles.
All fundamental forces (strong, weak, electromagnetic, gravity) have exchange particles, with the graviton theorized for gravity, but not yet observed due to gravity's weakness.

Electric field exists in the space around a charged object (source charge).
Another charged object (test charge) entering the field experiences a force.
The electric force is a vector, implying the electric field is also a vector. $\vec{F} = q \vec{E}$ or $\vec{E} = \frac{\vec{F}}{q}$
A source charge creates an electric field, and a test charge within it experiences an electric force.
The concept works best with point charges to avoid complications from the size and shape of charged objects.

SI units of electric field are Newtons per Coulomb (N/C), equivalent to Volts per meter (V/m).
Direction of the electric field is radially outwards from a positive point source charge and radially inwards to a negative point charge.

Calculating the electric field at the position of the electron in a hydrogen atom.
Given: a proton (positive charge) and the position of the electron.
Calculate the electric field magnitude: $E = k \frac{q}{r^2}$ where $k = 9 \times 10^9 \frac{N \cdot m^2}{C^2}$ , $q$ is the charge of the proton, and $r = 5.3 \times 10^{-11} m$ is the distance.
$E = (9 \times 10^9) \frac{1.6 \times 10^{-19}}{(5.3 \times 10^{-11})^2} = 5.14 \times 10^{11} N/C$
The direction is radially outwards from the proton (positive charge).
Calculate the force on the electron at this position: $F = qE$ , where $q = -1.6 \times 10^{-19} C$ (electron charge).
$F = (-1.6 \times 10^{-19}) (5.14 \times 10^{11}) = -8.224 \times 10^{-8} N$
The negative sign indicates the force is opposite to the field, i.e., inwards.

Calculating the electric field at the origin due to two charges: +4μC and -4μC, separated by 5mm.
The electric field due to charge 1: $E1 = k \frac{q1}{r^2}$ where $q_1 = 4 \times 10^{-6} C$ and $r = 2.5 \times 10^{-3} m$ .
$E_1 = (9 \times 10^9) \frac{4 \times 10^{-6}}{(2.5 \times 10^{-3})^2} = 5.76 \times 10^9 N/C$ , direction is outwards (positive x-axis).
The electric field due to charge 2: $E2 = k \frac{q2}{r^2}$ , where $q_2 = -4 \times 10^{-6} C$ and $r = 2.5 \times 10^{-3} m$ .
$E_2 = (9 \times 10^9) \frac{4 \times 10^{-6}}{(2.5 \times 10^{-3})^2} = 5.76 \times 10^9 N/C$ , direction is inwards (positive x-axis).
Total electric field: $E{total} = E1 + E_2 = 2 \times 5.76 \times 10^9 = 1.152 \times 10^{10} N/C$ along the positive x-axis.

Calculating the electric field due to a charged rod of length L with a charge density λ (charge per unit length) at a point P on the same axis.
Differential element: $dq = λ dx$ .
Differential electric field element: $dE = k \frac{dq}{x^2} = k \frac{λ dx}{x^2}$ .
Integrate to find the total electric field: $E = \int dE = \int{a}^{a+L} k \frac{λ dx}{x^2} = kλ \int{a}^{a+L} \frac{dx}{x^2}$ .
$E = kλ[-\frac{1}{x}]_{a}^{a+L} = kλ(\frac{1}{a} - \frac{1}{a+L})$ where $a$ is the distance from the end of the rod to point P.
If the point P is off-axis, the calculation becomes complicated, requiring vector components and more complex integration.

Pictorial representation of electric fields.
Tangent to the electric field vector.
Density is proportional to the magnitude of the electric field (more lines indicate a stronger field).
Field lines begin on positive charges and end on negative charges (or at infinity).
Demonstration: Van de Graaff generator charging hair, causing it to stand up radially due to repulsion of like charges.
Field lines must be straight and radial from a single point charge; otherwise, it violates Coulomb's law.

Electric dipole: a pair of equal and opposite charges separated by a distance.
Many biological molecules (e.g., HF, $H_2O$ ) behave as electric dipoles.
Dipole moment (p) is a vector from the negative to the positive charge.
Polar molecules (like water) can be manipulated with electric fields, causing rotation or vibration.
Carbon dioxide ( $CO_2$ ) is not polar because the dipole moments of the two C=O bonds cancel out.