Electric Charge and Coulomb's Law

Overview

Lecture covers second-semester physics topics: electric charge, Coulomb's law, electric force, electric field, electric potential energy, and electric potential (voltage).
Emphasis on conceptual understanding, vector addition, and problem-solving examples.

Force scales linearly with each charge: doubling one charge doubles F; doubling both quadruples F.
Force scales with inverse square of distance: doubling r reduces F by 1/4.

If F = 100 N at r = 1 m, doubling r to 2 m yields F = 100 × (1/2^2) = 25 N.
Combined changes example:
- q1 doubled (×2), q2 tripled (×3), r reduced by factor 3 (r → r/3).
- F multiplies by 2·3 / (1/3)^2 = 6 / (1/9) = 54.
- Original 500 N → new F = 500 × 54 = 27,000 N.

q1 = 200 μC = 200 × 10^-6 C; q2 = 300 μC = 300 × 10^-6 C; r = 80 cm = 0.80 m.
Use F = k q1 q2 / r^2; result ≈ 843.75 N (magnitude).
Direction: depends on sign of charges (attraction for opposite signs, repulsion for like signs).

One electron charge: -1.6 × 10^-19 C; one proton: +1.6 × 10^-19 C.
A net positive q means more protons than electrons; net negative means more electrons.
Converting charge to excess protons: N = Q / (1.6 × 10^-19 C).
- Example: +100 μC → 100 × 10^-6 / 1.6×10^-19 ≈ 6.25 × 10^14 excess protons.
To find total protons when electrons known: total protons = excess protons + electrons present.

Net charge q = (number of protons − number of electrons) × elementary charge magnitude.
Example: 2.5×10^16 protons and 4.8×10^16 electrons → Δ = 2.5−4.8 = −2.3 ×10^16 times 1.6×10^-19 → q ≈ −3.68×10^-3 C (−3.68 mC).

Net electric force on a charge equals vector sum of forces from each other charge.
Determine direction by:
- Finding individual force directions (attraction/repulsion).
- Considering magnitudes via Coulomb’s law (closer charges produce stronger forces).
- Adding vectors (head-to-tail or component method).
Example configurations:
- Three charges equal magnitude, equal spacing: resultant directions found by adding left/right and up/down components.
- For symmetric setups, some components cancel; remaining components determine net direction (e.g., southwest, northeast).

Electric field E defined as force per unit positive test charge: E = F / q (units N/C).
E gives force per coulomb; F on a test charge q_test is F = q_test·E.
Direction:
- Field points away from positive charges; towards negative charges.
- A positive test charge experiences force in same direction as E.
- A negative test charge experiences force opposite to E.

Electric field at a point is vector sum of fields from each charge: E_total = Σ k·qi / r_i^2 (direction considered).
Use relative distances and signs to compare magnitudes:
- Closer charges contribute larger fields.
- Positive charges contribute vectors away; negative charges contribute toward.
Use components to sum when directions are not collinear; symmetric components may cancel.

Between a positive left plate and negative right plate, field points from positive to negative (left → right).
A positive charge placed between plates accelerates in direction of E (toward negative plate).

Work done by force W = F · Δx (dot product); sign depends on orientation of F and displacement.
- If force and displacement parallel → W positive → KE increases → PE decreases.
- If anti-parallel → W negative → KE decreases → PE increases.
Analogy: ball falling (gravity does positive work, KE increases, PE decreases) vs ball thrown upward (gravity does negative work).

Electric potential energy (EPE) relates to configuration of charges; depends on position.
Electric potential V (voltage) defined as V = (electric potential energy) / q (unit: volt).
- 1 V = 1 J/C.
Only potential differences ΔV are physically measurable and relevant.
Voltage (ΔV) between two points tells work per unit charge when moving between them.

Quantity	Relation/Unit
V	Electric potential (V) = U/q; unit volt = J/C
ΔV	Potential difference (voltage) between points
W	Work by electric force moving charge q from a to b: W = −q·(Va − Vb)

Example: A 1 C charge moving from V=5 V to V=0 V gains 5 J of kinetic energy (electric field does +5 J work).
Scaling: a 2 C charge moving through ΔV = 5 V gains or loses 10 J (2×5).

Work done by electric force moving a charge from a to b: W = −q·(Va − Vb) = −q·ΔV (sign and order matter).
Example with negative charge:
- Charge q = −4 C moving from Va = 50 V to Vb = 0 V.
- W = −(−4 C)·(50 − 0) = +4·50 but careful with order: for a→b use W = −q·(Vb − Va) → W = −(−4)·(0−50) = −200 J.
- Result: work by electric force is −200 J (electric force did negative work while external force did +200 J).

Electric charge (q): intrinsic property, measured in coulombs (C).
Coulomb’s law: F = k q1 q2 / r^2.
Electric field (E): force per unit positive charge, E = F/q (N/C).
Electric potential energy (U or EPE): energy associated with charge position in field.
Electric potential (V): potential energy per unit charge, V = U/q (volt = J/C).
Voltage (ΔV): difference in electric potential between two points.
Work (W): W = F·d; for electric forces W = −q·ΔV (depending on direction/order).

Practice vector addition of force and field contributions for multiple-charge setups.
Solve Coulomb law problems with unit conversions (μC → C, cm → m).
Practice computing E from forces and computing forces from E.
Apply W = −q·ΔV on varied sign-charge and plate-potential problems.
Review analogies between gravitational potential/energy and electric potential/energy for sign and work interpretation.