Summary of "Electric Potential"

Core concepts and relationships

Work by a uniform force: W = F · d = F d cosθ. If force and displacement are in the same direction (θ = 0), W = F d.
Gravitational example: work done by gravity moving an object downward by distance D gives W = mgD = −mg Δy (since Δy = y_f − y_i is negative for downward displacement).
Electric force on a point charge in a uniform electric field: F = q E (direction: with E if q > 0, opposite if q < 0).
Work by an electric field for displacement D parallel to the field: W = q E D = −q E Δy, which leads to W = q (V_i − V_f) when potential is introduced.
Electric potential V (scalar) is introduced via W = q ΔV, where ΔV = V_f − V_i. For a uniform-field slab, ΔV = E Δy (or ΔV = −E d depending on sign conventions—use a picture to fix signs).
Potential of a point charge: V(r) = k q / r (scalar). For negative q, V is negative. Reference chosen so V(∞) = 0.
Equipotential surfaces: surfaces (or lines in 2D) where V is constant. For a point charge these are concentric spheres (circles in 2D); for a dipole they curve and are everywhere perpendicular to E. A charge moved along an equipotential experiences zero work.
Superposition principle for potential: total V is the algebraic sum of potentials from individual point charges (no vector addition needed).

Volt (V) = Joule per Coulomb (J/C).
Electron-volt (eV): energy gained by an electron accelerated through 1 V; 1 eV ≈ 1.6×10⁻¹⁹ J.
Capacitance unit: Farad (F) = C/V.

Deriving electric potential from analogy with gravity
- Start with gravitational work: W = mg D.
- Express displacement D in terms of Δy (D = −Δy because y_f < y_i when moving down).
- Rearrange: W = −mg Δy = mg (y_i − y_f).
- Replace mg with q E for the electric case (force on a charge in a uniform E).
- Rearrange and identify V_i = E y_i and V_f = E y_f, so W = q (V_i − V_f) → W = q ΔV.
- Conclude ΔV = E Δy and that V has units of volts.
Finding potential difference from kinetic energy given to a charge (example)
- Use the relation ΔK = q ΔV.
- Solve for ΔV: ΔV = ΔK / |q| (use magnitudes for algebra).
- Convert units as needed (J to eV if desired).
Calculating potential at the center of a symmetric charge arrangement (square)
- For point charges: V_total = Σ k q_i / r_i.
- Compute distances from geometry (e.g., center-to-corner distance for a square: r = L/√2).
- Sum contributions algebraically, including signs of charges.
Parallel-plate capacitor: field and charge calculations
- Field between large parallel plates (neglecting fringing): E = Q / (ε0 A) (or E = σ / ε0).
- Potential difference between plates: ΔV = E d.
- Capacitance: C = Q / V.
- To find Q for a desired E or ΔV: Q = ε0 A E or Q = C ΔV.
Electric field → required voltage for a spark gap (practical)
- Given E and gap d: ΔV = E d.
- This shows why ignition coils boost 12 V to several kilovolts to fire spark plugs.

Electron gaining kinetic energy: ΔK = 6.6×10⁻¹⁶ J → ΔV ≈ 4.1×10³ V.
Potential at center of a given square-charge arrangement (three +6 μC, one −3 μC, square side 2 cm): V_center ≈ 9.53×10⁶ V.
Required charge on each plate for E = 8.5×10⁵ V/m, area A = 45 cm², separation d = 2.45 mm: Q ≈ 3.4×10⁻⁸ C.
Spark-gap example: E = 2.8×10⁶ V/m, d = 0.75 mm → ΔV ≈ 2.1×10³ V (≈ 2.1 kV).

Electric potential V is a scalar; add potentials algebraically. This is simpler than adding the vector electric field E.
Equipotentials are always perpendicular to E; moving along an equipotential requires no work.
Use sketches and a test charge to determine sign and direction of forces and potential differences (like charges repel, opposites attract).
Pay attention to sign conventions and unit conversions—V = J/C, and 1 eV = 1.6×10⁻¹⁹ J.
Capacitors store charge; capacitance measures the ability to store charge per unit voltage. Real parallel-plate fields fringe at edges; the simple formulas assume plate area ≫ separation.