Short on time? This page is the recap. For the derivations, Figure 1 and the worked examples, use the full study guide. The integrals here are part of the calculus-based course.
Recap
- Electric potential V at a point is the potential energy per unit charge a test charge would have there. It depends only on the source charges.
- Potential difference ΔV = V_B − V_A is the change in potential energy per unit charge between two points.
- 1 V = 1 J/C, and 1 V/m = 1 N/C.
- V = 0 at infinity for charge of finite size; for an infinite line or cylinder use differences only.
- Equipotentials are perpendicular to field lines; no work is done moving along one.
- Batteries keep a potential difference between their terminals by separating charge chemically.
Key relationships
| Situation | Relationship |
|---|---|
| Definition | V = U/q; ΔU = qΔV |
| Point charge | V = kq/r |
| Several point charges | V = Σ kq_i/r_i (scalar sum) |
| Continuous charge | V = ∫ k dq/r |
| Ring on its axis | V = kQ/√(R² + x²) |
| Arc of any angle, at its centre | V = kQ/R |
| Rod on its line, d beyond one end | V = kλ ln[(d + L)/d] |
| Rod on its perpendicular bisector | V = kλ ln{[√(L²/4 + y²) + L/2]/[√(L²/4 + y²) − L/2]} |
| Long wire or outside a long cylinder | V(a) − V(b) = (λ/2πε₀) ln(b/a) |
| Field to potential | V_B − V_A = −∫_A^B E·dl; uniform field: ΔV = −EΔx |
| Potential to field | E_x = −dV/dx |
Mistakes to avoid
- Resolving V into components. It is a scalar.
- V = 0 ⇒ E = 0. E depends on the slope of V, not its value.
- 1/r² in V. Potential goes as 1/r.
- Dropping minus signs in ΔV = −∫E·dl and E_x = −dV/dx.
- V(∞) = 0 for an infinite line. It diverges.
- Centimetres in V/m. Convert to metres before dividing.
Quick self-check
- V at 0.30 m from +2.0 nC? (59.9 V)
- How much energy does a 9 V battery give to 2.0 C of charge passing through it? (18 J)
- Equipotentials 20 V apart are 0.050 m apart. Estimate E. (400 V/m, from high V to low V)
- Where does V = 0 between +q and −q? (Midway; E there is not zero)
Next: practice questions.