Overview
- Topic: Electric fields from point charges, direction, magnitude, and applications.
- Focus: Definitions, formulas, direction rules, worked examples, and problem-solving strategies.
- Units: Electric field in newtons per coulomb (N/C).
Key Equations and Constants
- Electric field definition: E = F / q (electric force divided by test charge).
- Coulomb's law (force between two point charges): F = k * q1 * q2 / r^2.
- Electric field of a point charge (magnitude): E = k * Q / r^2.
- Relation between force and field: F = q * E.
- Newton's second law: F_net = m * a.
- Kinematic relation used: v_f^2 = v_i^2 + 2 a d.
- Constants:
- k = 9.0 × 10^9 N·m^2/C^2.
- Electron mass = 9.11 × 10^-31 kg.
- Proton mass = 1.67 × 10^-27 kg.
- Elementary charge magnitude = 1.602 × 10^-19 C.
- g ≈ 9.8 m/s^2.
Direction Rules For Electric Field Vectors
- Field vectors point away from positive charges.
- Field vectors point toward negative charges.
- A positive test charge experiences force in the same direction as E.
- A negative test charge experiences force opposite to E.
Formula Derivations and Unit Check
- Derive E = kQ/r^2 by substituting Coulomb's law into E = F/q and cancelling the test charge.
- Units: k has N·m^2/C^2; after cancellations, E units become N/C.
Worked Examples — Summary Table
| Problem | Given / Setup | Key Steps / Formula | Result (magnitude & direction) |
| 1 | F = 100 N north on q = -20 μC | E = F / |q| | E = (100)/(20×10^-6) = 5.0×10^6 N/C, direction south (since q negative and force north) |
| 2 | q = +50 μC in E = 5.0×10^4 N/C upward; suspended | mg = qE => m = qE/g | m = (50×10^-6 × 5.0×10^4)/9.8 ≈ 0.255 kg (upward electric force balances weight) |
| 3 | Electron accelerates east at a = 4.0×10^6 m/s^2 | m a = q E => E = m a / |q| | E ≈ (9.11×10^-31 × 4.0×10^6)/(1.602×10^-19) = 2.27×10^-5 N/C, direction west (electron accelerated east → E is west) |
| 4a | Q = +40 μC at origin; point P at x = +5 m | E = kQ/r^2 | E = 9×10^9×40×10^-6 / 5^2 = 1.44×10^4 N/C, direction east |
| 4b | Same Q; point S at (3,4) m from origin | r = √(3^2+4^2)=5 m; E magnitude as 4a; θ = arctan(4/3) | E = 1.44×10^4 N/C, direction 53.1° north of east |
| 5a | Electron in E = 2.0×10^4 N/C west; mass m_e known | a = qE/m => a = (|q|E)/m | a ≈ (1.602×10^-19 × 2.0×10^4)/(9.11×10^-31) = 3.517×10^15 m/s^2 (electron accelerates east since q negative) |
| 5b | Same electron, distance d = 1 cm between plates | v_f = √(2 a d) | v_f ≈ √(2 × 3.517×10^15 × 0.01) ≈ 8.39×10^6 m/s |
| 6a | q1 = +200 μC at x=0; q2 = -300 μC at x=1 m; midpoint x=0.5 m | E_net = E1 + E2 (both point east at midpoint) | r = 0.5 m; E_net = k(200+300)×10^-6 / (0.5)^2 = 1.8×10^7 N/C east |
| 6b | Same charges; point B at x = 1 + 0.3 = 1.3 m from origin? (30 cm right of negative charge) | Compute E1 and E2 with respective distances and signs | E_net ≈ -2.89×10^7 N/C (negative → west), because E2 dominates |
| 7a | E at x = 2 m is 100 N/C; double Q | E ∝ Q | New E = 200 N/C |
| 7b | Double r (distance) | E ∝ 1/r^2 | New E = 100/4 = 25 N/C |
| 7c | Reduce r by factor 3 (r → r/3) | E increases by 3^2 = 9 | New E = 100 × 9 = 900 N/C |
| 7d | Triple Q and r → r/4 | Scale factor = 3 × (1/(1/4)^2) = 3×16 = 48 | New E = 100 × 48 = 4.8×10^3 N/C |
| 8a | Two identical charges +100 μC separated by 1 m | Symmetry: midpoint fields cancel | Net E = 0 at midpoint (between charges) |
| 8b | q2 doubles to 200 μC; find position x from q1 with E_net = 0 | Set k q1 / x^2 = k q2 / (1-x)^2; solve for x | x ≈ 0.414 m to the right of q1 (between q1 and midpoint) |
Problem-Solving Strategies
- Determine direction first using sign of source charge and test-charge behavior.
- For multiple charges, compute vector sum of fields (consider direction sign).
- Convert micro/nano/milli prefixes to standard scientific notation before algebra.
- Use symmetry whenever possible to identify zero-field locations.
- For points not on symmetry axes, compute components or use geometry (Pythagorean, arctangent).
- When charges differ, find location where magnitudes of fields equal and directions oppose.
Key Terms and Definitions
- Electric Field (E): Force per unit charge; vector field around charges.
- Test Charge: A very small positive charge used to probe E without disturbing it.
- Coulomb’s Law: Describes electrostatic force between two point charges.
- Uniform Electric Field: Field between parallel plates approximated as constant in magnitude and direction.
Common Numerical Values To Memorize
- k = 9.0 × 10^9 N·m^2/C^2.
- e (magnitude of electron/proton charge) = 1.602 × 10^-19 C.
- m_e = 9.11 × 10^-31 kg.
- m_p = 1.67 × 10^-27 kg.
- g ≈ 9.8 m/s^2.
Action Items / Next Steps
- Practice sign/direction reasoning with mixed-charge configurations.
- Solve vector addition problems by splitting fields into components.
- Memorize constants and prefix conversions (μ = 10^-6, n = 10^-9, m = 10^-3).
- Work additional problems involving motion of charges in uniform fields using F = qE and kinematics.*