Playground
Radial field lines emanate from a point charge. Adjust charge sign and magnitude to see field direction and strength change.
Variables
| Symbol | Name | SI | Dimension | Range |
|---|---|---|---|---|
| Electric field magnitudeoutput Strength of the electric field at distance r from the charge | V/m | M·L·T⁻³·I⁻¹ | 0 – 10000000 | |
| Source charge Point charge creating the field | C | I·T | -0.00001 – 0.00001 | |
| Distance from charge Radial distance from the source charge | m | L | 0.01 – 2 |
Deep dive
Derivation
Defined as F/q_test in the limit of vanishingly small test charge: E = lim(q_test→0) F/q_test = k_e·q/r². This definition ensures the test charge doesn't disturb the source distribution.
Experimental verification
Measured using electroscopes and later precision electrometers. Modern verification via Millikan-type experiments and atomic spectroscopy (Stark effect confirms the field's action on charges).
Common misconceptions
- The field exists whether or not a test charge is present — it is a property of space
- E points radially outward for positive charges, inward for negative
- The field from a point charge is not uniform — it varies with distance
Real-world applications
- Electron beam deflection in CRT displays and electron microscopes
- Electric field mapping in capacitor design
- Lightning rod placement and electrostatic shielding
- Particle accelerator beam steering
Worked examples
Field near a point charge
Given:
- q:
- 0.000004
- r:
- 0.3
Find: E
Solution
E = k_e × q / r² = 8.9875×10⁹ × 4×10⁻⁶ / 0.09 = 3.99×10⁵ V/m
Superposition of two point charges
Given:
- q1:
- 0.000002
- q2:
- -0.000002
- d:
- 0.4
- point:
- midpoint
Find: E at the midpoint
Solution
E = 2 × k_e × q / (d/2)² = 2 × 8.9875×10⁹ × 2×10⁻⁶ / 0.04 = 8.99×10⁵ V/m (pointing from + to −)
Scenarios
What if…
- scenario:
- What if the charge is negative?
- answer:
- Field magnitude is the same but direction reverses — field lines point inward toward the negative charge.
- scenario:
- What if you move to 10× the distance?
- answer:
- Field drops by 100×. At r = 3 m: E = 3.99×10³ V/m. The 1/r² falloff is dramatic.
- scenario:
- What if you place a test charge in the field?
- answer:
- Force on test charge q₀ is F = q₀E. A +1 nC test charge at 0.3 m feels F = 10⁻⁹ × 3.99×10⁵ = 3.99×10⁻⁴ N.
Limiting cases
- condition:
- r → 0
- result:
- E → ∞
- explanation:
- Field diverges at the location of the point charge — a singularity resolved by quantum theory.
- condition:
- r → ∞
- result:
- E → 0
- explanation:
- Field falls off as 1/r², becoming negligible at large distances.
- condition:
- q → 0
- result:
- E → 0
- explanation:
- No source charge produces no field.
Context
Michael Faraday · 1832
Faraday introduced the concept of 'lines of force' to visualize how electric influence propagates through space, replacing the action-at-a-distance view.
Hook
How does a charge 'know' another charge is nearby without touching it?
Find the electric field 0.3 m from a +4 μC point charge.
Dimensions: [E] = [k_e]·[q]·[r]⁻² → (N·m²·C⁻²)(C)(m⁻²) = N/C = V/m ✓
Validity: Valid for a single point charge in vacuum. For continuous distributions, integrate over the charge distribution. In dielectric media, the field is reduced by the relative permittivity εᵣ.