Electric field explorer
Compare electric fields and potentials for parallel plates and point charges by moving a probe and reading the graphs.
Learning goals
- Apply Coulomb's law to the force between point charges.
- Define electric field strength and calculate resultant fields due to point charges.
- Define electric potential and calculate potential due to point charges.
- Relate electric potential, potential energy and work for systems of point charges.
- Use the negative potential gradient and relate equipotentials to field lines.
- Calculate field strength and force in uniform electric fields.
- Analyse charged-particle motion in uniform electric fields.
Point charges: A +2.00 nanocoulombs, B −2.00 nanocoulombs. Field lines leave positive charges and end on negative ones; dashed equipotentials cross them at right angles. The probe reads E = 5754 newtons per coulomb and V = 0 volts.
- E at the probe
- 5754 N C⁻¹
- Direction of E (from +x)
- 0 °
- V at the probe
- 0 V
- Distance to nearest charge
- 7.21 cm
- V / d
- 12500 V m⁻¹
- Deflection leaving the plates
- 0.99 cm
Try this
0 of 4 donePut two like charges on the stage and find the point where E = 0. (not done yet)
Between like charges the two fields point opposite ways and cancel at a neutral point, nearer the smaller charge.
With a single charge, record E at three or more distances. (not done yet)
E against 1/r² is a straight line through the origin: E = kQ/r², so doubling r makes E a quarter.
Make the field between the plates 2.0 × 10⁴ V m⁻¹. (not done yet)
E = V/d: 400 V across 2.0 cm gives the same field as 800 V across 4.0 cm.
Make the beam leave the plates just clear of a plate. (not done yet)
Inside the plates the particle has a constant acceleration qE/m across the field, so its path is a parabola: y = qEL²/(2mv²).
You’ve tried every task.
Your readings
| # | r / cm | E / N C−1 | V / V | Remove |
|---|---|---|---|---|
| No readings yet. Set up a measurement, then record it. | ||||