Electric fields and equipotentials
Charges fill the space around them with a field that pushes on other charges. Field lines show which way, equipotentials show the voltage, and the two always cross at right angles.
Readouts
What's happening
A point charge Q sets up a field of strength E = kQ/r², where k = 1/(4πε₀) = 8.99 × 10⁹ N m² C⁻². E is a vector: the force per coulomb on a small positive test charge, so it points away from positive charges and towards negative ones. With several charges the fields simply add as vectors (superposition), which is what the probe measures. Field lines start on positive charges and end on negative ones, and they crowd together where the field is strong. The potential V = kQ/r is a scalar, the work done per coulomb to bring a positive charge in from far away, and it adds up without any directions. Lines of equal V are equipotentials: no work is done moving along one, so they always cross field lines at 90°, and the field is the steepest downhill slope of V, E = −dV/dr. Between two like charges there is a point where the fields cancel; between oppositely charged plates the field is almost uniform.
A-Level Physics (AQA, OCR A and B, Edexcel, WJEC): Coulomb's law, electric field strength, field lines, electric potential and equipotentials, uniform fields between parallel plates. First-year university electromagnetism.
Work through the numbers with Physics Formulas.
Challenge
Predict first: choose Single, so there is one +2 nC charge with the probe 5 cm away. What is |E| at the probe? Then guess what happens to E and to V when the probe moves out to 10 cm.
E = kQ/r² = 8.99 × 10⁹ × 2 × 10⁻⁹ / 0.05² = 7190 N/C. At 10 cm, E falls to a quarter, 1800 N/C, but V only halves, from 360 V to 180 V, because V goes as 1/r and E as 1/r².
FAQ
- What is the difference between electric field strength and electric potential?
- Field strength E is a vector, the force per coulomb on a positive test charge, measured in N/C (the same as V/m). Potential V is a scalar, the work done per coulomb bringing a positive charge from infinity, measured in volts. The field points down the steepest slope of the potential: E = −dV/dr.
- Why are equipotentials perpendicular to field lines?
- Moving a charge along an equipotential does no work, because V does not change. Work is force times distance moved along the force, so the force, and therefore the field, must have no component along the equipotential. That only happens if the two cross at right angles.
- Where is the electric field zero between two charges?
- For two like charges it is on the line between them, closer to the smaller one, where kq₁/r₁² = kq₂/r₂², so r₁/r₂ = √(q₁/q₂). For two unlike charges it is outside the pair, beyond the smaller charge. Equal and opposite charges have no zero point at a finite distance.
- Why is the field between parallel plates uniform?
- Each small patch of a plate pulls or pushes the test charge, and between two oppositely charged plates the sideways parts cancel while the parts across the gap add. Away from the edges the result hardly depends on where you are, so the field lines are parallel and equally spaced, with E = V/d.
- How far from a charge is the field a given strength?
- Rearrange E = kq/r² to r = √(kq/E). For a 5 nC charge and E = 1000 N/C, r = √(8.99 × 10⁹ × 5 × 10⁻⁹ / 1000) = 0.212 m. Pick r for a field in the Solve for panel to get it, with the steps, for any charge.