Field of a Finite Line of Charge

1 · Predict

A finite straight rod carries charge spread uniformly along its length. On the perpendicular bisector, does the field from a line of charge follow the same 1/r² law as a point charge?

• No — a finite line of charge has its own, more complex distance dependence, not a simple 1/r² law.
• Yes — from far enough away, it still looks like a point charge.
• The field doesn't depend on distance from an extended charge distribution.

2 · Set Up

1. Open the line-charge preset and press Reset. The rod carries a linear charge density of 2 nC/m over a 4.8 m length.
2. Enable the field-strength readout on the perpendicular bisector.
3. Set the perpendicular distance from the rod's midpoint for each trial and record the field strength.

3 · Collect Data

Plot |E_y| (y-axis) against 1/d (x-axis) and separately against 1/d² (x-axis) for your three trials. Which one looks more like a straight line at these distances?

4 · Analyze

1. For one trial, compute E_y = −2kλL_half/(d·√(L_half² + d²)) using k = 8.99×10⁹ N·m²/C², λ = 2 nC/m, L_half = 2.4 m. Compare to the table.
2. Explain why a finite line of charge's field doesn't reduce to a clean power law like 1/r² — it depends on both the distance d and the rod's half-length through the combination √(L_half² + d²).

5 · Extend

1. As distance d gets very small compared to the rod's length, the formula approaches E ≈ −2kλ/d (the classic 'infinite line of charge' result, which drops as 1/d not 1/d²). Explain, using your formula, why the finite rod behaves like an infinite one when you're close to it.
2. As distance d gets very large compared to the rod's length, the formula should approach the point-charge result E ≈ kQ_total/d² (with Q_total = λ·2L_half). Explain physically why an extended charge distribution looks like a point charge from far enough away.

The Physics Behind This Experiment

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