An Isolated Charged Sphere

1 · Predict

A single isolated conducting sphere, charged to some voltage, has a capacitance too — even with no second plate nearby. If you raise the sphere's voltage, does the field just outside it rise in proportion?

• Yes — the exterior field is directly proportional to the sphere's voltage.
• No — the field depends on voltage in a more complex way.
• The exterior field doesn't depend on the sphere's voltage at all.

2 · Set Up

1. Open the isolated-sphere preset and press Reset. The sphere's radius is fixed at 8 cm.
2. Enable the exterior field-strength readout.
3. Set the sphere's voltage for each trial and record the field strength measured at twice the sphere's radius from its center.

3 · Collect Data

Plot field strength E (y-axis) against voltage V (x-axis) for your three trials. Is the line straight through the origin?

4 · Analyze

1. For one trial, compute C = 4πε₀R using R = 0.08 m (same in every trial), then E = V/(2R) for the field at radius 2R from center. Compare both to the table.
2. Explain why, at a fixed measurement point (here, 2R from center), the exterior field scales linearly with the sphere's voltage — even though a point charge's field itself falls off as 1/r².

5 · Extend

1. A Van de Graaff generator charges an isolated metal sphere to very high voltage. Its capacitance C = 4πε₀R is quite small for a lab-sized sphere, which is exactly why even a modest amount of charge can drive it to extremely high voltage (V = Q/C).
2. The Earth itself behaves approximately like an isolated charged sphere with a small negative surface charge. Using C = 4πε₀R with Earth's radius (~6.4×10⁶ m), would you expect Earth's capacitance to be bigger or smaller than this experiment's 8 cm sphere?

The Physics Behind This Experiment

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