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Electric field of a disc

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Integrate rings to get the disc's field.

Integrate rings: E = (σ/2ε₀)(1 − x/√(x² + R²)). As R → ∞ it approaches the infinite sheet result σ/2ε₀.

Worked example

Close to the centre the disc acts like a plane.

Common mistake

Check the limits.

In more detail

On the axis of a uniformly charged disc, the field is E = (σ/2ε0)(1 − z/sqrt(z² + R²)). It approaches the infinite sheet result when the disc is large compared with the distance.

Real-world uses

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Frequently asked questions

What does the limit mean?

For points very close to a large disc, the field approaches σ/2ε0.

Is the field zero at the centre?

No, it is largest at the centre and decreases with distance.

Is the field of a disc the same as an infinite plane?

Only near the disc. Far away the field falls off like a point charge, while close to a large disc it approaches the plane value.

Related topics in Electrostatics

Electrostatic energy of a sphere · Earnshaw's theorem · Electric field of a ring

← All Electrostatics topics