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Electric field of a ring
Intermediate
On axis the field varies with distance.
E = kQx/(x²+R²)^{3/2}On the axis of a ring of charge Q and radius R, E = kQx/(x² + R²)^{3/2}; zero at the centre and ≈ kQ/x² far away.
Q = 1 nC, R = x = 5 cm: E = 9×10⁹·10⁻⁹·0.05/(0.005)^{3/2} ≈ 1.3 kV/m.
Perpendicular components cancel by symmetry.
In more detail
On the axis of a uniformly charged ring, the field is E = kQz/(z² + R²)^(3/2). It is zero at the centre and largest at a distance related to the ring radius.
Real-world uses
- Modelling charged rings in particle detectors.
- Calculating fields in ring-shaped electrodes.
- Engineers calculate the field on the axis of a charged ring to design sensors and particle detectors.
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Frequently asked questions
Why is the field zero at the centre?
Because the symmetric pulls from all parts of the ring cancel.
Where is the field largest?
At a distance of about R over the square root of two from the centre.
Why is the field zero at the centre of the ring?
By symmetry, the field from each part of the ring cancels at the centre.
Related topics in Electrostatics
Boundary conditions at interfaces · Electrostatic energy of a sphere · Earnshaw's theorem · Electric field of a disc