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Physics LibreTexts

1.6D: Field on the Axis of and in the Plane of a Charged Ring

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Field on the axis of a charged ring.

1.6.4.png

Ring, radius a, charge Q. Field at P from element of charge δQ=δQ4πϵ0(a2+z2). Vertical component of this =δQcosθ4πϵ0(a2+z2)=δQz4πϵ0(a2+z2)3/2. Integrate for entire ring:

Field E=Q4πϵ0z(a2+z2)3/2.

In terms of dimensionless variables:

E=z(1+z2)3/2.

where E is in units of Q4πϵ0a2, and z is in units of a.

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From calculus, we find that this reaches a maximum value of 239=0.3849 at z=1/2=0.7071.

It reaches half of its maximum value where z(1+z2)3/2=39.

That is, 372Z+9Z2+3Z2=0, where Z=z2.

The two positive solution are Z=0.041889 and 3.596267.

That is, z=0.2047 and 1.8964.

Field in the plane of a charged ring.

We suppose that we have a ring of radius a bearing a charge Q. We shall try to find the field at a point in the plane of the ring and at a distance r(0r<a) from the centre of the ring.

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Consider an element δθ of the ring at P. The charge on it is Qδθ2π. The field at A from this element of charge is

14πϵ0Qδθ2π1a2+r22arcosθ=Q4πϵ0.2πa2δθbccosθ,

where b=1+r2/a2 and c=2r/a. The component of this toward the centre is

Q4πϵ0.2πa2cosϕδθbccosθ.

To find the field at A due to the entire ring, we must express ϕ in terms of θ, r and a, and integrate with respect to θ from 0 to 2π (or from 0 to π and double it). The necessary relations are

p2=a2+r22arcosθ

cosϕ=r2+p2a22rp.

The result of the numerical integration is shown below, in which the field is expressed in units of Q/(4πϵ0a2) and r is in units of a.

I. page 14.png


This page titled 1.6D: Field on the Axis of and in the Plane of a Charged Ring is shared under a CC BY-NC 4.0 license and was authored, remixed, and/or curated by Jeremy Tatum via source content that was edited to the style and standards of the LibreTexts platform.

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