Question dealing with Physics, specifically working with spherical coordinates, uni level.

Question

The spherical coordinates of a circular ring are a, α. A sphere of relative capacitivity K and radius b has the origin as its center. If the ring carries a charge of linear density τ, find the potential between the sphere and ring. The final result will be accepted if it has a uncalculated but simplified series.

Answer 1

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To find the electric potential between the sphere and the charged circular ring in spherical coordinates, we'll start by expressing the charge distribution on the ring in terms of spherical coordinates and then calculate the electric potential due to this distribution.

Let's denote:

- \(a\) as the radius of the ring,

- \(\alpha\) as the polar angle that specifies the position of the ring on the sphere (measured from the positive z-axis),

- \(\tau\) as the linear charge density on the ring,

- \(K\) as the relative permittivity of the surrounding medium (not capacitivity),

- \(b\) as the radius of the sphere.

First, we need to express the differential charge element \(dq\) on the ring in spherical coordinates. This element can be expressed as:

\[dq = \tau \cdot dl\]

where \(dl\) is the differential length along the ring. In spherical coordinates, \(dl\) can be expressed as \(a \cdot d\alpha\) since it's along the curved portion of the ring.

Now, let's consider a point on the sphere with spherical coordinates \((b, \theta, \phi)\). The electric potential \(V\) at this point due to the differential charge \(dq\) is given by Coulomb's law:

\[dV = \frac{1}{4\pi K} \cdot \frac{dq}{r}\]

Where:

- \(r\) is the distance between \(dq\) and the point on the sphere, which can be expressed in spherical coordinates as \(r = \sqrt{a^2 + b^2 - 2ab\sin\alpha\sin\theta\cos\phi}\).

Now, we can calculate the electric potential \(V\) at the point on the sphere due to the entire ring by integrating \(dV\) over the ring:

\[V = \int dV = \frac{1}{4\pi K} \int \frac{\tau \cdot a \cdot d\alpha}{\sqrt{a^2 + b^2 - 2ab\sin\alpha\sin\theta\cos\phi}}\]

To simplify this integral, you can use the binomial expansion to approximate the denominator to first order (since \(a\) is small compared to \(b\)):

\[\sqrt{a^2 + b^2 - 2ab\sin\alpha\sin\theta\cos\phi} \approx \sqrt{b^2 - 2ab\sin\alpha\sin\theta\cos\phi}\]

Now, you can perform the integral, and the final result will involve a simplified series due to the binomial expansion.

The integral may not have a simple closed-form solution and may require numerical methods for evaluation. The final result will be expressed in terms of the spherical coordinates \((b, \theta, \phi)\) and will involve a series.