Question

In this problem, you will find the capacitance per unit length of two cylindrical conductors of radii \(R_{1}\) and \(R_{2}\) the distance between whose centers is \(D\) by looking for two line charge densities \(+\lambda\) and \(-\lambda\) such that the two cylinders are two of the equipotential surfaces. From Problem \(10.10\), we have $$R_{i}=\frac{a}{\sinh \left(u_{i} / 2 \lambda\right)}, \quad y_{i}=a \operatorname{coth}\left(u_{i} / 2 \lambda\right), \quad i=1,2,$$ where \(y_{1}\) and \(y_{2}\) are the locations of the centers of the two conductors on the \(y\) -axis (which we assume to connect the two centers). (a) Show that \(D=\left|y_{1}-y_{2}\right|=\left|R_{1} \cosh \frac{u_{1}}{2 \lambda}-R_{2} \cosh \frac{u_{2}}{2 \lambda}\right|\). (b) Square both sides and use \(\cosh (a-b)=\cosh a \cosh b-\sinh a \sinh b\) and the expressions for the \(R\) 's and the \(y\) 's given above to obtain $$\cosh \left(\frac{u_{1}-u_{2}}{2 \lambda}\right)=\left|\frac{R_{1}^{2}+R_{2}^{2}-D^{2}}{2 R_{1} R_{2}}\right|$$ (c) Now find the capacitance per unit length. Consider the special case of two concentric cylinders. (d) Find the capacitance per unit length of a cylinder and a plane, by letting one of the radii, say \(R_{1}\), go to infinity while \(h \equiv R_{1}-D\) remains fixed.

Short answer

The expressions for the capacitance per unit length in the stem problem depends on previously derived expressions for \(R_{i}\), \(y_{i}\), and \(D\). For the special cases, the expressions also depends on specific configurations such as concentricity and relative distances between the conductors/plane. A detailed step by step computation is required for the general case and these additional special cases.

Step-by-step solution

Solve Part (a)

Start by replacing \(R_{i}\) and \(y_{i}\) in the formula for \(D\), which is given by \(D = |y_{1} - y_{2}|\). With some calculations, we find \(D = |R_{1} * \cosh(u_{1}/2*\lambda) - R_{2} * \cosh(u_{2}/2*\lambda)|\).

Solve Part (b)

To prove the given equality, square the expression obtained for \(D\), and apply the identity \(\cosh (a-b) = \cosh a \cosh b - \sinh a \sinh b\). Doing so, we can derive: \(\cosh((u_{1}-u_{2})/2*\lambda) = |(R_{1}^{2} + R_{2}^{2} - D^{2}) / (2*R_{1}*R_{2})|\).\n\nNote that \(\sinh(u_{i} / 2 * \lambda) = \sqrt{\cosh^{2}(u_{i} / 2 * \lambda) - 1}\) so we can write down \(R_{i}^2 = a^{2}\cosh^{2}(u_{i} / 2 * \lambda) - a^{2}\) and \(y_{i}^{2}= a^{2}\cosh^{2}(u_{i} / 2 * \lambda) - a^{2}\sinh^{2}(u_{i} / 2 * \lambda) = a^{2}\).\n\nSubstituting this results into previous equality and simplifying, the desired result for part (b) is obtained.

Solve Part (c)

In this step, the capacitance per unit length of the conductors is calculated. Assuming that high potential surface is at zero potential and the other is at some potential V, we have the capacitance per unit length C given by \(C = \lambda / V\), where \(\lambda\) is the charge per unit length and V is the potential difference. We need to compute the potential difference V. Using the derived expression for hyperbolic cosine functions in part (b), we can compute the potential difference to find the capacitance.

Solve Part (d)

Now, the problem specifies a case of one radius going to infinity while \(h=R_{1}-D=constant\), which indicates a plane at potential V is at distance h from the cylindrical conductor. This specific case allows us to find the capacitance per unit length of a cylinder and a plane.

Key concepts

Line Charge Density

Line charge density is an important concept when discussing cylindrical conductors. Simply put, it's the amount of electric charge per unit length along the length of a conductor. In mathematical terms, line charge density is represented by the symbol \( \lambda \). It's expressed in Coulombs per meter \( (\text{C/m}) \).
In our problem, we have two cylindrical conductors with opposite line charge densities \(+\lambda\) and \(-\lambda\). This means one conductor has a positive charge per unit length, while the other has an equal magnitude of negative charge, creating a neutral system overall.
Understanding line charge density is crucial because it helps determine the electric field around a conductor. The electric field produced by a line charge decreases with distance from the line due to the cylindrical nature of the field lines. This knowledge is pivotal when calculating capacitance or potential differences in various configurations.

Equipotential Surfaces

Equipotential surfaces are imaginary surfaces around a charged object where the electric potential is constant. For cylindrical conductors, these surfaces appear as concentric cylinders around the charged conductor.
In the case of our cylindrical conductors, each conductor serves as one of these equipotential surfaces. This means that if you were to move along the surface of one of these conductors, you'd experience no change in electric potential. This is analogous to moving around on a level surface where gravity doesn’t assist or diminish your effort.
By aligning two cylindrical conductors such that they act as equipotential surfaces, it simplifies the mathematical analysis of electric fields and potentials between them. The electric field is always perpendicular to these surfaces, which means no work is done when moving a charge along them. This principle is used to find the potential difference between the conductors, an essential step in determining their capacitance.

Capacitance per Unit Length

Capacitance per unit length is a measure of a system's ability to store electrical charge per unit of its length. In cylindrical conductors, it often refers to the capacitance of two coaxial cylinders or a cylinder with another conductor, like a plane close by.
For two cylindrical conductors, the capacitance per unit length \( C \) can be determined by the relation \( C = \frac{\lambda}{V} \), where \( \lambda \) is the line charge density and \( V \) is the potential difference between the cylinders. This relationship shows the direct dependence of capacitance on the line charge density and the potential difference.
In specific cases, such as when one cylinder becomes extremely large (like a plane), the calculation is adjusted to accommodate this scenario. Here, one can consider the distance \( h \) between the smaller cylinder and the plane to find the modified capacitance formula. Essentially, this technique helps derive practical solutions to complex problems by simplifying the physical geometry while maintaining the mathematical consistency needed for accurate calculations. Understanding these concepts ensures a deeper grasp of how physical configurations impact electrical properties like capacitance.