Question

Consider a cylindrical capacitor, with outer radius \(R\) and cylinder separation \(d\). Determine what the capacitance approaches in the limit where \(d \ll R\). (Hint: Express the capacitance in terms of the ratio \(d / R\) and then examine what happens as the ratio \(d / R\) becomes very small compared to \(1 .)\) Explain why the limit on the capacitance makes sense.

Short answer

Answer: As the cylinder separation (d) approaches 0 compared to the outer radius (R), the capacitance of the cylindrical capacitor approaches infinity. This is because a smaller separation between the cylinders results in a stronger electric field and higher potential difference, ultimately leading to increased capacitance.

Step-by-step solution

Capacitance of a cylindrical capacitor

To find the capacitance of a cylindrical capacitor, we can use the following formula: C = \frac{2 \pi * \epsilon * L}{ln (\frac{R}{r})} where C is the capacitance, ε is the permittivity of the dielectric, L is the length of the capacitor, R is the outer radius, and r is the inner radius. In our problem, we are given the outer radius (R) and the cylinder separation (d) instead of r. We can find r using the given information: r = R - d Now, let's plug this into the capacitance formula: C = \frac{2 \pi * \epsilon * L}{ln (\frac{R}{R-d})}

Express the capacitance in terms of the ratio d/R

To express the capacitance in terms of the ratio d/R, we need to rewrite our formula for r, the inner radius, in terms of this ratio. We can set up a proportion as follows: \frac{d}{R} = \frac{R-r}{R} So, r = R(1-\frac{d}{R}) Now, we can substitute this into our capacitance formula: C = \frac{2 \pi * \epsilon * L}{ln (\frac{R}{R(1-\frac{d}{R})})} Simplifying the expression: C = \frac{2 \pi * \epsilon * L}{ln (\frac{1}{1-\frac{d}{R}})}

Find the limit as the ratio d/R approaches 0

As the ratio d/R approaches 0, we can observe the behavior of the capacitance: \lim_{\frac{d}{R} \rightarrow 0} C = \frac{2 \pi * \epsilon * L}{ln (\frac{1}{1-0})} The limit simplifies down to: \lim_{\frac{d}{R} \rightarrow 0} C = \frac{2 \pi * \epsilon * L}{ln (1)} Since the natural logarithm of 1 is 0, the limit equals: \lim_{\frac{d}{R} \rightarrow 0} C = C_{max} = \infty Thus, as the cylinder separation d becomes very small compared to the outer radius R, the capacitance approaches infinity.

Explain why this limit makes sense

The limit we found, C approaching infinity as d/R approaches 0, makes sense for the following reasons: 1) Smaller cylinder separation (d) implies that the charges on the inner and outer cylinders are closer together, which creates a stronger electric field between them. 2) A stronger electric field implies a higher potential difference between the capacitor plates, and thus an increased capacitance. 3) In the limit where d/R approaches 0, the capacitance becomes infinitely large, reflecting the fact that an infinitely small separation would result in an infinitely strong electric field and infinite potential difference. Hence, our obtained limit on the capacitance is logically consistent and makes sense in the context of a cylindrical capacitor.

Key concepts

Electric Field

In the context of a cylindrical capacitor, understanding the electric field is crucial. The electric field is a region where charged particles experience a force. In a capacitor, the electric field is produced between the capacitor plates, which in this case are co-axial cylinders.

The strength of the electric field depends on the amount of charge and the separation between the plates. A smaller separation between the plates leads to a stronger electric field because the charges are closer. For a cylindrical capacitor, the electric field is directed radially between the inner cylinder (radius r) and the outer cylinder (radius R).

The electric field's intensity plays a significant role in defining the potential difference across the plates, ultimately affecting the capacitance. It's important to understand that as the separation in this configuration decreases (i.e., as \(d \to 0\)), the field strength increases, which influences the other properties like the capacitance.

Capacitor Plates

Capacitor plates are the conductive surfaces that store charge in a capacitor. In a cylindrical capacitor, these plates are actually cylindrical surfaces. Each plate holds equal but opposite charges. This separation of charge creates the electric field that we're interested in.

The distance between these plates directly affects the capacitor's ability to store charge. As the plates move closer together, the interaction between charges on the plates increases, which affects the capacitance.

In our exercise, with the separation denoted as \(d\), its size relative to the cylinder's outer radius \(R\) is pivotal. When \(d \ll R\), the plate separation is so small that the field strength and thus the ability to store charge dramatically increase. That's why, as we calculated, the capacitance approaches infinity—because the plates are almost on top of each other, hypothetically leading to a very large amount of charge storage.

Permittivity

Permittivity is a key factor in defining a material's ability to support an electric field. It's a measure of how much electric potential is decreased inside a material compared to vacuum.

In the context of capacitors, permittivity (\(\epsilon\)) helps determine how much energy a capacitor can store. The higher the permittivity, the greater the capacitance, as the factor \(\epsilon\) is part of the formula for capacitance.

For our cylindrical capacitor, when we calculated the capacitance, the permittivity of the dielectric material between the plates was crucial in the equation \(C = \frac{2 \pi \cdot \epsilon \cdot L}{\ln \left(\frac{R}{r}\right)}\). Thus, selecting a material with a high permittivity enables more charge storage for any given physical size, making the capacitor more efficient. Understanding permittivity allows for optimally designing capacitors for various applications.