AP Physics C E&M: Cylindrical Capacitor

Cylindrical capacitors, particularly in the form of coaxial cables, are the backbone of modern telecommunications, carrying everything from cable TV signals to high-speed internet data with minimal interference. Mastering their capacitance derivation is not just an academic exercise—it sharpens your ability to apply Gauss's law and calculus to realistic, non-uniform geometries, a key skill for both the AP exam and future engineering work, from foundational concepts to the applied formula.

Capacitance Fundamentals and Cylindrical Geometry

Capacitance is defined as the ratio of stored charge to potential difference, $C=Q/V$. A cylindrical capacitor consists of two coaxial conductors: an inner cylinder of radius $a$ and an outer cylindrical shell of radius $b$, both with length $L$. Unlike parallel plates, the cylindrical symmetry means the electric field isn't constant, requiring integration to find the potential difference. In practice, such geometries are found in coaxial cables, where the inner conductor carries signals and the outer shield prevents interference. For analysis, we assume uniform charge distributions and ignore end effects by considering $L\gg a,b$, focusing on capacitance per unit length, $C/L$, which is more relevant for long cables.

You'll often encounter this setup with the inner cylinder holding charge $+Q$ and the outer cylinder $-Q$. The linear charge density $\lambda =Q/L$ simplifies calculations, as charge is spread uniformly along the length. Think of it like a long, thin pipe inside a larger pipe; the storage capacity depends on their sizes and the gap between them, analogous to how insulation thickness affects thermal retention.

Electric Field Using Gauss's Law

To find the electric field in the region between the cylinders ($a<r<b$), apply Gauss's law: $\oint \vec{E}\cdot d\vec{A}=Q_{\text{enc}}/{ϵ}_0$. Choose a cylindrical Gaussian surface coaxial with the conductors, with radius $r$ and length $l$ (a segment of the full length). By symmetry, the electric field points radially outward and has constant magnitude on this surface.

The flux through the curved surface is $E\cdot (2\pi rl)$, since the ends contribute no flux (field is perpendicular to area vectors). The enclosed charge is from the inner cylinder over length $l$: $Q_{\text{enc}}=\lambda l=(Q/L)l$. Substituting into Gauss's law:

$$E\cdot 2\pi rl=\frac{\lambda l}{{ϵ}_0}$$

Solving for $E$, the magnitude is:

$$E=\frac{\lambda }{2\pi {ϵ}_{0}r}$$

This $1/r$ dependence shows the field weakens as you move outward, unlike the constant field in parallel plates. For example, if you double the radius, the field halves, which influences how voltage builds up across the gap.

Potential Difference Through Integration

The potential difference $V$ between the inner and outer cylinders is the line integral of the electric field. Since $V=-\int \vec{E}\cdot d\vec{l}$, and for a radial path $d\vec{l}=dr\hat{r}$, we compute the magnitude from $a$ to $b$:

$$V={\int }_a^{b}Edr={\int }_a^b\frac{\lambda }{2\pi {ϵ}_{0}r}dr$$

Factor out constants:

$$V=\frac{\lambda }{2\pi {ϵ}_0}{\int }_a^b\frac{dr}{r}$$

The integral $\int dr/r=\ln r$, so:

$$V=\frac{\lambda }{2\pi {ϵ}_0}[\ln r{]}_a^b=\frac{\lambda }{2\pi {ϵ}_0}(\ln b-\ln a)=\frac{\lambda }{2\pi {ϵ}_0}\ln \left(\frac{b}{a}\right)$$

Substitute $\lambda =Q/L$:

$$V=\frac{Q/L}{2\pi {ϵ}_0}\ln \left(\frac{b}{a}\right)$$

This result shows that voltage scales linearly with charge but logarithmically with the radius ratio, a key insight for cable design.

Deriving Capacitance per Unit Length

Now, use $C=Q/V$ to find capacitance. Substitute the expression for $V$:

$$C=\frac{Q}{V}=\frac{Q}{\frac{Q/L}{2\pi {ϵ}_0}\ln (b/a)}=\frac{Q\cdot 2\pi {ϵ}_0}{(Q/L)\ln (b/a)}=\frac{2\pi {ϵ}_{0}L}{\ln (b/a)}$$

Thus, the capacitance per unit length is:

$$\frac{C}{L}=\frac{2\pi {ϵ}_0}{\ln (b/a)}$$

This is the core formula: capacitance per unit length depends only on ${ϵ}_0$ and the natural logarithm of the radius ratio $b/a$. The logarithmic dependence means that even large changes in radii yield modest capacitance changes. For instance, if $b/a$ increases from 2 to 4, the capacitance per unit length decreases only slightly because $\ln (4)\approx 1.386$ compared to $\ln (2)\approx 0.693$, illustrating the logarithmic dependence. This principle is crucial in designing real coaxial cables, where minimizing signal loss requires optimizing the radius ratio within practical constraints.

Common Pitfalls

A frequent mistake is to assume a constant electric field, as in parallel plates, leading to incorrect integration. Also, confusing linear charge density $\lambda$ with total charge $Q$ can derail the derivation. Always remember that $V$ is found by integrating $E$ from $a$ to $b$, not the other way around, and ensure consistent use of units when applying the formula to real-world cables.

Summary

• Cylindrical capacitors consist of coaxial conductors with inner radius $a$ and outer radius $b$, commonly used in coaxial cables for telecommunications.
• The electric field between them is derived using Gauss's law: $E=\frac{\lambda }{2\pi {ϵ}_{0}r}$, showing a $1/r$ dependence.
• The potential difference is integrated to yield $V=\frac{\lambda }{2\pi {ϵ}_0}\ln (b/a)$, highlighting logarithmic scaling with radius ratio.
• Capacitance per unit length is $C/L=\frac{2\pi {ϵ}_0}{\ln (b/a)}$, a key result for AP Physics C and engineering applications.
• This formula applies directly to real coaxial cables, where the logarithmic dependence means capacitance changes slowly with adjustments to $b/a$, aiding in design for minimal interference.