AP Physics C: Electricity and Magnetism

Electric Potential and Capacitance

The Potential of a Cylinder

Learning Objectives

By the end of this lesson, students should be able to:

• Determine the electric field of an infinitely long charged cylinder.

• Derive the electric potential associated with cylindrical charge distributions.

• Understand the relationship between electric field and electric potential in cylindrical symmetry.

• Calculate potential differences between two points near a charged cylinder.

• Analyze conducting cylindrical shells.

• Apply cylindrical potential equations to AP Physics C problems.

Introduction

Why Study Cylindrical Potentials?

Many practical electrical systems possess cylindrical symmetry.

Examples include:

• Coaxial cables

• Transmission lines

• Cylindrical conductors

• Particle accelerator components

Unlike spherical charge distributions, cylindrical charge distributions produce electric fields that vary as:

$$
\frac{1}{r}
$$

rather than

$$
\frac{1}{r^2}
$$

As a result, the electric potential for cylinders involves logarithmic functions.

Review: Electric Potential Difference

Fundamental Relationship

Electric potential difference is related to electric field by:

$$
\Delta V=\int \vec{E}\cdot d\vec{r}
$$

This equation applies to all charge distributions.

The specific form depends on the electric field.

Electric Field of an Infinite Line Charge

Linear Charge Density

For a line of charge:

$$
\lambda=\frac{Q}{L}
$$

where:

• (\lambda) = linear charge density

• (Q) = charge

• (L) = length

Units:

$$
C/m
$$

Electric Field

Using Gauss’s Law:

$$
E(2\pi rL)=\frac{\lambda L}{\varepsilon_0}
$$

Solving:

$$
E=\frac{\lambda}{2\pi\varepsilon_0 r}
$$

This electric field decreases as:

$$
E\propto\frac{1}{r}
$$

Potential Difference Near a Line Charge

Starting with the Definition

Substitute the electric field into:

$$
\Delta V=\int \vec{E}\cdot d\vec{r}

$$

Assuming radial motion:

$$
\Delta V=\int_{r_1}^{r_2}\frac{\lambda}{2\pi\varepsilon_0 r}dr
$$

Performing the Integration

The integral becomes:

$$
\Delta V=\frac{\lambda}{2\pi\varepsilon_0}=\int_{r_1}^{r_2}\frac{dr}{r}
$$

Since:

$$
\int\frac{dr}{r}=\ln r
$$

we obtain:

$$
\Delta V=\frac{\lambda}{2\pi\varepsilon_0}\ln\left(\frac{r_2}{r_1}\right)
$$

Alternative Form

The same equation may be written as:

$$
\Delta V=\frac{\lambda}{2\pi\varepsilon_0}\ln\left(\frac{r_1}{r_2}\right)
$$

Both forms are equivalent.

Important Concept

No Natural Zero at Infinity

For point charges:

$$
V(\infty)=0
$$

is commonly chosen.

For an infinite line charge:

$$
V(\infty)
$$

cannot be defined because the logarithm diverges.

Therefore only potential differences have physical meaning.

This is an important distinction from spherical charge distributions.

Conducting Cylinder

Charge Distribution

Consider a conducting cylindrical shell of radius (R).

At electrostatic equilibrium:

• Charge resides on the outer surface.

• The electric field inside the conductor is zero.

$$
E=0
$$

for

$$
r<R
$$

Electric Field Outside

Outside the cylinder:

$$
E=\frac{\lambda}{2\pi\varepsilon_0 r}
$$

for

$$
r\ge R
$$

Potential Inside a Conducting Cylinder

Equipotential Property

Since:

$$
E=0
$$

inside the conductor,

the potential must remain constant.

Every point inside the conductor has the same potential as the surface.

Thus:

$$
V=\text{constant}
$$

for

$$
r<R
$$

Physical Meaning

A conductor in electrostatic equilibrium is an equipotential.

Therefore:

$$
\Delta V=0
$$

between any two points within the conducting material.

Potential Difference Between Two Cylindrical Surfaces

General Equation

For radii (r_1) and (r_2):

$$
V_2-V_1=\frac{\lambda}{2\pi\varepsilon_0}\ln\left(\frac{r_2}{r_1}\right)
$$

This equation is one of the most important results for cylindrical symmetry.

Dependence on Radius

Notice that the potential depends on:

$$
\ln r
$$

rather than

$$
\frac{1}{r}
$$

This logarithmic behavior is unique to cylindrical charge distributions.

Coaxial Cylinders

Physical Setup

A common application involves two concentric conducting cylinders.

Inner radius:

$$
a
$$

Outer radius:

$$
b
$$

The potential difference between them is:

$$
\Delta V=\frac{\lambda}{2\pi\varepsilon_0}\ln\left(\frac{b}{a}\right)
$$

Importance

This equation is used extensively in:

• Coaxial cables

• Cylindrical capacitors

• High-voltage transmission systems

Example 1

Potential Difference

A line charge has:

$$
\lambda=4.0\times10^{-6}C/m
$$

Find the potential difference between:

$$
r_1=0.10m
$$

and

$$
r_2=0.30m
$$

Solution

Use:

$$
\Delta V=\frac{\lambda}{2\pi\varepsilon_0}\ln\left(\frac{r_2}{r_1}\right)
$$

Substitute:

$$
\Delta V=\frac{4.0\times10^{-6}}{2\pi(8.85\times10^{-12})}\ln(3)
$$

Evaluating:

$$
\Delta V=-7.9\times10^4V
$$

Answer

$$
\Delta V=-7.9\times10^4V
$$

Example 2

Conducting Cylinder

A conducting cylindrical shell has radius:

$$
R=0.50m
$$

If the surface potential is:

$$
200V
$$

what is the potential at:

$$
r=0.20m
$$

inside the cylinder?

Solution

Because:

$$
r<R
$$

the point lies inside the conductor.

The conductor is an equipotential.

Therefore:

$$
V=200V
$$

Answer

$$
V=200V
$$

Comparison with a Sphere

Electric Field

Sphere:

$$
E\propto\frac{1}{r^2}
$$

Cylinder:

$$
E\propto\frac{1}{r}
$$

Electric Potential

Sphere:

$$
V\propto\frac{1}{r}
$$

Cylinder:

$$
V\propto\ln r
$$

This distinction is commonly tested conceptually on AP exams.

Common AP Exam Mistakes

Mistake 1

Using the spherical potential equation:

$$
V=k\frac{Q}{r}
$$

for cylindrical systems.

This equation only applies to spherical symmetry.

Mistake 2

Forgetting the logarithm.

For cylindrical symmetry:

$$
V\propto\ln r
$$

not

$$
\frac{1}{r}
$$

Mistake 3

Assuming potential is zero inside a conductor.

Inside a conductor:

$$
E=0
$$

but

$$
V\neq0
$$

The potential remains constant.

AP Free-Response Strategy

Start with Gauss’s Law

For cylindrical symmetry:

$$
\oint \vec{E}\cdot d\vec{A}=\frac{Q_{enc}}{\varepsilon_0}
$$

This immediately gives the electric field.

Use the Potential Integral

After finding the field:

$$
\Delta V=\int \vec{E}\cdot d\vec{r}
$$

Substitute the electric field and integrate carefully.

Most AP derivations involving cylindrical potentials follow this process.

Summary

Key Takeaways

• An infinite line charge produces an electric field:

$$
E=\frac{\lambda}{2\pi\varepsilon_0 r}
$$

• Electric potential difference is obtained from:

$$
\Delta V=\int \vec{E}\cdot d\vec{r}
$$

• For cylindrical symmetry:

$$
\Delta V=\frac{\lambda}{2\pi\varepsilon_0}\ln\left(\frac{r_2}{r_1}\right)
$$

• Cylindrical potentials depend on logarithmic functions.

• Conducting cylinders are equipotential objects.

$$
\Delta V=0
$$

inside the conductor.

• Cylindrical systems form the foundation for understanding coaxial cables and cylindrical capacitors.