Problem 51 - University Physics with Modern Physics

Question

A very long cylinder of radius 2.00 \(\mathrm{cm}\) carries a uniform charge density of 1.50 \(\mathrm{nC} / \mathrm{m}\) . (a) Describe the shape of the equipotential surfaces for this cylinder. (b) Taking the reference level for the zero of potential to be the surface of the cylinder, find the radius of equipotential surfaces having potentials of \(10.0 \mathrm{V}, 20.0 \mathrm{V},\) and 30.0 \(\mathrm{V}\) . (c) Are the equipotential surfaces equally spaced? If not, do they get closer together or farther apart as \(r\) increases?

Step-by-Step Solution

Answer

Equipotential surfaces are concentric cylinders and get farther apart with increased radius.

Step 1: Calculate the Electric Field

The electric field, for a long charged cylinder, can be found using Gauss's law. The electric field outside a cylinder with charge density \( \lambda \) is given by:\[ E = \frac{\lambda}{2\pi \varepsilon_0 r} \]where \( r \) is the radial distance from the axis of the cylinder, \( \varepsilon_0 \) is the permittivity of free space, and \( \lambda = 1.50 \times 10^{-9} \mathrm{C/m} \) is the charge density.

Step 2: Find the Electric Potential

The electric potential \( V \) at a distance \( r \) from a charged cylinder is found by integrating the electric field from the surface of the cylinder \( r = R_0 = 2.00 \mathrm{cm} = 0.02 \mathrm{m} \) to a distance \( r \). The potential \( V \) at a distance \( r \) from the surface is:\[ V = - \int_{R_0}^{r} E \cdot dr = - \int_{R_0}^{r} \frac{\lambda}{2\pi \varepsilon_0 r} \, dr \]Solving this, we get:\[ V(r) = - \frac{\lambda}{2\pi \varepsilon_0} \ln \left( \frac{r}{R_0} \right) \]

Step 3: Solve for Equipotential Radii

With the potential function, solve for the radii \( r \) where the potentials are \( 10.0 \, \mathrm{V}, 20.0 \, \mathrm{V}, \) and \( 30.0 \, \mathrm{V} \):For each \( V \):- Solve \( V = - \frac{\lambda}{2\pi \varepsilon_0} \ln \left( \frac{r}{R_0} \right) \) for \( r \).For \( 10.0 \, \mathrm{V}: \)- \( r = R_0 e^{-\left(\frac{2\pi \varepsilon_0 V}{\lambda}\right)} \)- Calculate \( r \).Repeat for \( 20.0 \, \mathrm{V} \) and \( 30.0 \, \mathrm{V} \).

Step 4: Determine Equipotential Spacing

Examine the expression used to find \( r \) and note whether the solution for \( r \) gets larger or smaller as \( V \) increases. The radius grows exponentially with potential, hence the equipotentials are not equally spaced, and they get farther apart as \( r \) increases.

Step 5: Describe the Equipotential Surfaces

Equipotential surfaces around a cylinder are cylindrical shells. As charge distributions in long cylinders cause the electric field lines to spread radially outward, the equipotential surfaces take the form of concentric cylinders.

Key Concepts

Electric Field

The electric field is a crucial concept when discussing charged objects. It describes the force per unit charge exerted by a charge or distribution of charges on a test charge placed within the field.
The electric field strength around a long, uniformly charged cylinder can be determined using Gauss's Law. Gauss's Law relates the electric field to the charge enclosed by a closed surface, simplifying calculations for symmetrical charge distributions.
For our cylinder with a uniform charge density of \( \lambda = 1.50 \times 10^{-9} \, \mathrm{C/m} \), the electric field outside the cylinder, at a radial distance \( r \), is given by:
\[E = \frac{\lambda}{2\pi \varepsilon_0 r} \]
This formula shows that the electric field decreases with increasing radial distance \( r \) from the cylinder's axis. The field's direction is radially outward, perpendicular to the cylinder's surface. When studying charged cylinders, it is important to note how the electric field behaves as it determines other related properties such as electric potential.

Equipotential Surfaces

Equipotential surfaces are imaginary surfaces where every point on them represents the same electric potential. When it comes to a charged cylinder, these surfaces are particularly interesting.
For our long charged cylinder, the equipotential surfaces take the shape of cylindrical shells surrounding the charged cylinder. This is because the electric field lines radiate outward horizontally, making constant potential contours (shells) concentric around the cylinder.
If you imagine slicing the cylinder like bread, each slice would show a cross-section with rings representing different equipotential surfaces.
This configuration makes it easier to visualize and calculate changes in potential energy in the system. It also highlights that moving along an equipotential surface requires no work, as the electric potential remains constant at all points on that surface.

Electric Potential

Electric potential, often simply called the potential, is a measure of the potential energy a unit charge would have at a specific point in an electric field. It allows us to determine the potential energy changes as charges move within an electric field.
For a point outside a charged cylinder, the electric potential \( V \) at a distance \( r \) can be found using the integral of the electric field:
\[V(r) = - \frac{\lambda}{2\pi \varepsilon_0} \ln \left( \frac{r}{R_0} \right)\]
Here, \( R_0 \) is the reference radial position, which is the surface of the cylinder. This formula reveals that the potential decreases logarithmically as you move away from the cylinder.
The negative sign indicates that as you move away from the reference point (cylinder surface), the potential decreases. Understanding electric potential helps predict how charges will behave when placed in the vicinity of charged objects like our cylinder.

Charge Density

Charge density is a measure of how much electric charge is accumulated over a given area or volume. In our scenario, we look at linear charge density, given by \( \lambda \). It tells us how much charge exists per meter length of the cylinder. In this exercise, charge density is \( 1.50 \, \mathrm{nC/m} \).
Having a uniform charge density means every segment of the cylinder has the same charge per unit length. This uniformity simplifies calculations, especially when using Gauss's Law to find the electric field or potential.
A key point to remember is that higher charge densities correspond to stronger electric fields and higher potentials near the surface of the charged object.
Charge density plays a central role in understanding the behavior of electric fields and potentials in systems involving charged objects. It gives us a foundation to predict how a system will behave under the influence of electric forces.