Problem 29 - University Physics with Modern Physics

Question

22.29. An infinitely long cylindrical conductor has radius \(R\) and uniform surface charge density \(\sigma\) . (a) In terms of \(\sigma\) and \(R\) , what is the charge per unit length \(\lambda\) for the cylinder? (b) In terms of \(\sigma\) , what is the magnitude of the electric field produced by the charged cylinder at a distance \(r>R\) from its axis? (c) Express the result of part (b) in terms of \(\lambda\) and show that the electric field outside the cylinder is the same as if all the charge were on the axis. Compare your result to the result for a line of charge in Example 22.6 (Section \(22.4 ) .\)

Step-by-Step Solution

Answer

(a) \( \lambda = \sigma \times 2\pi R \). (b) \( E = \frac{\sigma R}{\varepsilon_0 r} \). (c) \( E = \frac{\lambda}{2\pi \varepsilon_0 r} \), same as a line charge.

Step 1: Determine Charge Per Unit Length (λ)

First, consider the surface charge density \( \sigma \), which is the charge per unit area on the cylinder's surface. Since the cylinder is infinitely long, we focus on a unit length of the cylinder, say 1 meter. The surface area of the cylindrical section with radius \( R \) and unit length is \( 2\pi R \times 1 = 2\pi R \). Therefore, the charge per unit length \( \lambda \) is given by \( \lambda = \sigma \times 2\pi R \).

Step 2: Find Electric Field at Distance r (r > R)

Use Gauss's Law to find the electric field \( E \) at a distance \( r \geq R \) from the axis of the cylinder. Choose a cylindrical Gaussian surface of radius \( r \) and length \( L \), co-axial with the charged cylinder. The electric flux through the curved surface is \( E \times 2\pi r L \). According to Gauss's Law, the enclosed charge \( \lambda L \) gives \[ E \times 2\pi r L = \frac{\lambda L}{\varepsilon_0} \]. Solving for \( E \), we get \( E = \frac{\lambda}{2\pi \varepsilon_0 r} \).

Step 3: Express Electric Field in terms of σ

Substitute \( \lambda = \sigma \times 2\pi R \) into the expression for \( E \): \[ E = \frac{\sigma \times 2\pi R}{2\pi \varepsilon_0 r} = \frac{\sigma R}{\varepsilon_0 r} \].

Step 4: Check Equivalence with a Line Charge

Compare this with the field of an infinite line charge \( E = \frac{\lambda}{2\pi \varepsilon_0 r} \), realizing that the field outside the cylinder behaves as if it were from a line charge because it depends only on \( \lambda \) and \( r \). Therefore, this shows the electric field outside the cylinder is the same as a line charge along the axis.

Key Concepts

Gauss's Law

Gauss's Law is a powerful tool in electrostatics to calculate electric fields when there is a high degree of symmetry. Essentially, it relates the electric flux through a closed surface to the charge enclosed by that surface. The law is often stated mathematically as:

• \[ \Phi_E = \oint \mathbf{E} \cdot d\mathbf{A} = \frac{Q_{\text{enc}}}{\varepsilon_0} \]

For cylindrical conductors, Gauss's Law can simplify the calculation of the electric field. You use a Gaussian surface that matches the symmetry of the charge distribution. If the conductor is infinitely long and cylindrical, a co-axial cylindrical surface is an ideal choice. This makes it simple to evaluate the electric field at any point outside the surface because the integral becomes a product due to constant electric field over the surface.

Electric field

The electric field is a vector field around charged particles that represents the force exerted per unit charge at any point in space. In the context of an infinitely long charged cylindrical conductor, the electric field at a distance greater than the cylinder's radius behaves interestingly. Thanks to Gauss's Law, calculating this is more straightforward.

For points outside the cylinder, the electric field (\(E\)) only depends on the linear charge density (\(\lambda\)) and the distance from the axis (\(r\)). According to the derived formula, the electric field is:

• \[ E = \frac{\lambda}{2\pi \varepsilon_0 r} \]

This formula implies that as you go farther from the cylinder, the electric field decreases inversely with distance (\(r\)). It is crucial to understand that the linear density simplifies this, whether considering the surface density \(\sigma\) or directly using \(\lambda\).

Surface charge density

Surface charge density (\(\sigma\)) is a measure of how much electric charge is accumulated on a surface, defined as charge per unit area. In the context of a cylindrical conductor, the surface charge density is constant, leading to uniform charge distribution over its entire curved surface.

When calculating the total charge per unit length (\(\lambda\)) for a cylinder, multiply \(\sigma\) by the curved surface area of the cylinder per unit length. Mathematically, it’s expressed as:

• \[ \lambda = \sigma \times 2\pi R \]

This shows how surface properties relate to linear properties in electrostatic problems. The entire methodology bridges the gap between surface phenomena and how they translate to linearly charged systems, crucial for further calculations and understanding.

Cylindrical conductors

Cylindrical conductors, especially when considered infinite, provide a simplified model that helps derive important relationships in electrostatics. These include evaluating electric fields and understanding charge distribution characteristics.

In practical applications, a cylindrical conductor's infinite nature is an abstraction, often used to simplify calculations. This model becomes important for producing uniform electric fields, making them ideal for study purposes. Key points include:

• Fields inside are zero due to the conductor's shielding effect.
• For outside fields, charges appear concentrated along the axis.
• The mathematical representation transforms complex calculations into manageable expressions.

This simplified approach helps in engineering applications and in understanding electrostatic shielding and transmission characteristics. In all electric and magnetic field problems, this balance of realism and idealization is crucial.