Problem 31 - Fundamentals of Physics

Question

Two long, charged, thin-walled, concentric cylindrical shells have radii of 3.0 and \(6.0 \mathrm{~cm} .\) The charge per unit length is \(5.0 \times 10^{-6} \mathrm{C} / \mathrm{m}\) on the inner shell and \(-7.0 \times 10^{-6} \mathrm{C} / \mathrm{m}\) on the outer shell. What are the (a) magnitude \(E\) and (b) direction (radially inward or outward) of the electric field at radial distance \(r=4.0 \mathrm{~cm} ?\) What are (c) \(E\) and (d) the direction at \(r=8.0 \mathrm{~cm} ?\)

Step-by-Step Solution

Answer

At \( r=4.0 \) cm, the field is \( 2.25 \times 10^{6} \) N/C, radially outward; at \( r=8.0 \) cm, it's \( 4.5 \times 10^{5} \) N/C, radially inward.

Step 1: Identify the Problem

We have two concentric cylindrical shells with given radii and charge per unit length. We need to find the electric field and its direction at specified radial distances using Gauss's Law.

Step 2: Apply Gauss's Law

Gauss's Law states that \( \Phi_E = \oint \mathbf{E} \cdot d\mathbf{A} = \frac{Q_{enc}}{\varepsilon_0} \). For a cylindrical surface around the cylinders, the electric field \( E \) is constant over the surface. Hence, \( E(2\pi rl) = \frac{Q_{enc}}{\varepsilon_0} \).

Step 3: Calculate Electric Field at r = 4.0 cm

At \( r = 4.0 \) cm, only the inner shell contributes to the electric field as the cylindrical Gaussian surface does not enclose the outer shell. \( Q_{enc} = 5.0 \times 10^{-6} \times l \). Applying Gauss's Law, \( E(2\pi \times 0.04 \times l) = \frac{5.0 \times 10^{-6} \times l}{8.85 \times 10^{-12}} \), which simplifies to \( E \approx 2.25 \times 10^{6} \) N/C, pointing radially outward.

Step 4: Evaluate Electric Field Direction at r = 4.0 cm

Since the charge on the inner shell is positive, the electric field at \( r = 4.0 \) cm is directed radially outward from the inner shell.

Step 5: Calculate Electric Field at r = 8.0 cm

At \( r = 8.0 \) cm, the cylindrical Gaussian surface encloses both shells. The total charge enclosed is \( Q_{enc} = (5.0 \times 10^{-6} - 7.0 \times 10^{-6}) \times l = -2.0 \times 10^{-6} \times l \). Applying Gauss's Law, \( E(2\pi \times 0.08 \times l) = \frac{-2.0 \times 10^{-6} \times l}{8.85 \times 10^{-12}} \), which simplifies to \( E \approx -4.5 \times 10^{5} \) N/C, indicating the magnitude is \( 4.5 \times 10^{5} \) N/C.

Step 6: Evaluate Electric Field Direction at r = 8.0 cm

Since the net charge is negative, the electric field at \( r = 8.0 \) cm is directed radially inward towards the outer shell.

Key Concepts

Electric Field

The electric field is a fundamental concept in electromagnetism, describing the force a charged particle experiences per unit charge at a given location. It is a vector field, meaning it has both a magnitude and a direction. The electric field produced by a charge extends infinitely in space but decreases in strength with distance.
In this exercise, we are interested in the electric field generated by cylindrical shells. Gauss's Law is particularly useful for finding electric fields when dealing with symmetrical charge distributions like cylindrical, spherical, or planar symmetry. By using Gauss's Law, which relates the electric flux through a closed surface to the charge enclosed, we're able to calculate the electric field around these cylindrical shells.

In mathematical terms, Gauss's Law can be expressed as: \[ \Phi_E = \oint \mathbf{E} \cdot d\mathbf{A} = \frac{Q_{enc}}{\varepsilon_0} \]where \( \Phi_E \) is the electric flux, \( \mathbf{E} \) is the electric field, \( d\mathbf{A} \) is a differential area on the closed surface, \( Q_{enc} \) is the charge enclosed by the surface, and \( \varepsilon_0 \) is the permittivity of free space.
Understanding how the electric field behaves and how it can be calculated is crucial for interpreting many physical phenomena in electromagnetism.

Cylindrical Shells

Cylindrical shells are a type of geometric shape with its own set of properties when it comes to charge distribution and electric fields. In the context of our problem, we have two coaxial or concentric cylindrical shells. The inner shell has a smaller radius compared to the outer shell.

When charges are placed uniformly along the axis of these cylindrical shells, the system maintains cylindrical symmetry. This symmetry is key because it allows for simplifications when applying Gauss's Law to find the electric field.

• The inner shell in our exercise has a positive charge per unit length of \( 5.0 \times 10^{-6} \) C/m.
• The outer shell carries a negative charge per unit length of \(-7.0 \times 10^{-6} \) C/m.

With cylindrical symmetry, the direction of the electric field is always perpendicular to the surface of the shell and along the radial direction. This greatly simplifies the calculations, as the electric field remains constant over the Gaussian surface chosen. Only when the surface encloses the charge on the shell does it contribute to the electric field.

Charge Distribution

Charge distribution refers to how charge is spread over a given surface or volume. It critically affects the resulting electric fields due to distribution properties.

In this exercise, both the inner and outer cylindrical shells have their charges distributed uniformly over their surfaces. Each shell’s charge affects the electric field at different radial distances.

• Inside the inner shell: no electric field exists because we're within a conductor.
• Between the inner and outer shell: only the inner shell's charge affects the electric field. Thus, the electric field is radially outward due to the positive charge of the inner shell.
• Outside the outer shell: both shells affect the electric field, with the net effect determined by their total charges. Here, the net charge is negative, causing the electric field to point inward.

Gauss's Law helps to clearly define these electric field characteristics based on charge distribution. By integrating this concept with symmetry, we can analyze electric fields in complex setups in simpler terms.