Problem 38 - University Physics with Modern Physics

Question

A long line carrying a uniform linear charge density +50.0 \(\mu\)C/m runs parallel to and 10.0 cm from the surface of a large, flat plastic sheet that has a uniform surface charge density of -100 \(\mu\)C/m2 on one side. Find the location of all points where an \(a\) particle would feel no force due to this arrangement of charged objects.

Step-by-Step Solution

Answer

The no-force point is 15.9 cm from the line charge and 5.9 cm from the sheet.

Step 1: Understand the Problem

We are given a long line with a linear charge density of +50.0 \( \mu \)C/m and a plastic sheet with a surface charge density of -100 \( \mu \)C/m². We need to find a point where an \( \alpha \) particle (positively charged) experiences no net electric force.

Step 2: Calculate Electric Field by the Line Charge

The electric field due to an infinite line charge with linear charge density \( \lambda \) at a distance \( r \) is given by \( E_{line} = \frac{\lambda}{2\pi\varepsilon_0 r} \). Here, \( \lambda = +50.0 \, \mu C/m \), so \( E_{line} = \frac{50.0 \times 10^{-6}}{2\pi \times 8.85 \times 10^{-12} \times 0.10} \approx 8.988 \times 10^6 \; N/C \). This field points away from the line.

Step 3: Calculate Electric Field by the Sheet

The electric field due to an infinite plane with surface charge density \( \sigma \) is \( E_{sheet} = \frac{\sigma}{2\varepsilon_0} \). For \( \sigma = -100 \, \mu C/m^2 \), we find \( E_{sheet} = \frac{-100 \times 10^{-6}}{2 \times 8.85 \times 10^{-12}} \approx -5.65 \times 10^6 \, N/C \). This field points towards the sheet on the side with charge density.

Step 4: Setting Up Equality of Forces

Since we want the \( \alpha \) particle to experience no net force, the electric field due to the line and the sheet must cancel each other out. Thus, \( E_{line} + E_{sheet} = 0 \). Substituting the calculated values: \( 8.988 \times 10^6 + (-5.65 \times 10^6) = 0 \), which simplifies to the correct balance when \( r = x \) from the line.

Step 5: Solve for Position

Given \( E_{line} \) must equal \( E_{sheet} \) in magnitude: \( \frac{50.0 \times 10^{-6}}{2\pi\varepsilon_0 x} = \frac{100 \times 10^{-6}}{2\varepsilon_0} \). Solving for \( x \): \( x = \frac{50.0 \times 10^{-6}}{100 \times 10^{-6} \times \pi} = \frac{1}{2\pi} \approx 0.159 m \).

Step 6: Conclusion

The point where an \( \alpha \) particle would feel no force is located approximately 15.9 cm from the line charge and 5.9 cm from the sheet, on the side closer to the line charge.

Key Concepts

Linear Charge Density

Linear charge density refers to the amount of electric charge per unit length along a line of charge. It is denoted by the symbol \( \lambda \) and measured in coulombs per meter (C/m). In the given problem, the linear charge density is +50.0 \( \mu \)C/m, meaning there are 50 microcoulombs of charge for every meter along the line. This charge distribution creates an electric field that radiates outward perpendicular to the line at every point.
Understanding linear charge density is crucial when working with line charges, as it directly influences the strength and direction of the electric field they produce. The electric field (\( E_{line} \)) due to an infinite line of charge at a distance \( r \) from it is given by the formula: \( E_{line} = \frac{\lambda}{2\pi\varepsilon_0 r} \). Here, \( \varepsilon_0 \) is the permittivity of free space, which is a constant value approximately equal to \( 8.85 \times 10^{-12} \) F/m.
When solving problems like this, considering the symmetry of the system simplifies calculations, as the infinite line charge produces a consistent field pattern along its length.

Surface Charge Density

Surface charge density is the amount of electric charge per unit area on a surface. It is denoted by the symbol \( \sigma \) and is measured in coulombs per square meter (C/m²). In the exercise provided, the plastic sheet has a surface charge density of -100 \( \mu \)C/m², indicating a uniform distribution of negative charge over its area.
This characteristic is important when predicting how the electric field behaves near the surface. Unlike point or line charges, which have radially symmetric fields, a large sheet generates a consistent electric field that is directed perpendicularly away from the charge distribution.
The magnitude of the electric field (\( E_{sheet} \)) from an infinite sheet of charge with surface charge density \( \sigma \) is given by: \( E_{sheet} = \frac{\sigma}{2\varepsilon_0} \). This indicates that the field strength is constant close to the sheet, regardless of the distance (\( r \)) from its surface, which is particularly useful when determining equilibrium points for charged particles among multiple charged objects.

Alpha Particle

An alpha particle is a type of atomic particle that consists of two protons and two neutrons bound together, identical to the nucleus of a helium atom. It bears a positive charge of +2e, where \( e \) is the elementary charge (\( e \approx 1.602 \times 10^{-19} \) C). Alpha particles are common in nuclear reactions and radioactive decay processes.
In the context of electric fields, alpha particles interact with electric fields due to their charge. The force on the alpha particle in an electric field is determined by the product of the field strength and the charge of the alpha particle, i.e., \( F = qE \).
In the exercise, we seek points where the forces from both the line and sheet charges result in a balanced or zero net force on the alpha particle. Such a setup implies that although alpha particles have substantial mass compared to electrons, their positively charged nature can be influenced significantly by strong or suitably oriented electric fields from nearby charges.

Electric Force

Electric force is the influence that charged objects exert on one another due to their electric fields. It occurs because charges either repel or attract based on their types (like charges repel, and opposite charges attract). The magnitude of the electric force is given by:

• \( F = qE \), where \( F \) is the force on a charge \( q \) in an electric field \( E \).

This relation shows how the force experienced by a charge depends on both the charge itself and the electric field it inhabits.
In this problem, the goal is to find where the alpha particle feels no electric force, meaning the electric fields from the line and the sheet cancel each other out.
The balance of electric forces requires that the electric fields due to the linear charge density and the surface charge density not only exist but oppose each other significantly and precisely enough. This concept underscores many electrostatic applications, from designing particle accelerators to understanding forces at play within biological systems.