Problem 36 - University Physics with Modern Physics

Question

22.36. A long line carying a uniform linear charge density \(+50.0 \mu C / m\) runs parallel to and 10.0 \(\mathrm{cm}\) from the surface of a large, flat plastic sheet that has a uniform surface charge density of \(-100 \mu \mathrm{C} / \mathrm{m}^{2}\) on one side. Find the location of all points where an \(\alpha\) particle would feel no force due to this arrangement of charged objects.

Step-by-Step Solution

Answer

The point is 25.9 cm from the plane, on the opposite side of the line charge.

Step 1: Understand the Problem

We need to find the points where an alpha particle experiences no net electric force due to a line charge and a plane charge. This involves finding where the electric fields due to these charges cancel each other out.

Step 2: Electric Field Due to Line Charge

The electric field at a distance \( r \) from an infinite line charge with linear charge density \( \lambda \) is given by the formula: \[E_{line} = \frac{\lambda}{2\pi\varepsilon_0 r}\]. Here, \( \lambda = 50.0 \, \mu C/m \) and \( r \) is the distance from the line charge to the point of interest.

Step 3: Electric Field Due to Plane Charge

The electric field due to an infinite plane with surface charge density \( \sigma \) is \[E_{plane} = \frac{\sigma}{2\varepsilon_0}\]. Here, \( \sigma = -100 \, \mu C/m^2 \). The field is constant and directed perpendicular to the surface.

Step 4: Set the Conditions for Zero Net Force

For an \( \alpha \) particle to experience no net force, the electric fields due to the line charge \( E_{line} \) and the plane charge \( E_{plane} \) must cancel: \[E_{line} + E_{plane} = 0\]. Substitute the expressions for the electric fields to solve for \( r \).

Step 5: Solve for \( r \)

Using the equations from previous steps:\[\frac{\lambda}{2\pi\varepsilon_0 r} = \frac{-\sigma}{2\varepsilon_0}\].Cancel \( \varepsilon_0 \) and rearrange:\[\frac{\lambda}{2\pi r} = -\frac{\sigma}{2}\], solve for \( r \): \[ r = \frac{\lambda}{\pi \sigma} \approx \frac{50.0 \times 10^{-6}}{\pi \times -100 \times 10^{-6}} \]. Solve this calculation to find that \( r \) is positive.

Step 6: Calculate and Interpret the Result

Calculate \( r \) to find the precise distance from the line charge where the electric fields cancel: \[ r = \frac{50.0 \times 10^{-6}}{-3.1416 \times 100 \times 10^{-6}} \approx -0.159 \, m \].This implies the point of zero net force is 10 cm (given distance) + 15.9 cm (the positive value found) from the plane, on the side away from the plane.

Key Concepts

Linear Charge Density

Linear charge density is a measure of the quantity of charge per unit length along a charged line. It is particularly relevant in situations involving long, charged lines, like wires or rods. This concept is expressed in units of coulombs per meter - (\( \mu C/m \)).A higher linear charge density implies a stronger electric field around the line, influencing nearby charged particles.
In the problem, the linear charge density is given as - \( 50.0 \mu C/m \),which means that every meter of the line contains - \( 50.0 \mu C \) of charge.The electric field created by an infinite line charge is radial and its strength decreases with distance. It is mathematically expressed by the formula \( E_{line} = \frac{\lambda}{2\pi\varepsilon_0 r} \), where \( \lambda \) is the linear charge density, \( \varepsilon_0 \) is the vacuum permittivity, and \( r \) is the distance from the line.

Surface Charge Density

Surface charge density refers to the amount of electric charge per unit area located on a surface. It is expressed in units of coulombs per square meter - (\( \mu C/m^2 \)).Commonly associated with flat surfaces, such as planes or sheets, surface charge density affects the electric field emitted above and below the surface.
In this particular problem, the plastic sheet has a surface charge density of - \( -100 \mu C/m^2 \),implying an excess of negative charge. This results in a uniform electric field perpendicular to and away from the surface.The electric field due to an infinite plane is constant and its strength is determined by the formula \( E_{plane} = \frac{\sigma}{2\varepsilon_0} \).Here, \( \sigma \) represents the surface charge density, and \( \varepsilon_0 \) is the vacuum permittivity. The electric field is directed either towards or away from the surface, depending on the sign of \( \sigma \).

Alpha Particle

An alpha particle is a type of subatomic particle consisting of two protons and two neutrons. It is typically released during alpha decay of certain radioactive substances and carries a positive charge - (twice that of the proton's charge). Alpha particles are relatively heavy and positively charged compared to other types of radioactive emissions, such as beta or gamma particles. In the context of electric fields, alpha particles are significantly influenced by the presence of charged objects due to their positive charge. They move in response to forces exerted by electric fields, seeking positions where they experience net zero force.
The problem is centered on finding such a point where the alpha particle feels no electric force, achieved by the careful cancellation of fields produced by the linear and surface charge densities.

Electric Force

Electric force is a manifestation of the electric field's influence on a charged particle. When a charged particle is placed in an electric field, it experiences a force proportional to its charge and the electric field's strength, as described by Coulomb's law.
The equation for electric force \( F = qE \) illustrates that it depends on the magnitude of both the charge - \( q \) and the electric field - \( E \).In our scenario, an alpha particle is subject to two components of electric force due to:- the positive linear charge density of the long line,- the negative surface charge density of the plastic sheet. The overall objective is to determine where these electric forces balance out to zero, indicating the alpha particle experiences no net electric force and thus remains stationary. Achieving this balance involves setting equal magnitudes of electric forces from opposing fields.

Charge Distribution

Charge distribution represents how electric charge is spread out in space. There are various forms of charge distribution, including linear, surface, and volume distributions, corresponding respectively to charges spread along a line, across a surface, or throughout a volume.
Charge distribution affects the electric field pattern and intensity created around these charged objects.In the exercise example, the presence of linear charge density - (\( +50.0 \mu C/m\)) and surface charge density - (\( -100 \mu C/m^2\)) produces distinct fields. The challenge lies in determining where these fields integrate to create a zero-force condition for the alpha particle.Understanding the spatial effects of these charge distributions helps predict how materials and particles will behave in those fields. Calculating exact points of cancellations and equilibriums, like the task at hand, involves leveraging these distributions' fundamental properties and applying mathematical equations as seen in Gauss's law and superposition principle.