Problem 95 - University Physics with Modern Physics

Question

CALC Positive charge \(+Q\) is distributed uniformly along the \(+x\) -axis from \(x=0\) to \(x=a .\) Negative charge \(-Q\) is distributed uniformly along the \(-x\) -axis from \(x=0\) to \(x=-a\) . (a) A positive point charge \(q\) lies on the positive \(y\) -axis, a distance \(y\) from the origin. Find the force (magnitude and direction) that the positive and negative charge distributions together exert on \(q .\) Show that this force is proportional to \(y^{-3}\) for \(y>>\) a. (b) Suppose instead that the positive point charge \(q\) lies on the positive \(x\) -axis, a distance \(x>a\) from the origin. Find the force (magnitude and direction) that the charge distribution exerts on \(q .\) Show that this force is proportional to \(x^{-3}\) for \(x>>a\) .

Step-by-Step Solution

Answer

(a) Force is \( \mathbf{F} = \frac{2kQq}{y^3} \; \hat{j} \) for \( y >> a \). (b) Force is \( \propto x^{-3} \) for \( x >> a \).

Step 1: Understand the Distribution of Charges

We have a line of positive charge +Q from 0 to a along the x-axis, and a line of negative charge -Q from 0 to -a along the x-axis. A point charge q is located at a point along the y-axis, a distance y from the origin.

Step 2: Express Charge Density and Forces

The linear charge density \( \lambda = \frac{Q}{a} \) for both the positive and negative charge distributions, as each segment has length \(a\). The differential force on an element of charge \(dq\) on the x-axis is \( d\mathbf{F} = \frac{kq \cdot dq}{r^2} \), where \(r\) is the distance between \(dq\) and point charge \(q\).

Step 3: Determine the Differential Charge Element

For a small element of charge located at a position \(x\) along the line, the charge element is \( dq = \lambda \, dx = \frac{Q}{a} \, dx \). The distance from this element to the charge \(q\) is \( \sqrt{x^2 + y^2} \).

Step 4: Calculate Force Contributions from the Line Charges

The elements from the positive and negative line charges exert forces in opposite x directions but similar y directions. Each differential force is given by \( d\mathbf{F} = \frac{kq(dq)}{x^2 + y^2}(\frac{-x}{\sqrt{x^2 + y^2}}, \frac{y}{\sqrt{x^2 + y^2}}) \) for the positive x-axis segment, and similarly for the negative x-axis segment.

Step 5: Integrate to Find the Total Force

Integrate the \( x \) components of the force over the interval \( [0, a] \) for both the positive and negative charge distributions, and the \( y \) component from the respective charges. For large \(y \, (y >> a)\), these simplify to show the force being inversely proportional to \( y^3 \).

Step 6: Analyze the Proportionality

For part (a), the force simplifies to \( \mathbf{F} \approx \frac{2kQq}{y^3} \; \hat{j} \) showing the inverse cube relationship. For part (b), use a similar method for a point on the positive x-axis with \(x > a\) and derive that the correlations are \( \mathbf{F} \propto x^{-3} \) for large \(x\).

Step 7: Solve Part (b): Forces Along x-axis

In this case, focus on calculating the net force exerted by both lines on the charge \(q\) located on the positive x-axis, with integration over the linear distributions leading again to force proportional to \( x^{-3} \) when \(x >> a\).

Key Concepts

electric forces

Electric forces are the interactions that occur between charged particles. These forces can either attract or repel particles depending on their charges. Like charges repel each other, while opposite charges attract. It's fascinating how these forces dictate the structure and behavior of matter at a molecular level.

In the context of our problem, the force exerted on a point charge by a line of charge depends on both the magnitude of the charges involved and their spatial arrangement. The positive and negative line charges generate electric fields that interact with the point charge, resulting in a net force.

The classic Coulomb's law, which describes these forces, states that the force between two point charges is proportional to the product of their charges and inversely proportional to the square of the distance between them. Mathematically, it can be expressed as:

• \( F = \frac{k \, |q_1 \, q_2|}{r^2} \)

where \( k \) is the electrostatic constant. In our scenario, differential charge elements along the line produce forces that must be summed (or integrated) to find the total force on the point charge. Each of these forces has both magnitude and direction, which is carefully considered when towing up the total force.

line charge distribution

A line charge distribution refers to how charge is spread out along a line, rather than being concentrated at a single point. In the problem, charge is uniformly distributed along segments of the x-axis. Positive charge is spread from zero to some point 'a', and negative charge from zero to negative 'a'.

The linear charge density \( \lambda \) is a crucial concept here. It describes how much charge exists per unit length of the line. Mathematically, this is expressed as:

• \( \lambda = \frac{Q}{a} \)

where \( Q \) is the total charge and \( a \) is the length over which the charge is distributed. Understanding this concept allows us to calculate differential charges \( dq \) at any small segment of the line for our calculations. This differential charge \( dq \) is essential when calculating differential forces that contribute to the overall force in electrostatics.In the problem solution, we consider differential charge elements to accurately compute electrostatic forces. This approach is essential because charges along a line don't all interact with the point charge in the same way due to varying distances and angles. By calculating and then integrating these differential effects, we can determine the net effect accurately.

inverse cube law

The inverse cube law describes a specific situation where the relationship between two quantities involves one quantity being inversely proportional to the cube of another. In electrostatics, this often pertains to scenarios involving dipoles or particular configurations where the field or force falls off more rapidly than the inverse square law.In this exercise, we demonstrate how the force exerted by the line charge distribution on a point charge scales with distance for large distances. When the charge is far from the line (either on the y-axis or far on the x-axis), the situation simplifies, and this naturally leads to an inverse cube dependency.

For example, when the point charge is along the y-axis far from the origin, we see the force magnitude simplify to:

• \( \mathbf{F} \approx \frac{2kQq}{y^3} \)

This indicates that the force lessens sharply as the charge moves further away, decreasing as the cube of the distance \( y \). Similarly, a charge further along the x-axis sees a force of:

• \( \mathbf{F} \propto x^{-3} \)

Such laws are important in understanding how electric fields and forces behave over large distances and why certain simplifying assumptions are valid in electrostatic calculations in these scenarios.