Problem 95 - University Physics with Modern Physics

Question

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 the force is proportional to \(y^{-3}\) for \(y \gg a\) . (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 \(x \gg a\) .

Step-by-Step Solution

Answer

The force on the point charge is proportional to \( y^{-3} \) on the y-axis and \( x^{-3} \) on the x-axis for large distances.

Step 1: Understand the Problem

We have two uniformly charged lines: positive charge "+Q" from x=0 to x=a on the +x-axis, and negative charge "-Q" from x=0 to x=-a on the -x-axis. We need to find the forces these distributions exert on a point charge "q" at different positions.

Step 2: Express Linear Charge Density

Calculate the linear charge density \( \lambda \). Since the charge is uniformly distributed along a length \( a \), the linear charge density is \( \lambda = \frac{Q}{a} \).

Step 3: Determine Force on Charge at y

Consider the charge \( q \) placed on the positive y-axis at distance \( y \). Each charge element \( dQ \) from the lines exerts a Coulomb force on \( q \). The force due to a small positive charge element \( dQ = \lambda dx \) at \( x \) is given by:\[ dF_+ = \frac{k \cdot q \cdot dQ}{x^2 + y^2} \cos \theta = \frac{k \cdot q \cdot \lambda dx}{(x^2 + y^2)^{3/2}} \cdot \frac{y}{\sqrt{x^2 + y^2}} \]The net force from x=-a to x=a and considering symmetry yields only a force in the y-direction.

Step 4: Integrate to Find Total Force on y-axis

Integrate the force along the x-axis from \( -a \) to \( a \):\[ F_y = \int_{-a}^{a} \frac{k \cdot q \cdot \lambda y}{(x^2 + y^2)^{3/2}} \, dx \]This integral evaluates to a proportionality relationship \( F_{y} \) for \( y \gg a \), which simplifies to \( F_{y} \propto \frac{1}{y^3} \).

Step 5: Force on Charge at x

For charge \( q \) placed on the +x-axis at \( x > a \), consider the effect of the two charge distributions as point charges. The total force arises due to their symmetrically opposite positions relative to \( q \).

Step 6: Show Force Proportion for x

Derive the force expression for the position of charge \( q \) on the +x-axis. Due to symmetry, the resulting force simplifies to an expression:\[ F_x \propto \frac{1}{x^3} \]This progression involves substituting \( x \gg a \) for simplification.

Key Concepts

Coulomb's Law

Coulomb's Law is a fundamental principle used to calculate the electrostatic force between two point charges. According to this law, the force between two charges is directly proportional to the product of their magnitudes and inversely proportional to the square of the distance between them. The mathematical expression for Coulomb's Law is given by:

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

where \( F \) is the magnitude of the force, \( k \) is Coulomb's constant \( (8.99 \times 10^9 \, \text{N m}^2/\text{C}^2) \), \( q_1 \) and \( q_2 \) are the charges, and \( r \) is the distance between the charges.
In this exercise, Coulomb's Law helps us calculate forces on a specific charge due to continuous charge distributions along the axes. Since charge is distributed uniformly along lines, each tiny element of charge exerts a Coulomb force on the point charge \( q \).
By integrating the contributions from these infinitesimal charges, we determine the net force experienced by \( q \) due to the linear distributions on both the positive and negative x-axes.

Charge Distribution

Charge distribution refers to how charge is spread along a given length, area, or volume. In this problem, we consider a linear charge distribution, which is charge spread out along a line. Two distributions are given:

• A positive charge \(+Q\) uniformly distributed from \(x=0\) to \(x=a\)
• A negative charge \(-Q\) uniformly distributed from \(x=0\) to \(x=-a\)

For a linear charge distribution, we use linear charge density \( \lambda \), which is calculated as the total charge \( Q \) divided by the length over which it is distributed. This is expressed as:

• \( \lambda = \frac{Q}{a} \)

The charge density allows us to express each infinitesimal segment of charge along the line, \( dQ = \lambda \, dx \), making it easier to apply integration techniques to solve for the total electrostatic effects on a point charge located at specified positions relative to these distributions.

Electric Field

An electric field is a region around a charged object within which other charges experience a force. The strength and direction of the electric field depend on the charge distribution causing it and the position where it is measured.
When dealing with continuous charge distributions, such as linear charge distributions along axes, the electric field can be considered the result of the sum of the fields produced by each infinitesimal charge element.
The electric field \( E \) created by an infinitesimal charge \( dQ \) at a distance \( r \) from the point of interest is given by:

• \( dE = \frac{k \, dQ}{r^2} \)

By integrating this expression over the entire length of the charge distribution, we can find the total electric field at any point influenced by the distribution. This integral involves accounting for the direction of each small electric field vector and often utilizes trigonometric components to resolve the overall field effect in a specific direction.

Linear Charge Density

Linear charge density \( \lambda \) is a measure of how much charge is distributed along a line. It plays a crucial role in problems involving linear charge distributions, as it allows us to quantify the amount of charge per unit length. The formula for linear charge density is:

• \( \lambda = \frac{Q}{a} \)

here, \( Q \) is the total charge and \( a \) is the length over which it is distributed.
This concept simplifies the calculations of forces and electric fields due to a linear distribution because it lets us consider tiny elements of charge divided across the entire length.
By using \( \lambda \), we can efficiently perform integrations to determine the total forces acting on a point charge from continuous and uniformly distributed charges, thereby connecting the continuity of charge with the discrete nature of forces and fields acting at specified points beyond the distribution.