Problem 89 - University Physics with Modern Physics

Question

Positive charge \(Q\) is distributed uniformly along the \(x\) -axis from \(x=0\) to \(x=a\) A positive point charge \(q\) is located on the positive \(x\) -axis at \(x=a+r,\) a distance \(r\) to the right of the end of \(Q\) (Fig. 21.47\()\) . (a) Calculate the \(x\) - and \(y\) -components of the electric field produced by the charge distribution \(Q\) at points on the positive \(x\) -axis where \(x>a\) . (b) Calculate the force (magnitnde and direction) that the charge distribution \(Q\) exerts on \(q .\) (c) Show that if \(r \gg a,\) the magnitude of the force in part \((b)\) is approximately \(Q q / 4 \pi \epsilon_{0} r^{2} .\) Explain why this result is obtained.

Step-by-Step Solution

Answer

Electric field: \(E_x = \frac{k_e Q}{r(r+a)}\), \(E_y=0\); Force on \(q\): \(F = \frac{k_e Q q}{r(r+a)} \approx \frac{Q q}{4\pi\epsilon_0 r^2}\) for \(r \gg a\).

Step 1: Understand the problem and setup

We have a positive linear charge distribution \(Q\) along the \(x\)-axis from \(x=0\) to \(x=a\), and a positive point charge \(q\) at \(x=a+r\). The task is to find the electric field components due to \(Q\) at location \(x>a\) and the force exerted on \(q\).

Step 2: Define the linear charge density

The total charge \(Q\) is uniformly distributed over the distance \(x=0\) to \(x=a\). The linear charge density \(\lambda\) will be \(\lambda = \frac{Q}{a}\).

Step 3: Calculate the electric field component along the x-axis

Consider an infinitesimal segment \(dx\) at a point \(x\) on the distribution. The charge of this segment is \(dQ = \lambda \, dx\). The electric field due to this segment at a point \(x=a+r\) is \(dE = \frac{k_e \, dQ}{(a+r-x)^2}\). Since field components other than in the direction of x-axis from this linear distribution will cancel due to symmetry, we just consider x-components. Integrate from 0 to \(a\):\[E_x = \int_0^a \frac{k_e \, \lambda \, dx}{(a+r-x)^2}\].

Step 4: Integrate to find the electric field

Solve the integral from the previous step: \[ E_x = \frac{k_e \lambda}{r(r+a)} \left[ a \right] = \frac{k_e Q}{r(r+a)}.\]

Step 5: Confirm the lack of y-component

Due to symmetry of the charge distribution along the x-axis, the y-components of the electric fields created by symmetric charge elements will cancel each other, making \(E_y = 0\).

Step 6: Calculate the force on the charge q

The force on \(q\) due to the electric field created by \(Q\) is \( F = qE = \frac{k_e Q q}{r(r+a)}\).

Step 7: Simplify the force equation for r >> a

For \(r \gg a\), \(r(r+a) \approx r^2\), so \(F \approx \frac{k_e Q q}{r^2}\). Since \(k_e = \frac{1}{4\pi\epsilon_0}\), we find \[F \approx \frac{Q q}{4\pi\epsilon_0 r^2}.\] This is the force expression for a point charge, showing distant distributions act like point charges.

Key Concepts

Charge Distribution

In physics, understanding how charge is distributed over a certain region is crucial for analyzing electric fields and forces. A charge distribution provides a roadmap for where charges are located and how they interact with their surroundings. In our problem, the charge distribution occurs along the \(x\)-axis between \(x = 0\) and \(x = a\). Here, the positive charge \(Q\) is spread uniformly along this line.
The density of this spread is described by the linear charge density \(\lambda\). This is determined by dividing the total charge \(Q\) by the length of the distribution, giving us \(\lambda = \frac{Q}{a}\). This concept is essential, as it allows us to calculate the effects of the entire charge line on other charges located nearby. Understanding charge distribution helps us simplify many complex interactions into manageable calculations.

Coulomb's Law

Coulomb's Law is fundamental to understanding electric forces between charges. It's a principle that quantifies the amount of force between two point charges. According to Coulomb's Law, the electric force \(F\) between two point charges is directly proportional to the product of the magnitudes of charges \(q_1\) and \(q_2\) and inversely proportional to the square of the distance \(r\) between them. The equation is \(F = \frac{k_e \cdot q_1 \cdot q_2}{r^2}\), where \(k_e\) is Coulomb's constant \(8.9875 \times 10^9 \text{ N m}^2/ \text{C}^2\).
While Coulomb's Law is traditionally applied to point charges, we use an integrated form here to consider a continuous charge distribution, helping us model real-world problems more accurately. It's the backbone of analyzing electric interactions and is especially useful for calculating fields due to linearly distributed charges.

Force Calculation

The step of calculating the force in problems involving charge distributions is crucial and can be broken down methodically. First, we consider the electric field at the location of the point charge \(q\) due to the entire distribution \(Q\). The total force experienced by the charge \(q\) is then the product of this electric field and the magnitude of the charge \(q\).
In our scenario, the force \(F\) exerted on \(q\) by the charge distribution \(Q\) is given by \(F = qE\). After integrating to find the electric field \(E\) on \(q\), we get \(F = \frac{k_e Q q}{r(r+a)}\). It confirms how integrated electric fields can be used to determine the force impacts on singular charges beside charge distributions.

Linear Charge Density

Linear charge density is a measure of how much charge is distributed per unit length along a conductor or a line segment. It provides clarity on the intensity of the charge distribution over a line, which is essential for accurately calculating electric effects.
This quantity becomes especially important when dealing with charges spread out over a finite distance, such as the \(x\)-axis segment from \(x=0\) to \(x=a\) in our exercise. The linear charge density \(\lambda\) here is defined as \(\lambda = \frac{Q}{a}\), characterizing how \(Q\) is spread over the distance \(a\). This helps in determining small increments of charge \(dQ\) as \(dQ = \lambda \, dx\), making integral calculations about electric fields and forces achievable.