Problem 79 - University Physics with Modern Physics

Question

Electric charge is distributed uniformly along a thin rod of length \(a,\) with total charge \(Q .\) Take the potential to be zero at infinity. Find the potential at the following points (Fig. P23.79): (a) point \(P,\) a distance \(x\) to the right of the rod, and (b) point \(R,\) a distance \(y\) above the right- hand end of the rod. (c) In parts (a) and \((\mathrm{b}),\) what does your result reduce to as \(x\) or \(y\) becomes much larger than \(a\) ?

Step-by-Step Solution

Answer

(a) At point P: \( V = \frac{kQ}{a} \ln\left( \frac{x+a}{x} \right) \); (b) At point R: \( V = \frac{kQ}{a} \ln\left( \frac{y + \sqrt{a^2 + y^2}}{y} \right) \); (c) For large distances \( V \approx \frac{kQ}{x} \) or \( V \approx \frac{kQ}{y} \).

Step 1: Define Linear Charge Density

The linear charge density \( \lambda \) is defined as the total charge \( Q \) divided by the length \( a \) of the rod: \[\lambda = \frac{Q}{a}.\]

Step 2: Setup the Potential at Point P using Integration

To find the potential \( V \) at point \( P \), located at distance \( x \) from the measured end of the rod, we use:\[dV = \frac{k \cdot dQ}{r},\]where \( dQ = \lambda \, dx' \) and \( r = x + x' \). This gives:\[V = \int_0^a \frac{k \lambda \, dx'}{x + x'}.\]

Step 3: Evaluate the Integral for Point P

Perform the integration for \( V \) at point \( P \):\[V = k \lambda \int_0^a \frac{dx'}{x + x'} = k \lambda \ln(x + a) - k \lambda \ln(x).\]This simplifies to:\[V = k \lambda \ln\left( \frac{x+a}{x} \right).\] Substituting \( \lambda = \frac{Q}{a} \) gives:\[V = \frac{kQ}{a} \ln\left( \frac{x+a}{x} \right).\]

Step 4: Setup the Potential at Point R using Integration

For point \( R \), a distance \( y \) above the rod, calculate \( V \) using:\[dV = \frac{k \cdot dQ}{r},\]where \( dQ = \lambda \, dx' \) and \( r = \sqrt{x'^2 + y^2} \). This gives:\[V = \int_0^a \frac{k \lambda \, dx'}{\sqrt{x'^2 + y^2}}.\]

Step 5: Evaluate the Integral for Point R

The integral becomes:\[V = k \lambda \int_0^a \frac{dx'}{\sqrt{x'^2 + y^2}}.\] The result of this standard integral is:\[V = k \lambda \ln\left( y + \sqrt{a^2 + y^2} \right) - k \lambda \ln(y).\] Simplified, it becomes:\[V = \frac{kQ}{a} \ln\left( \frac{y + \sqrt{a^2 + y^2}}{y} \right).\]

Step 6: Analysis for Large Distances

For part (c), as \( x \to \infty \) or \( y \to \infty \), both results reduce to:\[ V \approx \frac{kQ}{x} \quad \text{or} \quad V \approx \frac{kQ}{y} \] respectively, which is the potential due to a point charge.

Key Concepts

Linear Charge Density

When dealing with electric charges distributed along a line, such as a rod, linear charge density becomes a critical concept. Linear charge density, denoted as \( \lambda \), quantifies how much charge is concentrated along the line per unit length.
For a rod with total charge \( Q \) extending over a length \( a \), the linear charge density is defined as:

• \( \lambda = \frac{Q}{a} \)

This simplification means that any portion of the rod can be thought of as containing a charge equal to \( \lambda \times \) its length. Understanding linear charge density helps set the stage for calculations involving electric fields and potentials across objects with distributed charges. It also streamlines our mathematical approach when dealing with seemingly complex geometrical configurations of charges.

Integration

Integration is a vital mathematical tool used to calculate quantities that vary continuously. In the context of electrostatics, it is often used to sum up infinitesimally small contributions from a distributed charge to determine properties like electric potential.
To find the potential at a point due to a charged rod, we integrate small elements of charge \( dQ \). The expression for a small piece of the potential \( dV \) is:

• \( dV = \frac{k \cdot dQ}{r} \)

where \( dQ = \lambda \, dx' \) and \( r \) is the distance from the charge element to the point of interest.
The integral:

• \( V = \int_0^a \frac{k \lambda \, dx'}{r} \)

calculates the entire potential by summing the contributions from each section of the rod. Integrating along the length of the rod allows us to see how the potential changes based on the geometry and position of the charge distribution.

Potential Due to Point Charge

The concept of electric potential due to a point charge is foundational in electrostatics. It describes how much work is needed to bring a unit charge from infinity to a point in an electric field.
The potential \( V \) due to a point charge \( Q \) at a distance \( r \) is given by:

• \( V = \frac{kQ}{r} \)

where \( k \) is the electrostatic constant. When considering distributed charge systems becoming infinitely distant, their potential nature approaches that of a point charge.
In the problem, as distances \( x \) or \( y \) become significantly larger than the rod's length \( a \), the distributed charge appears as a single point charge. Thus, the potential simplifies to the expression above, highlighting how distant charge distributions mimic point charges in their electric field influence.

Electrostatics Problem Solving

Solving electrostatics problems requires understanding how charge distributions influence electric fields and potentials. These problems often demand a systematic approach.
The typical steps in electrostatics problem solving include:

• Identifying the type of charge distribution (point, line, surface, or volume).
• Defining appropriate charge density expressions for the system.
• Employing integration to calculate electric fields or potentials.
• Analyzing the symmetry of the problem to simplify calculations when possible.
• Understanding asymptotic behaviors, such as how results change at large distances.

In the exercise, after defining the linear charge density, the potential is found using integration to sum up contributions from each element of the charged rod. Finally, examining the results at large distances reinforces the connection to simpler point charge models. This approach aids in grasping the full picture of electric potential in complex charge systems.