Problem 21 - University Physics with Modern Physics

Question

A positive charge \(q\) is fixed at the point \(x = 0, y = 0\), and a negative charge \(-2_q\) is fixed at the point \(x = a, y = 0\). (a) Show the positions of the charges in a diagram. (b) Derive an expression for the potential \(V\) at points on the \(x\)-axis as a function of the coordinate \(x\). Take \(V\) to be zero at an infinite distance from the charges. (c) At which positions on the \(x\)-axis is \(V = 0\)? (d) Graph \(V\) at points on the \(x\)-axis as a function of \(x\) in the range from \(x = -2a\) to \(x = +2a\). (e) What does the answer to part (b) become when \(x \gg a\)? Explain why this result is obtained.

Step-by-Step Solution

Answer

(a) Diagram has charges at (0,0) and (a,0). (b) \( V(x) = \frac{kq}{x} - \frac{2kq}{x-a} \). (c) \( x = 2a \). (d) Graph shows crossing at \( x = 2a \). (e) For \( x \gg a \), \( V(x) \approx \frac{-kq}{x} \), dominated by the negative charge.

Step 1: Diagram the Charge Configuration

To show the positions of the charges, draw a horizontal line to represent the x-axis. Place a positive charge labeled \( q \) at the origin \((0, 0)\). Place a negative charge labeled \( -2q \) at the point \((a, 0)\) on the x-axis.

Step 2: Expression for Potential on the x-axis

The electric potential due to a point charge \( q \) at a distance \( r \) is given by \( V = \frac{kq}{r} \), where \( k \) is Coulomb's constant. For a point \( (x,0) \) on the x-axis, the potential due to the positive charge at \((0,0)\) is \( V_1 = \frac{kq}{x} \). The potential due to the negative charge at \((a,0)\) is \( V_2 = \frac{-2kq}{x-a} \). The total potential at \( x \) is the sum: \[ V(x) = \frac{kq}{x} + \frac{-2kq}{x-a} \].

Step 3: Determine where V = 0

To find where the potential \( V(x) \) is zero, set the equation from Step 2 to zero: \[ \frac{kq}{x} + \frac{-2kq}{x-a} = 0 \].Simplifying, this gives \( \frac{1}{x} = \frac{2}{x-a} \). Cross-multiply and solve for \( x \) to find the point where the potential is zero. This simplifies to:\[ x^2 - 3ax + 2a^2 = 0 \].Solve this quadratic equation for \( x \) to get:\( x = \frac{3 \pm \sqrt{1}}{2}a \), yielding the solutions \( x = a \) and \( x = 2a \). However, \( x = a \) is not valid as it is the position of a charge, leaving \( x = 2a \).

Step 4: Graph the Potential

Plot the function \( V(x) = \frac{kq}{x} + \frac{-2kq}{x-a} \) over the range \(-2a \leq x \leq 2a \). Note the point where the curve crosses the x-axis at \( x = 2a \), indicating where \( V = 0 \). The graph will show positive potential close to the positive charge and negative potential close to the negative charge.

Step 5: Consider the case when x is much greater than a

When \( x \gg a \), the expression for \( V(x) \) can be approximated. Since both \( \frac{1}{x} \) and \( \frac{1}{x-a} \) are very small, consider: \[ V(x) \approx \frac{kq}{x} - \frac{2kq}{x} \approx \frac{-kq}{x} \]. "The approximation \( V(x) = \frac{-kq}{x} \) shows that the negative charge significantly influences the potential as the distance increases, reducing the potential as if it were a single negative charge."

Key Concepts

Point Charge

A point charge is an idealized model of an electrical charge which is located at a single point in space. This is a foundational concept in electrostatics and is used to simplify complex charge distributions. Point charges are treated as particles with a certain amount of charge, denoted usually as

• Positive point charge: +q
• Negative point charge: -q

In the exercise, we consider two point charges along the x-axis. A positive charge located at the origin (0,0) and a negative charge placed at some distance, a, along the x-axis. These charges create an electric field in their surroundings, impacting the potential energy at any point on the x-axis. Their separation and polarity dictate the nature of the electric potential equation.

Coulomb's Constant

Coulomb's constant, denoted as k, is a crucial part of the formulas used in electrostatics, specifically when dealing with point charges. It is a proportional constant in the electric force and electric field calculations, and its value is approximately \( k = 8.9875 \times 10^9 \text{ N m}^2/\text{C}^2 \).
This constant helps determine the strength of the interaction between charges. It can be seen in the electric potential equation that the potential, V, at a point due to a charge is given by:\[ V = \frac{kq}{r} \]where:

• q is the charge magnitude,
• r is the distance from the charge to the point of interest.

Coulomb's constant links the electric potential directly to the charge's magnitude and the distance from it, thus playing a fundamental role in calculating electric potentials.

Electric Potential Equation

Electric potential, often simply called "potential," is a measure of the potential energy per unit charge at a particular position in an electric field. The electric potential due to a point charge is given by the formula\[ V = \frac{kq}{r} \]where V is the electric potential, k is Coulomb's constant, q is the charge, and r is the distance from the charge to the point where we want to find the electric potential.
In our scenario, we have two charges:

• A positive charge, q, at the origin contributes \( V_1 = \frac{kq}{x} \) to the potential on the x-axis.
• A negative charge, -2q, at position x = a, contributes \( V_2 = \frac{-2kq}{x-a} \).

The total electric potential at any point x is simply the sum of individual potentials:\[ V(x) = V_1 + V_2 = \frac{kq}{x} + \frac{-2kq}{x-a} \]This equation is central to determining how potential changes along the x-axis and is used to calculate where potential is zero.

Graphing Electric Potential

Graphing the electric potential as a function of position provides visual insight into how the potential varies across different points. Here, we focus on the electric potential along the x-axis for two point charges placed at specific positions.
This involves plotting the equation \[ V(x) = \frac{kq}{x} + \frac{-2kq}{x-a} \]for a range of x values from -2a to 2a.

• Near the positive charge, the potential is positive and decreases as you move away.
• Near the negative charge, the potential is negative, with the strength diminishing as you go farther.
• Where the graph crosses the x-axis, potential is zero, meaning the contributions from both charges cancel out each other.

The graph effectively depicts the effects of both charges on the electric potential, clearly showing where the influence of one charge dominates over the other. Analyzing this graph helps in understanding the behavior of electric fields and potentials around point charges.