Problem 23 - University Physics with Modern Physics

Question

A positive charge \(+q\) is located at the point \(x=0, y=-a\) and a negative charge \(-q\) is located at the point \(x=0, y=+a\) . (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 coondinate \(x\) . Take \(V\) to be zero at an infinite distance from the charges. (c) Graph \(V\) at points on the \(x\) -axis as a function of \(x\) over the range from \(x=-4 a\) to \(x=+4 a\) . (d) What is the answer to part (b) if the two charges are interchanged so that \(+q\) is at \(y=+a\) and \(-q\) is at \(y=-a ?\)

Step-by-Step Solution

Answer

Potential along the x-axis is zero for both configurations.

Step 1: Diagram Construction

To represent the positions of the charges, draw a straight vertical line where both charges have a coordinate of \(x=0\). Place the positive charge \(+q\) at \(y=-a\), and the negative charge \(-q\) at \(y=+a\). The charges should be symmetrically located along the y-axis.

Step 2: Deriving the Potential Expression

To find the potential \(V\) at a point on the x-axis a distance \(x\) from \(x=0\), consider the contributions from both charges. The potential due to a point charge \(q\) at a distance \(r\) is \(V = \frac{kq}{r}\), where \(k\) is Coulomb's constant. The distance from the positive charge \(+q\) to a point on the x-axis is \(\sqrt{x^2+a^2}\), and similarly for the negative charge. Therefore, the net potential at the point is given by: \[ V(x) = \frac{kq}{\sqrt{x^2+a^2}} - \frac{kq}{\sqrt{x^2+a^2}} = 0 \] This simplifies to zero because the contributions cancel out.

Step 3: Graphing the Potential

Since the net potential calculated in Part (b) is zero for all positions on the x-axis, graphing \(V(x)\) for the range \(x=-4a\) to \(x=+4a\) would show a horizontal line at \(V=0\). This implies that regardless of the position on the x-axis, the potential is neutral.

Step 4: Interchanging Charges

If the charges are interchanged, with \(+q\) at \(y=+a\) and \(-q\) at \(y=-a\), the situation is symmetric to the original configuration. Therefore, when recalculating the potential, the same cancellation occurs along the x-axis. Thus the potential at any point on the x-axis remains zero: \[ V'(x) = \frac{kq}{\sqrt{x^2+a^2}} - \frac{kq}{\sqrt{x^2+a^2}} = 0 \].

Key Concepts

Point Charges

In the realm of electromagnetism, a point charge is a very small charged particle. Imagine it as an idealized model allowing us to simplify complex charge distributions by focusing on discrete particles with negligible size compared to the distances involved. Point charges can be either positive or negative and are used extensively in calculations involving electric fields and potentials to predict the behavior of more complicated systems.

When discussing point charges, it's essential to remember that these charges can attract or repel each other depending on their nature. Opposite charges (+ and -) attract, while like charges (+/+ or -/-) repel. This behavior is foundational in understanding how electric fields and potentials behave.

In the given problem, two point charges are strategically placed. A positive point charge, denoted as +q, is at one location on the vertical y-axis, while a negative point charge, -q, is placed symmetrically on the other side of the origin. This specific arrangement of charges helps us understand symmetries in electric fields and potentials, which simplifies many calculations.

Coulomb's Law

Coulomb's Law is central to electromagnetism and provides a formula for the force between two point charges. According to the law, the electric force (\( \vec{F} \)) between two point charges is given by:

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

where \( q_1 \) and \( q_2 \) are the magnitudes of the charges, \( r \) is the distance between the charges, \( \hat{r} \) is the unit vector pointing from one charge to the other, and \( k \) is Coulomb's constant, approximately \( 8.99 \times 10^9 \text{ N m}^2/\text{C}^2 \).

Coulomb's Law elaborates that the magnitude of the force is directly proportional to the product of the absolute values of the charges and inversely proportional to the square of the distance separating them. The direction of the force follows intuitive physical laws: charges repel if they are alike and attract if they are opposite.

This fundamental concept is illuminated in the given exercise: though each charge acts independently, their mutual existence results in cancellation effects along the symmetry line (here, the x-axis), rendering the net electric potential as zero.

Electric Field

The electric field is a vector field that surrounds electric charges, representing the force per unit charge exerted on a positive test charge placed in the field. It is expressed mathematically as \( \vec{E} = \frac{\vec{F}}{q} \), where \( \vec{F} \) is the force experienced and \( q \) is the test charge.

For a single point charge, the electric field (\( \vec{E} \)) can be described by the equation:

• \( \vec{E} = k \frac{q}{r^2} \hat{r} \)

Here, \( q \) is the charge creating the field, \( r \) is the distance from the charge, and \( \hat{r} \) points radially away from positive charges and towards negative charges.

When we have multiple charges, the resulting electric field at any point in space is the vector sum of the individual fields from each charge. In this exercise, the symmetry of the arrangement causes the electric fields due to the two charges \(+q\) and \(-q\) to cancel each other out along the x-axis. Thus, along this axis, not only does the potential simplify to zero, but the electric field also results in combined effects that are neutral.