Problem 48 - University Physics with Modern Physics

Question

A positive point charge \(q\) is placed at \(x=a,\) and a negative point charge \(-q\) is placed at \(x=-a\) . (a) Find the magnitude and direction of the electric field at \(x=0\) .(b) Derive an expression for the electric field at points on the \(x\) - axis. Use your result to graph the \(x\) -component of the electric field as a function of \(x\) , for values of \(x\) between \(-4 a\) and \(+4 a\) .

Step-by-Step Solution

Answer

At \(x=0\), the electric field magnitude is \(2\frac{kq}{a^2}\), directed left. The field expression is \(E_x = \frac{-4kqax}{(x^2-a^2)^2}\).

Step 1: Understanding the Problem Context

We have two point charges: a positive charge \(q\) at \(x=a\) and a negative charge \(-q\) at \(x=-a\). We need to find the electric field at specific locations on the x-axis due to these charges.

Step 2: Electric Field Definition

The electric field \(E\) due to a point charge is given by the formula: \(E = \frac{k \, |q|}{r^2}\), where \(k\) is the electrostatic constant, \(q\) is the charge, and \(r\) is the distance from the charge.

Step 3: Calculating Electric Field at \(x=0\)

At \(x=0\), look at the contributions from both charges. The distance from each charge to \(x=0\) is \(a\). The magnitude of the electric field generated by each charge is \(E_1 = \frac{kq}{a^2}\). Both fields are directed toward the negative charge (leftward), so they add up: \(E = 2\frac{kq}{a^2}\).

Step 4: Direction of Electric Field at \(x=0\)

As both charge-generated fields point toward the negative charge at \(x=-a\), the resultant electric field at \(x=0\) is directed toward the negative charge (i.e., left along the x-axis).

Step 5: General Expression for Electric Field on the x-axis

At any point \(x\), the electric field due to \(q\) at \(x=a\) is \(E_+ = \frac{kq}{(x-a)^2}\) to the left, and due to \(-q\) at \(x=-a\) is \(E_- = \frac{kq}{(x+a)^2}\) to the right. The total electric field \(E_x\) is \( E_x = \frac{kq}{(x+a)^2} - \frac{kq}{(x-a)^2}\).

Step 6: Simplifying the Expression

Combine the terms to express the electric field as a function of \(x\):\[ E_x = \frac{kq[(x-a)^2 - (x+a)^2]}{(x^2-a^2)^2} = \frac{-4kqax}{(x^2-a^2)^2} \].

Step 7: Graphing the Electric Field

This expression can now be graphed for \(x\) between \(-4a\) and \(+4a\). As \(x\) approaches \(\pm a\), the field intensifies and changes direction sharply. At \(x=0\), the field points left. The graph will be symmetrical about the origin.

Key Concepts

Superposition Principle

The superposition principle is a pivotal concept when dealing with electric fields from multiple charges. It states that the total electric field created by several point charges is the vector sum of the fields produced by each charge separately. This means if you have several charges influencing a point in space, you determine the effect of each charge independently, then combine these effects to find the total electric field. This principle simplifies complex problems, making them manageable.

In contexts like the original exercise, where both a positive and a negative charge create fields, we apply the superposition principle to calculate the net electric field. For a point on the x-axis, each charge contributes its own field magnitude and direction. By summing these individual contributions, taking care to respect their directions, we successfully determine the combined field at any given point.

• Calculate each charge's electric field separately.
• Add vectorially to find total effect.
• Respect direction: Fields in opposite directions may cancel out.

The superposition principle makes it possible to break down complex field calculations into more manageable bits, as seen by considering the influence of point charges separately before calculating the resultant field.

Coulomb's Law

Coulomb's law is fundamental when discussing the interactions between point charges. This law establishes that the electric force between two charges is directly proportional to the product of their magnitudes and inversely proportional to the square of the distance between them. Mathematically, this is represented by:\[F = \frac{k |q_1 q_2|}{r^2}\]where \(k\) is Coulomb's constant, \(q_1\) and \(q_2\) are the charges, and \(r\) is the distance between the charges.

For electric fields, which are the focus in this example, Coulomb's law allows us to determine the field created by any given point charge at a specific distance:\[E = \frac{k |q|}{r^2}\]where \(E\) is the electric field, \(k\) is the electrostatic constant, \(q\) is the charge, and \(r\) is the distance from the charge. The direction of the electric field is away from positive charges and toward negative charges.

• Specify the charge creating the field.
• Determine the distance to the point of interest.
• Apply the formula to find the field's magnitude.

Understanding Coulomb's law is crucial for predicting how charges will interact and influence each other's fields.

Electric Field Graph

Graphing the electric field is an excellent way to visualize how the field changes across different points in space. In the original exercise, the goal is to derive and display the electric field along a section of the x-axis, influenced by two opposite charges.

The derived expression for the electric field, \(E_x = \frac{-4kqax}{(x^2-a^2)^2}\), gives insights into how the field behaves between the two point charges, from \(-4a\) to \(+4a\). This graph shows:

• A sharpening field intensity as \(x\) nears each charge.
• A direction shift at \(x = 0\), reflecting the influence of the negative charge.
• Symmetrical behavior around the origin due to opposite charges.

Plotting this equation allows visualization of these changes: as you move along the line, the field's strength and direction continuously vary, creating a dynamic representation.
By studying the graph, one sees how proximity to the charges amplifies field strength, while direction reversals illustrate their relative influences. This helps in understanding and predicting field behavior in physical spaces more intuitively.