Problem 10 - Physics

Question

A charge \(+q\) is located at the origin, while an identical charge is located on the \(x\) axis at \(x=+0.50 \mathrm{~m}\). A third charge of \(+2 q\) is located on the \(x\) axis at such a place that the net electrostatic force on the charge at the origin doubles, its direction remaining unchanged. Where should the third charge be located?

Step-by-Step Solution

Answer

Place the third charge at \(x = +0.707 \text{ m}\).

Step 1: Understand the Problem

We are given two charges: a charge \(+q\) at the origin and another identical \(+q\) at \(x=+0.50\, \text{m}\) on the x-axis. A third charge \(+2q\) needs to be placed such that the net force on the charge at the origin \(+q\) doubles but remains in the same direction. We need to find the location of the third charge.

Step 2: Analyze the Initial System

The initial electrostatic force on the charge at the origin due to the charge at \(x=+0.50\, \text{m}\) is directed along the positive x-axis. Using Coulomb's law: \[ F = k \frac{q^2}{0.5^2} \] where \(k\) is the electrostatic constant.

Step 3: Determine Required Force

The force on the charge at the origin must double. Therefore, the new force should be:\[ 2F = 2\left( k \frac{q^2}{0.5^2} \right) \] which gives us the new magnitude of the force without altering its direction (still in the positive x-direction).

Step 4: Consider the Third Charge's Effect

Let the third charge \(+2q\) be placed at a distance \(x\) from the origin. It will exert an electrostatic force on the charge at the origin given by:\[ F_{2q} = k \frac{2q^2}{x^2} \] This force should combine with the original force to achieve the required double force, still aiming to the positive x-direction.

Step 5: Solve for x

For the force direction to remain unchanged, the third charge must be on the positive x-axis, so its force adds directly to the force from the second charge:\[ k \frac{q^2}{0.5^2} + k \frac{2q^2}{x^2} = 2\left(k \frac{q^2}{0.5^2}\right) \]Simplifying, we get:\[ k \frac{2q^2}{x^2} = k \frac{q^2}{0.5^2} \]Canceling terms and solving for \(x\), we find:\[ \frac{2}{x^2} = \frac{1}{0.25} \]\[ 2 = \frac{x^2}{0.25} \]\[ x^2 = 0.5 \]\[ x = \sqrt{0.5} \approx 0.707 \text{ m}\] The third charge is placed at \(x = +0.707 \text{ m}\).

Step 6: Conclusion

Placing the third charge at \(x = +0.707 \text{ m}\) on the positive x-axis ensures the net force on the charge at the origin doubles while maintaining the same direction.

Key Concepts

Coulomb's law

Coulomb's law serves as a fundamental principle in electrostatics, describing the force between two electric charges. To understand this better, imagine two charges exerting force on one another, much like little magnetic fields interacting. The magnitude of this force is dependent on a few factors:

• The amount of charge each has, denoted by their absolute values.
• The distance between the charges, as force decreases with increased separation.
• A constant factor known as the electrostatic constant, symbolized by the letter “k”.

The formal expression of Coulomb's law is given by:\[ F = k \frac{q_1 q_2}{r^2} \]where:

• \( F \) is the magnitude of the force.
• \( q_1 \) and \( q_2 \) are the magnitudes of the charges.
• \( r \) is the distance between the charges.

The constant \( k \) in this equation reflects the permittivity of the medium between charges and is approximately \( 8.99 \times 10^9 \, \text{N m}^2/\text{C}^2 \) in a vacuum. This law explains how charges repel or attract each other, mirroring the problem's situation where charges influence each other by adding forces along an axis.

Electric charge distribution

Electric charge distribution refers to the arrangement of electric charges in a given space. In electrostatics, the distribution determines how electrostatic forces act on a localized charge. This exercise deals with point charges arranged along the x-axis, which is a simple but insightful configuration. For example, if you have a series of charges lined up on an axis, the way these charges are spaced influences the resulting force felt by each. Identical charges distributed at specific intervals create predictable patterns in force distribution. When a charge experiences a force, it's due directly to this distribution. Here, altering positions slightly can drastically change how forces align and balance. In our problem, the specific placement of the third charge was critical. This positioning was essential to ensure that the net force experienced by a charge at the origin doubled, keeping the direction constant. By understanding charge distribution and how charges impact and are impacted by one another, predicting and manipulating the results becomes intuitive.

Vector addition in electrostatics

Vector addition in electrostatics is crucial for accurately predicting how forces from multiple charges combine. In simple terms, vectors have direction and magnitude. When forces act on a charge, they are represented as vectors, often depicted with arrows indicating direction and length proportional to magnitude. In electrostatics, each charge can exert a force in a specific direction. To find the overall force on a particular charge, like the one at the origin in our exercise, we add these vectors together. This process considers both the direction and magnitude:

• Add forces pointing in the same direction by combining their lengths.
• Resolve situations where forces have different directions by using components, thus breaking them down into simpler, manageable pieces.

In the exercise, adding the forces due to the two surrounding charges used this principle of vector addition. Despite the introduction of a third charge, the goal was to ensure the resultant force remains in a stable direction, still pointing "up" on the x-axis but with doubled magnitude. This clever balancing act highlights the power and necessity of understanding vector addition in handling electrostatic problems.