Force on 3rd Charge: Examining x-Component vs. x

In summary, the conversation discusses the net force exerted on a third positive charge located at the origin on the x-axis between two positive point charges of equal magnitude located on the y-axis at points y=+a and y=-a. The net force is calculated using the formula F= k(q1)(q3)/(r^2) and its components, resulting in F=0 at the origin and (2kx q^2)/ ((x^2)+(a^2))^(3/2) in the +x-direction. To sketch the graph of the x-component of the net force as a function of x for values of x between +4a and -4a, the x-axis represents the values of x (-
  • #1
eil2001
13
0
Here's a question from my textbook:
Two positive point charges, each of magnitude q, are located on the y-axis at points y=+a and y=-a. A third positive charge of the same magnitude is located at some point on the x-axis.
(a) What is the net force exerted on the third charge when it is at the origin?
(b) What are the magnitude and direction of the net force on the third charge when its coordinate is x?
(c) Sketch a graph of the x-component of the net force on the third charge as a function of x for values of x between +4a and -4a. Plot forces to the right upward and forces to the left downward.

I got (a) and (b) using F= k(q1)(q3)/(r^2) and components. (a) F=0, (b) F= (2kx q^2)/ ((x^2)+(a^2))^(3/2) in the +x-direction.

However, I don't really understand (c). The x-component of the net force is F_x = (2kx q^2)/ ((x^2)+(a^2))^(3/2) , but then do I plug in 4a, 3a, ... , -4a for values of x to get the graph? What are the axes? Are the "4a, 3a, ..., -4a" on the x-axis and F_x on the y-axis?

Thanks!
 
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  • #2
eil2001 said:
...
However, I don't really understand (c). The x-component of the net force is F_x = (2kx q^2)/ ((x^2)+(a^2))^(3/2) , but then do I plug in 4a, 3a, ... , -4a for values of x to get the graph? What are the axes? Are the "4a, 3a, ..., -4a" on the x-axis and F_x on the y-axis?

Thanks!

Yes, that's right.

On the Y-axis to plot the F_x in multiples of kq^2/a^2 (ie : let each division be equal to kq^2/a^2)

On the X-axis you have different values of x, namely -4a, -3a, ..., 3a, 4a (ie : multiples of 'a' from -4 to +4)
 
  • #3


Your understanding of (a) and (b) is correct. For (c), you are correct in plugging in values of x from +4a to -4a to get the graph. The x-axis will be the values of x, and the y-axis will be the x-component of the net force (F_x). So, for example, when x=4a, the x-component of the net force will be (2k(4a) q^2)/ ((4a)^2)+(a^2))^(3/2), and so on for the other values of x. Then, you can plot these values on a graph with x on the x-axis and F_x on the y-axis. This will give you a visual representation of how the x-component of the net force changes as the position of the third charge changes along the x-axis. I hope this helps clarify the concept for you.
 

1. What is the force on the third charge in a system?

The force on the third charge in a system is the result of the interaction between the third charge and the other charges in the system. It is a vector quantity, meaning it has both magnitude and direction. The force on the third charge can be calculated using Coulomb's Law, which states that the force is directly proportional to the product of the charges and inversely proportional to the square of the distance between them.

2. How is the x-component of the force on the third charge calculated?

The x-component of the force on the third charge is calculated by using trigonometric functions to break down the force vector into its horizontal and vertical components. The x-component specifically refers to the horizontal component of the force, which is parallel to the x-axis. This calculation is important in understanding the direction and magnitude of the force acting on the third charge.

3. What is the significance of examining the x-component of the force on the third charge?

Examining the x-component of the force on the third charge allows us to understand how the force is acting in a specific direction. This is important in determining the overall motion and equilibrium of the system. By analyzing the x-component, we can also determine the net force on the third charge, which is crucial in understanding the resulting acceleration.

4. How does the x-component of the force on the third charge differ from the x-component of the force on other charges in the system?

The x-component of the force on the third charge may differ from the x-component of the force on other charges in the system due to variations in charge magnitude and distance. The x-component of the force is dependent on the individual charges and their relative positions, so it may differ from charge to charge within the system.

5. What are some real-life applications of examining the x-component of the force on the third charge?

The x-component of the force on the third charge has many real-life applications, including in electrical circuits, where it is important to understand the forces acting on individual charges to ensure proper functioning. It is also important in understanding the stability of structures, such as bridges, where the x-component of the force must be considered to prevent structural failures. Additionally, it is crucial in the study of celestial mechanics, where the x-component of the force on celestial bodies helps us understand their orbits and interactions.

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