Is the Calculation of Linear Density k for a Uniformly Charged Bar Correct?

In summary: The first point is fairly simple. If you integrate the function from 0 to a given length, then the result is a quantity that has the same dimensions as the original function, but is called a "linear density" because it is proportional to the square of the distance from the point of integration.The second point is more complicated, but essentially it involves using a mathematical formula to calculate the electric potential of a point in space. This potential is determined by the distance to other points in space, and is used to determine the Electric field at that location.In summary, the two points discuss how linear density is proportional to the distance from a point of integration, and how the electric potential at a location is determined by the distance
  • #1
ToucanFodder
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Homework Statement
A charge Q is distributed on a insulating bar of lenght L with linear density λ, expressed in C/m. λ=kx where k is a constant and x the distance of the generic point P expressed in meters from the origin of the bar O.
1)Calculate k
2)Calculate the electric potential at the point A positioned perpendicularly from O at a distance R
Relevant Equations
λ=kx
I attached a drawing of the problem for a better understanding and my attempted solutions.

The first point is fairly simple but there's something that I can't figure out.

dq=λdx=kxdx

Q=∫ k x dx from 0 to L -> Q=k[x^2/2]0-L -> Q=(L^2/2)k -> k=2Q/L^2

This is what I came up with. I integrated on the entire bar and calculated k but I'm not quite sure that's correct honestly. I feel like it makes sense mathematically but not physically? Linear density in general is C/m and here I have something that will look like this C/m=(C/m^2)x. Is that fair? I don't understand but I'd really love to.

Point 2 wasn't too bad I just repeated a similar reasoning this time using the electric potential formula. I attached my calculations since writing them down in this format doesn't provide a great result. I think I got it right but I'd love for some feedback if I missed the point.

Also sorry for some mistakes, English is not my first language and scientific terms and expressions can be hard.
 

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  • #2
The way that ##k## is defined it will have units of ##C/m^2##. In which case ##\lambda## has units of ##C/m##.

The potential looks correct.
 
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  • #3
Thanks a lot!
 
  • #4
"linear density λ, expressed in C/m. λ=kx where k is a constant and x the distance of the generic point P expressed in meters"

I dislike questions that prescribe units for unknowns. It should be enough to say that x is a distance, independently of any units one might choose to express its value in.
The difficulty, of course, is that if λ=kx then k has dimension of a surface density, but that feels awkward because there is no surface here. That could have been solved merely by stating that its dimension is QL-2.
 
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1. What is the linear density of a bar?

The linear density of a bar is the mass per unit length of the bar. It is typically denoted by the symbol μ (mu) and has units of kilograms per meter (kg/m).

2. How is the linear density of a bar calculated?

The linear density of a bar is calculated by dividing the total mass of the bar by its length. This can be represented by the equation μ = m/L, where μ is the linear density, m is the mass, and L is the length.

3. What factors affect the linear density of a bar?

The linear density of a bar is affected by the material of the bar, its length, and its cross-sectional area. The density of the material and the length of the bar directly impact the linear density, while the cross-sectional area indirectly affects it through the mass of the bar.

4. How does the linear density of a bar relate to its stiffness?

The linear density of a bar is directly related to its stiffness. A bar with a higher linear density will be stiffer, meaning it will resist deformation more than a bar with a lower linear density. This is because a higher linear density indicates a greater mass per unit length, which requires more force to deform.

5. Can the linear density of a bar change?

Yes, the linear density of a bar can change if the mass or length of the bar changes. For example, if a bar is cut in half, its length will decrease, causing the linear density to increase. Similarly, if additional mass is added to the bar, its linear density will increase.

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