- #1
ToucanFodder
- 2
- 0
- 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.
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.