Thin rod symbolic questions based on Image.

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The discussion revolves around calculating the electric field due to a uniformly charged thin rod with charge -Q. For part (a), the charge per unit length is determined to be λ = -Q/(2A), correcting earlier misconceptions about the rod's length. In part (b), the amount of charge dQ on a small segment of length dx is expressed as dQ = λ * dx, confirming the relationship between charge density and segment length. Participants clarify the total length of the rod and its implications for calculations. The user successfully resolves the problem with assistance from others in the thread.
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Homework Statement


A thin rod lies on the x-axis with one end at -A and the other end at A, as shown in the diagram. A charge of -Q is spread uniformly over the surface of the rod. We want to set up an integral to find the electric field at location ‹ 0, y, 0 › due to the rod.

16-050-Erod_setup.jpg


Use the following as necessary: x, y, dx, A, Q. Remember that the rod has charge -Q.

(a) In terms of the symbolic quantities given above and on the diagram, what is the charge per unit length of the rod?
λ = ?

(b) What is the amount of charge dQ on the small piece of length dx?
dQ = ?

Homework Equations



Electric field of a uniformly charged Rod: 1/4πε0 * 2(Q/L) / r

The Attempt at a Solution


PartA: -ΔQ: I tried this because since I'm dividing the rods by a based amount, the charge Q, should be written by -ΔQ.
-ΔQ/L: I input this wrong, because it's suppose to be A, since that's the length of the rod. I was thinking of -ΔQ/A, because since I'm dividing the charge by a certain amount, I'm also dividing the length of the rod as well. Haven't tried it yet.

PartB: -dx/A: Well I think similarly like part a, we're dividing the rod again, should be divided by A. Was thinking of Q (dx/A) if I'm trying to find the dQ

Just need help on these parts. From there I'll attempt the rest of the parts on my own.
 
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Christian121 said:
I input this wrong, because it's suppose to be A, since that's the length of the rod.
The rod goes from x = -A to x = A. How long is it?
Well I think similarly like part a, we're dividing the rod again, should be divided by A. Was thinking of Q (dx/A) if I'm trying to find the dQ
You've got the right idea, just the wrong length.
 
tms said:
The rod goes from x = -A to x = A. How long is it?

I would think Part A: is λ = -ΔQ/-Δx, if it's being uniformly distributed.

You've got the right idea, just the wrong length.

Ok thanks , er dq = λ*dx? :P. I'll figure it out eventually.
 
In the first part, the charge density is the total charge divided by the total length. The total length is not A.

Yes, dq = \lambda \:dx.
 
tms said:
In the first part, the charge density is the total charge divided by the total length. The total length is not A.

Yes, dq = \lambda \:dx.

Ok got part A I believe now. It's -Q/2*A.

Thanks for your help. I got the rest of the parts on this problem right as well.
 
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