Thin rod symbolic questions based on Image.

[Christian121]

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.

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.

 

[tms]

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.

 

[Christian121]

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.

 

[tms]

In the first part, the charge density is the total charge divided by the total length. The total length is not A.

Yes, $dq = \lambda \:dx$.

 

[Christian121]

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.