Finding Electric Field on rod or ring?

1. Dec 13, 2005

Its from an example in the book, and it doesnt seem to make sense,

A rod of length l has a uniform positive charge per unit length (lambda) and a total charge Q. Calculate the electric field at a point P that is located along the long axis of the rod and a distance a from one end.

So then the example takes a small part of the rod, dx which has charge of dq, and is distance x from point P.

dq = (lambda)dx and dE = ke dq/x^2 or ke lambda dx / x^2

fine so far.

Now the example says we must sum up the contributions of all the segments.

it becomes an integral E = (integral) from a to l+a of ke lambda dx/x^2

the example breaks the dq component out into ke lambda [ - 1/x ] from a to l+a Im somewhat confused now.

then it goes on, ke lambda(1/a - 1/l+a) = keQ/a(l+a) ??? what??!!

okay it kind of makes sense, the total charge divided by length, but the last part there is a divide by zero error to my thought process, multiplying an item with example values: Lambda(1/a - 1/b) or Q/l(1/a - 1/b) might give (Q/l * 1/a) - (Q/l * 1/b) if doing the same thing for the actual values, should give Q/la - Q/la + l^2 ?? no?

reducing it down to Q/a(a+l) ? the book doesnt explain how it arrived at this.

Can anyone give an example of calculating the Field from a charged rod? apparently its the same in a ring from the x axis but using vectors, but this concept seems tough, thanks for any explanations

2. Dec 13, 2005

Staff: Mentor

I assume you understand and agree that that's the total field at the point in question.

What's the anti-derivative of $1/x^2$? That's where the $1/x$ comes from.

$$k\lambda (\frac{1}{a} - \frac{1}{(l + a)}) = k\lambda (\frac{l+a}{a(l+a)} - \frac{a}{a(l+a)}) = k\lambda l \frac{1}{a(l+a)} = \frac{k Q}{a(l+a)}$$

Last edited: Dec 13, 2005