Finding Electric Field on rod or ring?

[Brad_1234]

Its from an example in the book, and it doesn't 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 I am 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 doesn't 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

 

[Doc Al]

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

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

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

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

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

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

 

[quicknote]

I can understand your confusion with the example provided in the book. Let me try to clarify the process for finding the electric field at a point on a charged rod.

First, it is important to understand that the electric field is a vector quantity, meaning it has both magnitude and direction. In this case, we are looking for the electric field at a point P located along the long axis of the rod. This means that the direction of the electric field will be along the long axis of the rod, pointing towards or away from the rod depending on the charge distribution.

Now, let's break down the steps for finding the electric field at point P:

1. Divide the rod into small segments: In order to calculate the electric field at a point, we need to consider the contribution of each small segment of the rod. This is why the example in the book takes a small part of the rod, dx, with charge dq.

2. Calculate the electric field contribution from each segment: As you mentioned, the electric field contribution from a small segment can be calculated using the equation dE = ke dq/x^2. This equation takes into account the distance between the segment and the point P.

3. Sum up the contributions from all segments: Now, we need to consider the contribution from all the segments of the rod. This is where the integral comes in. The integral essentially adds up all the contributions from each segment, giving us the total electric field at point P.

4. Simplify the integral: Here is where the example in the book may have caused confusion. In order to simplify the integral, we can use the fact that the charge distribution on the rod is uniform. This means that lambda, the charge per unit length, is constant throughout the rod. Thus, we can take it out of the integral and write it as ke lambda times the integral of dx/x^2.

5. Evaluate the integral: The integral of dx/x^2 can be evaluated as -1/x, giving us the expression ke lambda times (-1/x) from a to l+a.

6. Simplify further: This is where the book may have used some algebraic manipulations to simplify the expression. We can rewrite -1/x as 1/a - 1/(a+l). This is because the distance x is changing from a to l+a, and we can break it down into two parts - from a to x and from x to l+a. So,