# Strength of Electric Field

1. Feb 3, 2009

### glennpagano44

1. The problem statement, all variables and given/known data

Halliday and Resnick edition 7 chapter 22 number 29

In Fig 22-45, positive charge q = 7.81 pC is spread uniformly along a thin nonconducting rod of length L = .145m. The y distance between the point and rod is R = .06m. What is the magnitude and direction of the electric field produced at point P.

I will attempt to describe the figure:

There is a thin nonconducting rod in the x dimension with length L and a point P is directly above the center of the rod in the y dimension (R = .06m).

2. Relevant equations

E = (1/4$$\pi$$$$\epsilon$$)(q/r$$^{2}$$)

3. The attempt at a solution

$$\int$$dE = 1/4$$\pi$$$$\epsilon$$$$\int$$$$\lambda$$Rdx/(R$$^{2}$$+x$$^{2}$$)$$^{3/2}$$
I then took lamba and R out of the integral since they are constants
(4 is not rasied to ($$\pi$$$$\epsilon$$)

After taking the integral I got:

($$\lambda$$/4$$\pi\epsilon$$)(x/R(R$$^{2}$$+x$$^{2}$$)$$^{1/2}$$

When I plug in for the varibles I do not get the correct answer, I get 7.48 n/C but the correct answer is 12.4 N/C

I plug in L for x and R for R and for $$\lambda$$ I use q/L
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Feb 4, 2009

### tiny-tim

Hi glennpagano44!

(have a lambda: λ and an epsilon: ε and a pi: π and a square-root: √ )
No … how can you still have an x when you've just eliminated x by integrating over it?

And where did that extra R come from?

3. Feb 4, 2009

### glennpagano44

I went back to class today and I figured it out thanks alot. I just integrated it wrong, I had to do some trig subsitutions. (that is where the extra R came from)