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Find the Y component of Electric field at a point

  • #26
Most likely you will.

You will first have to come up with a linear charge density and will need to set up an integral.
omfg I hate doing trig sub thought I was done with that after calc 2 im really pissed right now
 
  • #28
linear algebra test tomorrow and 2 programming assignments due theres no way ill get this frikkin degree I don't even have a job right now and I cant even keep up with the work
 
  • #29
Im thinking I can use a double integral so that I don't have to do the frikking trig substitution.. plus I took calc 3 in the summer so I should be using double integrals
double integral.JPG




does this look good?
 

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  • #30
SammyS
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Im thinking I can use a double integral so that I don't have to do the frikking trig substitution.. plus I took calc 3 in the summer so I should be using double integrals
[ ATTACH=full]230927[/ATTACH]

does this look good?
I can't imagine how a double integral can even be used here. Don't even be concerned about Trig substitution at this point in solving the problem. That integral doesn't correspond to this problem at all.

Let me remind you of the advice haruspex gave you in post #8
Consider an element of charge with length dy at distance y up the rod. What is the charge on it? What field does it generate at P? What is the x component of that?
To use this idea, you will need to use the linear charge density, usually represented by the variable λ. You can get a value for λ from the given information, but there's no need bother with that yet.

( By the way: It looks like you will not need to use Trig. substitution for finding the y-component. )
 
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  • #31
I just did the integration and got -140677.0588 and that was incorrect.
I had my integral limits from 0 to 0.05 and ∫(kqy(dy)/(a(x^2+y^2)3/2))

used U substitution and was a pretty straight forward integral. not sure what I did wrong here
 
  • #32
SammyS
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I just did the integration and got -140677.0588 and that was incorrect.
I had my integral limits from 0 to 0.05 and ∫(kqy(dy)/(a(x^2+y^2)3/2))

used U substitution and was a pretty straight forward integral. not sure what I did wrong here
It's much more helpful to do the integration using symbols. Plug in the numerical values later. There is virtually no way for anyone to look at the result, −140677.0588 and tell what might be wrong with it. It doesn't even have units.

That is pretty close to being the correct integral.
Having ##\ q\ ## there isn't correct. That should be ##\ \lambda\ ##, the linear charge density.
What is ' a ' ? (It's in the denominator.)
For integration limits, 0 to L .​

What do you get for the integral, prior to substituting the integration limits?.
 
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  • #33
It's much more helpful to do the integration using symbols. Plug in the numerical values later. There is virtually no way for anyone to look at the result, −140677.0588 and tell what might be wrong with it. It doesn't even have units.

That is pretty close to being the correct integral.
Having ##\ q\ ## there isn't correct. That should be ##\ \lambda\ ##, the linear charge density.
What is ' a ' ? (It's in the denominator.)
For integration limits, 0 to L .​

What do you get for the integral, prior to substituting the integration limits?.
I had 'a' as the length of the entire rod, I will try again
 
  • #34
This is really frustrating but im staying calm like a monk
 
  • #35
SammyS
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I had 'a' as the length of the entire rod, I will try again
Well in that case, that does give you a factor of q/L, which is correct and it's needed because, q/L is λ .

Making that replacement, makes your integral:
∫(kqy(dy)/(L(x2+y2)3/2)) with limits from 0 to L .

In more readable form, using LaTeX that can be written:
##\displaystyle \frac{kq}{L} \int_0^L \frac {y\,dy}{(x^2+y^2)^{3/2}} \,,\ ## where I factored out the constant factors.​
.
 
  • #36
ok I got the answer it is -8700 rounded with sig figs. Thank you @haruspex @SammyS. now I will attempt the x component since I got lucky with that one
 

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