Electricity question that I thought I had but I don't

  • #1
Ok, the question is: you have a uniformly charged rod of length 16.6cm with charge -22.3 microCouloumbs, determine the magnitude of the electric field along the axis of the rod at a point 18.9358 cm from the center of the rod, answer in units of N/C(k is given as 8.98755 x 10^9)

Soooo then, I converted everything(hopefully correctly)16.6cm =.166m, -22.3 microC=-2.23x10^-5, 18.9358cm=.189358m. Yay. Using the formula dE=k(dQ/r^2), I found dQ to be Q/L(dl), or charge/length(dl), which I found to be -1.34337x10^-4(which I'll say is &) times dl, so taking the integral and taking out all the constants, I got k&/r^2 S(dl), where S is the integral sign, and I took it over the length of the rod which was .166, so ultimately I get k&.166/r^2 and I'll spare the long number typing. Now, to get r I subtracted .083 from .189358, I ALSO just did the given distance when I got the wrong answer(online hw, yay)and both haven't worked. The answers that have failed have been 1.77176x10^7, and the negative of that, and 5.58956x10^6(which I got using the dist from the center as given)I haven't tried the negative yet because I lose points everytime I get it wrong, and was wondering if that was possibly right? I was pretty sure I was doing it right the first way but I've quadruple checked everything and it's still wrong
 

Answers and Replies

  • #2
learningphysics
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Hmmm... I'm not understanding what you did. Are you given the radius of the rod? Also I want to check.... Total charge on the rod is -22.3 microCouloumbs? It's not microcoulombs/m^3 by any chance?
 
  • #3
r was just the variable I was using to represent either the length from the rod itself to the point, or the length from the center of the rod to the point, both of which I tried as I mentioned. And no, it was given as (little u sign for micro)C

Oh, and this is early in our study on electricity, so assume the rod has a negligible thickness and all that
 
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  • #4
Ohhh, ok, hang on. I was assuming it was a rod lying along the x-axis, and the point was a point also on the x-axis, I worded that problem pretty much as it was worded, so should I assume the point is on a line perpendicular to the rod(which I think is what you're saying)? If so that would explain everything
 
  • #5
learningphysics
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schattenjaeger I'm really sorry. Please ignore my posts so far... I misunderstood the problem. I'll re-read it and try to help!
 
  • #6
Oh, no problem. I really should be sleeping but then this'll bug me allll day
 
  • #7
learningphysics
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Ok... I see now I believe. Your integral is done wrong...

I'll call & the charge density like you did.

So what you need to integrate is

k(&ds)/[(0.272358-s)^2]

as s goes from 0 to 0.166

&ds is dq. r= 0.272358 -s. where s is the distance from the beginning of the rod. Since 0.189 is the distance from the center of the rod 0.189+0.083 is the distance from the beginning of the rod = 0.272

Does this make sense? Also let me know if I've screwed up somewhere.

EDIT: Had to change 0.189 to 0.272
 
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  • #8
I think the problem with that is that k, &, and all that are constants so you can't integrate them, per se. I'm not really sure, I'll look into it tomorrow and let us know, thanks for the help!
 
  • #9
learningphysics
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schattenjaeger said:
I think the problem with that is that k, &, and all that are constants so you can't integrate them, per se. I'm not really sure, I'll look into it tomorrow and let us know, thanks for the help!

Yea, they're constants so just take k& out of the integral and you're left with
the integral of:

ds/[(0.272-s)^2]
 
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  • #10
learningphysics
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The thing to remember is that each dq has a different r... so the r has to go under the integral.
 
  • #11
You were absolutely right! Finalllllly

I just learned a valuable lesson/reminder about doing integrals in actual applications. I knew something was logically wrong with what I was doing, but I couldn't quite put my finger on it 'till you brought it up, thanks! FYI the answer is 6918880.704 if anyone wants to practice
 
  • #12
learningphysics
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schattenjaeger said:
You were absolutely right! Finalllllly

I just learned a valuable lesson/reminder about doing integrals in actual applications. I knew something was logically wrong with what I was doing, but I couldn't quite put my finger on it 'till you brought it up, thanks! FYI the answer is 6918880.704 if anyone wants to practice

Cool! :smile: Glad I was able to help! BTW are you a Gabriel Knight fan? (I'm guessing from the name schattenjaeger)...
 
  • #13
Yup. Got GK3 for my sister's b-day years ago, she's not exactly computer savvy so I pretty much ended up playing it for her and I liked it. After that, through a mighty pulling of strings and computer manipulation, I managed to play the original on my computer which was pretty awesome
 

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