Find the Y component of Electric field at a point

  • #1

Homework Statement


The figure below shows a thin, vertical rod of length L with total charge Q. The indicated point P is a horizontal distance x from the one end of the rod. What is the electric field at point P. Express your answer in component notation in the two blanks below.

L = 5.0 cm, Q = 3.0 nC, and x = 3.0 cm.
physs2.jpg

Homework Equations



kq/r^2

The Attempt at a Solution


I found the x component to be 15000N/C

theta = arctan(5/3) = 59 deg
total charge = [(8.99x10^9)(3x10^-9)]/((0.03)^2) = 29966
x component = 29966cos59=15417 with sig figs is 15000

so now the y component should just be 29966sin59 = 25685 rounded to 26000 N/C but that doesn't make sense because the y component should be facing down
 

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Answers and Replies

  • #2
even when I did theta = -59 degrees the answer was incorect
 
  • #3
haruspex
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total charge = [(8.99x10^9)(3x10^-9)]/((0.03)^2) = 29966
Not all the charge is at distance 3cm from P.
 
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  • #6
Ok I think I see what youre saying. I got the distance from the top of the rod as
[(8.99x10^9)(3x10^-9)]/((0.058^2))=7932
7932sin59 = 6800

incorrect

@haruspex
 
  • #7
I tried generalizing from the middle

[(8.99x10^9)(3x10^-9)]/((0.039^2))=17685
(new angle arctan(2.5/3) = 39 deg
17685sin(39) = 11321

incorrect
 
  • #8
haruspex
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Ok I think I see what youre saying. I got the distance from the top of the rod as
[(8.99x10^9)(3x10^-9)]/((0.058^2))=7932
7932sin59 = 6800

incorrect

@haruspex
Consider an element 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?
 
  • #9
Consider an element 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?
so the charge on dy is kq/r^2 at point P the x component is (kq/r^2)cos(theta)
 
  • #10
my x component is correct tho, I thought the y should be simple just sin of the angle @haruspex
 
  • #11
haruspex
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so the charge on dy is kq/r^2
That s not a charge.
What is the charge on a section of length dy?
If it is distance y up the rod, how far is it from point P?
 
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  • #12
That s not a charge.
What is the charge on a section of length dy?
Of it is distance y up the rod, how far is it from point P?
would it be dy/3? im really not understanding. dy is sqrt(y^2+9) away
 
  • #13
haruspex
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would it be dy/3?
No.
What is the charge density along the rod? If the whole rod length L carries charge Q, how much charge is there on section length dy?
dy is sqrt(y^2+9) away
Units! But better still, don't plug in numbers yet, just call it x2, not 9.
 
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  • #14
No.
What is the charge density along the rod? If the whole rod length L carries charge Q, how much charge is there on section length dy?
Is it Q*dy????
 
  • #16
GOT IT!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! >=[
 
  • #17
haruspex
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Is it Q*dy????
The charge Q is spread out along a rod length L, what is the charge density (charge per unit length)?
 
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  • #18
The charge Q is spread out along a rod length L, what is the charge density (charge per unit length)?
Then I think it is dQ*dL

I got the answer partial correct 1.8/2 points with the answer -7900N/C
 
  • #19
Im confused why im getting partial credit

[(8.99x10^9)(3x10^-9)]/((sqrt(.05^2+.03^2)^2) = 7932 taking the negative because +/- sqrt = -7932

1.8/2 points
 
  • #20
SammyS
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Is it Q*dy????
You seem to be guessing, rather than answering the specific question haruspex asked you.

To ask basically the same thing:
If the length of the rod is 5.0 cm, and the charge on the rod is 3.0 nC, how much charge is on each 1 cm of the rod?
How would you write that using symbols: L for length and Q for the charge?
 
  • #21
SammyS
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Im confused why im getting partial credit

[(8.99x10^9)(3x10^-9)]/((sqrt(.05^2+.03^2)^2) = 7932 taking the negative because +/- sqrt = -7932

1.8/2 points
Slow down. You're nowhere near ready to solve this. The solution will require that you use Calculus to do integration.
 
  • #22
Slow down. You're nowhere near ready to solve this. The solution will require that you use Calculus to do integration.
Oh god.. I have to go to physics class right now. im getting frustrated awith this anyway I have to step away
 
  • #23
SammyS
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So did I just get lucky with the first answer??
You mean to ask if you got lucky with getting the correct numerical answer for the x component of the leectric field (which is in your "other" thread.)

The answer is absolutely yes, you got lucky.

You used cosine, when what was required there was to use sine.

Both components require the use of integral calculus.
 
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  • #25
SammyS
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hey @SammyS do I have to use trig substitution?
Most likely you will.

You will first have to come up with a linear charge density and will need to set up an integral.
 
  • #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.​
.
 

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