Find the Y component of Electric field at a point

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SUMMARY

The discussion focuses on calculating the electric field's Y component at point P due to a charged vertical rod with length L = 5.0 cm and total charge Q = 3.0 nC. The initial attempts yielded incorrect results due to misunderstanding the charge distribution and the need for integration. The correct approach involves using linear charge density and integral calculus to derive the electric field components accurately. The final correct answer for the Y component was determined to be approximately -8700 N/C after proper integration.

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isukatphysics69
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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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even when I did theta = -59 degrees the answer was incorect
 
isukatphysics69 said:
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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haruspex said:
Not all the charge is at distance 3cm from P.
So did I just get lucky with the first answer??
 
im confused
 
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
 
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
 
isukatphysics69 said:
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?
 
haruspex said:
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
isukatphysics69 said:
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
haruspex said:
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? I am really not understanding. dy is sqrt(y^2+9) away
 
  • #13
isukatphysics69 said:
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?
isukatphysics69 said:
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
haruspex said:
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?
 
  • #15
Help?
 
  • #16
GOT IT! >=[
 
  • #17
isukatphysics69 said:
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
haruspex said:
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 I am 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
isukatphysics69 said:
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
isukatphysics69 said:
Im confused why I am 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
SammyS said:
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. I am getting frustrated awith this anyway I have to step away
 
  • #23
isukatphysics69 said:
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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  • #24
hey @SammyS do I have to use trig substitution?
 
  • #25
isukatphysics69 said:
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
SammyS said:
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 I am really pissed right now
 
  • #27
im just going to sleep
 
  • #28
linear algebra test tomorrow and 2 programming assignments due there's no way ill get this frikkin degree I don't even have a job right now and I can't 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
isukatphysics69 said:
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
haruspex said:
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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