E-field from one end of an infinite non-conducting rod

[arturo]

Homework Statement

Homework Equations

dE= k dq/r²

The Attempt at a Solution

[/B]
I started off taking a derivative of q(x).

dq = -q_o/l ⋅ e^-x/ldx

Then, I decided that r was the distance x along the rod + .02m. r=(.02+x)

Following that, I plugged everything into the formula:

dE = k⋅q_o/l ∫₀^∞ e^-x/l / (.02 + x)dx

Integrating that I resulted in:

E = -kq_o/l ⋅(-e⁰/(.02)²)

E = kq_o/l(.02)²

Plugging in values leaves me with 2.3*10⁹ N/C, which is incorrect. I'm not quite sure where I am going wrong here, any advice or clues to lead me in the right direction would be appreciated!
Thanks!

 

[SammyS]

Homework Statement

View attachment 238278

Homework Equations

dE= k dq/r²

The Attempt at a Solution

[/B]
I started off taking a derivative of q(x).

dq = -q_o/l ⋅ e^-x/ldx

Then, I decided that r was the distance x along the rod + .02m. r=(.02+x)

Following that, I plugged everything into the formula:

dE = k⋅q_o/l ∫₀^∞ e^-x/l / (.02 + x)dx

Integrating that I resulted in:

E = -kq_o/l ⋅(-e⁰/(.02)²)

E = kq_o/l(.02)²

Plugging in values leaves me with 2.3*10⁹ N/C, which is incorrect. I'm not quite sure where I am going wrong here, any advice or clues to lead me in the right direction would be appreciated!
Thanks!

What value did you use for $\ q_0 \$ and how did you determine that value ?

 

[BvU]

$dq = -q_0/l ⋅ e^{-x/l}\;dx$

meaning it is negative ?

 

[BvU]

dE = k⋅qo/l ∫0∞ e-x/l / (.02 + x)dx

what happened to the square in the denominator in your relevant equation ?

 

[arturo]

What value did you use for $\ q_0 \$ and how did you determine that value ?

I was using the value from the problem statement, 2.9 microC. I did try something else where I integrated q(x) and set it equal to 2.9μC. That resulted in an incorrect value too, but perhaps I did it wrong, I’ll type up what I did when I am no longer on mobile. Is this the correct approach?

what happened to the square in the denominator in your relevant equation ?

that’s just a typo. If you look, it comes back in the next line.

 

[arturo]

Okay, I tried solving for q_o via:
dq = -lq_oe^-x/l
2.9μC = ∫₀^∞-lq_oe^-x/l
2.9*10^-6 = lq_o
q_o = 2.9*10^-6/(.0286)

 

[BvU]

I started off taking a derivative of q(x).
dq = -qo/l ⋅ e-x/ldx

$q(x)$ is a charge distribution, so your $dq = q(x) dx$, not the derivative.
You have a a distribution function and a total charge given, so you can calculate the $q_0$.

[edit]I see you already did the latter.

 

[arturo]

Okay, I think I see why that's the case, but I might have to think about it more. Trying that out:

dq = q(x)dx
dq = q_oe^-x/l
q_o = 2.9*10^-6/l (from above comment)
E = k * 2.9*10^-6/l ∫₀^∞ e^-x/l / (.02 + x)²
E = k * 2.9*10^-6 / (l * .02²)
E = 2.3*10⁹ N/C

Does this look correct?

 

[BvU]

Plugging in values leaves me with 2.3*10⁹ N/C, which is incorrect

So from post #1 I'd say no

I agree with $$E = {kQ\over l}\int_0^\infty {e^{-x\over l}\over (\Delta+x)^2}\;dx$$
How do you work out the integral ? Dimension-wise length^-3 doesn't look good...

 

[arturo]

You are right, I didn’t even notice, haha!

For the integral, at infinity, e goes to zero and the denominator goes to infinity so I say that is 0. At 0, e^-x/l becomes one and the denominator becomes .02².

If k is N*m² / C² we would get N/(C*m) not N/C which is what we want... so I suppose I’m getting an extra m term in there somewhere...

 

[BvU]

Repeat: how do you work out the integral? Show steps, please...

 

[arturo]

Okay, so I realized that I was just evaluating without actually integrating.
∫₀^∞e^-x/l/(x²+.04x + .02²)
I’m not sure I actually know how to evaluate this integral by hand.

 

[vela]

You may need to evaluate it numerically.

 

[BvU]

partial integration is the keyword

 

[arturo]

Update: even after partial integration, you end up with an exponential integral— something I’m not familiar with yet. Professor assumed we would use an integral calculator apparently. Thank you for your help, I feel like I understand the setup of these problems more than before.

 

[SammyS]

Update: even after partial integration, you end up with an exponential integral— something I’m not familiar with yet. Professor assumed we would use an integral calculator apparently. Thank you for your help, I feel like I understand the setup of these problems more than before.

As @vela said, evaluate it numerically, which is likely what an integral calculator does.

Regarding the attempt at evaluating the integral, $\displaystyle \int_0^\infty {e^{-x\over L }\over (\Delta+x)^2}\, dx \,, ~$ by hand, the method to use is called "Integration by Parts", not partial integration.

After two applications of integration by parts, you will have the following integral to evaluate, along with two other terms and a coefficient.

$\displaystyle \int_0^\infty {\ln (\Delta+x)} ~ e^{-x/L} \, dx \,, ~$​

This also must be evaluated numerically, so there's not much reason to us integration by parts for this problem.

 

[BvU]

the method to use is called "Integration by Parts", not partial integration.

Yeah, I used the original dutch wording that unfortunately was damaged in translation