Find the Y component of Electric field at a point

[isukatphysics69]

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

 

[SammyS]

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?.

 

[isukatphysics69]

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

 

[isukatphysics69]

This is really frustrating but I am staying calm like a monk

 

[SammyS]

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.​

.

 

[isukatphysics69]

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