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.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
does this look good?
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.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?
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.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
I had 'a' as the length of the entire rod, I will try againIt'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?.
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 .I had 'a' as the length of the entire rod, I will try again