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lgbasallote Group TitleBest ResponseYou've already chosen the best response.0
welcome to the club
 2 years ago

richyw Group TitleBest ResponseYou've already chosen the best response.1
my solutions manual goes from \[\vec{E}=k_eq\left( \frac{1}{(xa)^2}\frac{1}{(x+a)^2}\right)\hat{i}\]to\[\vec{E}=k_eq\frac{4ax}{(x^2a^2)^2}\hat{i}\] and I can't quite remember how to get that!
 2 years ago

richyw Group TitleBest ResponseYou've already chosen the best response.1
I can get to \[\vec{E}=k_eq\left(\frac{4ax}{(xa)^2(x+a)^2}\right)\hat{i}\] so basically I have forgotten how to deal with the denominator :O
 2 years ago

richyw Group TitleBest ResponseYou've already chosen the best response.1
without expanding it all out of course...
 2 years ago

ganeshie8 Group TitleBest ResponseYou've already chosen the best response.1
\(\vec{E}=k_eq\left(\frac{4ax}{(xa)^2(x+a)^2}\right)\hat{i} \) \(\vec{E}=k_eq\left(\frac{4ax}{((xa)(x+a))^2}\right)\hat{i} \)
 2 years ago

richyw Group TitleBest ResponseYou've already chosen the best response.1
i need to sleep haha
 2 years ago

ganeshie8 Group TitleBest ResponseYou've already chosen the best response.1
haha i agree ;p
 2 years ago

richyw Group TitleBest ResponseYou've already chosen the best response.1
thanks a lot
 2 years ago
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