Electric field at a point from three equal positive charges
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flyingpig
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tiny-tim said:we seem to be completely misunderstanding each other
i'm saying that < -√3/2, 3/2 > is not a unit vector, so it shouldn't be part of the answer …
why do you keep writing it? what is it for?
But 1/√3<-√3/2, 3/2> is
I had a √3 beside a d^2
Do you see?
flyingpig
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tiny-tim said:yeees …
but, as i said, why do you keep writing < -√3/2, 3/2 > as part of the final answer?
it's not a unit vector, so what is it there for?
But it isn't <-√3/2, 3/2>, it is 1/√3<-√3/2,3/2>
flyingpig
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Are we going in circles?
flyingpig
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tiny-tim said:yes … isn't that the flyingpig's natural method of travel?
i can't force you to write a unit vector!
This is bugging me, if I switch it to i and j notations
Namely, my final solution giving me 1.6N(-√3/2i + 3/2j) isnot a unit vector, but if 1.6N/√3(-√3/2i + 3/2j) then is a unit vector. So how do I deal with the magnitude? Should I take √3(1.6N) or a magnitude of one from my unit vector?
flyingpig
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tiny-tim said:the way to convert 1.6(-√3/2i + 3/2j) into a magnitude and a unit vector is:
1.6 √3 (-1/2i + √3/2j)
You multiply and divided the vector by its magnitude, is that what you did? If so, then what was I doing? I can see you destributed the 1/√3 into the unit vector, but you left the top √3 beside 1.6N
flyingpig
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Let me add another question to clarify the question I just asked.
Isn't 1/√3(-√3/2i + 3/2j) a unit vector? At least in my Linear Algebra Book
Isn't 1/√3(-√3/2i + 3/2j) a unit vector? At least in my Linear Algebra Book
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flyingpig said:
i only left the √3 separate because i didn't want to bother to calculate 1.6 √3 !
flyingpig said:
yes of course …
but it's a very strange way of writing it, when (-1/2i + √3/2j) is so much simpler, and you'll certainly lose a mark for it in the exam
flyingpig
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tiny-tim said:i only left the √3 separate because i didn't want to bother to calculate 1.6 √3 !
yes of course …
but it's a very strange way of writing it, when (-1/2i + √3/2j) is so much simpler, and you'll certainly lose a mark for it in the exam
Why is that (-1/2i + √3/2j) simpler? If we are just talking about numbers here where 1.6 is a scalar (talking only about the numbers for now) then, 1.6/√3 <-√3/2, 3/2> is a unit vector, but that is not the same as 1.6√3<-1/2, √3/2>.
(sorry for switching notations all of a sudden, I have no idea what i was thinking)
In 1.6/√3 <-√3/2, 3/2>, I needed (wel, you needed) to multiply √3 again to get 1.6√3<-1/2, √3/2>
But noting that 1.6√3 is a scalar that will EXTEND the unit vector<-1/2, √3/2>, so how is that different?
flyingpig
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tiny-tim said:hi flyingpig!
because it's a unit vector!
no it isn't!
I think I got confused with the outside scalar being inside the vector itself. In that case, why don't I just distribute the 1.6N?
flyingpig
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My book says "Leave your answers in Fxi + Fyj
flyingpig
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Never mind, if i need to use Fxi and Fyj I will just convert it to a unit vector, that's all I really need to do.
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