Coulomb's Law: Calculating Electric Field Due to Multiple Charges

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SUMMARY

The forum discussion focuses on calculating the electric field and force due to multiple positive charges using Coulomb's Law. Participants discuss the vector representations of electric fields, specifically how to express them in terms of unit vectors and magnitudes. Key equations mentioned include \(\vec{E_{21}} = k\frac{q_2}{d^2}\) and \(\vec{F_{21}} = k\frac{q^2}{d^2}<0,1>\). The conversation emphasizes the importance of correctly identifying angles and the relationship between unit vectors and magnitudes in the context of electrostatics.

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  • Coulomb's Law for electric fields and forces
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  • #31
we seem to be completely misunderstanding each other :redface:

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? :confused:
 
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  • #32
tiny-tim said:
we seem to be completely misunderstanding each other :redface:

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? :confused:

But 1/√3<-√3/2, 3/2> is

I had a √3 beside a d^2

Do you see?
 
  • #33
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? :confused:
 
  • #34
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? :confused:

But it isn't <-√3/2, 3/2>, it is 1/√3<-√3/2,3/2>
 
  • #35
Are we going in circles?
 
  • #36
flyingpig said:
Are we going in circles?

yes … isn't that the flyingpig's natural method of travel? :wink:

i can't force you to write a unit vector!
 
  • #37
tiny-tim said:
yes … isn't that the flyingpig's natural method of travel? :wink:

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?
 
  • #38
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) :wink:​
 
  • #39
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) :wink:​

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
 
  • #40
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
 
  • #41
flyingpig said:
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

i only left the √3 separate because i didn't want to bother to calculate 1.6 √3 ! :rolleyes:
flyingpig said:
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

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 :redface:
 
  • #42
tiny-tim said:
i only left the √3 separate because i didn't want to bother to calculate 1.6 √3 ! :rolleyes:


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 :redface:

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?
 
  • #43
hi flyingpig! :smile:
flyingpig said:
Why is that (-1/2i + √3/2j) simpler?

because it's a unit vector! :redface:
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

no it isn't!
 
  • #44
tiny-tim said:
hi flyingpig! :smile:


because it's a unit vector! :redface:


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?
 
  • #45
My book says "Leave your answers in Fxi + Fyj
 
  • #46
flyingpig said:
My book says "Leave your answers in Fxi + Fyj

No, it doesn't! …

in post #1, you provided a photo of the book, it clearly says "What is the magnitude and direction … Leave your answer in algebraic form" …

you need to state the magnitude and direction !
 
  • #47
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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