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  • DLS

Find the force on any vertice on a cube because of the rest.All edges have same charge i.e q.

Physics
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  • DLS
|dw:1364748658135:dw|
Use coulombs law to find the Required Answer...
  • DLS
thanks

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|dw:1364749278793:dw|
and ON the surface of cube?
  • DLS
please leave it till here,i want to try it.
knock your self out :)
  • DLS
yeah!
but, your question says, "find the charge"?
  • DLS
that place u assumed as 0,0,0 we have to find there due to rest 7 charges
find what? force?
  • DLS
yeah!
no. force at any point on the cube. so, a point P on the cube.
  • DLS
on the edge
well, off you go then.
answer mila kya?! :P
  • DLS
will try in smtym
and i didn't get the question.. is it to find the electric field at any point on the cube?!
  • DLS
on any edge
  • DLS
all edges have a charge
there is a difference between vertices and edge you know :P.. all charges are on the VERTICES.. so you need to find electric field at any point on an edge?!
  • DLS
yeah
its 3d vector analysis man :-/
  • DLS
i suckat it
its not easy.. you ll have to go brutal math here!
yea then its piece of cake.. just use i j k form.. !
  • DLS
:/
  • DLS
ata hoga usse yr
@shubhamsrg Who else? :')
E at any point on the edge ? this can't be the question, please recheck.
  • DLS
on vertice
|dw:1364751725488:dw| Please circle/mark the position where you want to find the electric field
  • DLS
|dw:1364751777632:dw|
a charge is kept there? and we have to find E on that? -_-
hmm, so you want to find the net force acting on the charge.
  • DLS
force
force nikalni hai? lol
  • DLS
yes :/
  • DLS
tume samaj agaya ques :")
  • DLS
@yrelhan4 ata hai??!!
Hmm shayad ho jayega. but abhi nhi kr rha mai. biomolecules kr rha hun.
Dls.. you seriously need to get your terminologies right :-/.. !!!!
^ best response.
It does not matter at what corner you find the net force. use the 3D distance vectors. The individual forces would be way easier that way.
yea... just use position vectors as marked.. and keep your Forces in i j k format.. and then finally add them up and get the magnitude !!
|dw:1364753502863:dw| and so on
they'd all have a common factor of \[k{q^2\over a^2}\]
just be careful with the signs.. since the forces will be directed along the position vectors but in opposite direction
  • DLS
  • DLS
  • DLS
doubt!
shoot it.
@DLS yes?
  • DLS
|dw:1364821535025:dw| Naming them like that for meanwhile. Now what I have is, \[\large F_{DA}=\frac{kq^2}{a^2}\] \[\large F_{CA}=\frac{kq^2}{\sqrt{2a}^2}cos45^0\] Fca=Fba am i correct with the magnitudes till here?help me determining the directions too! i want to clarify everything tilll this point first.
  • DLS
in the end i meant Fba=Fda* sorry
  • DLS
if you mean \[F_{CA}=\frac{kq^2}{a\sqrt{2}}\] then you are correct. since all charges are alike, the forces are repulsive and act along the distance vectors from other charge to A.\\ in my above diagram I give two such directions. look it up
  • DLS
yes i meant a root 2 only sorry!
  • DLS
I didn't get the direction part,where will the force act?
that cos 45 should not be there!
  • DLS
why?:|
how did you get it to begin with?
  • DLS
IDK when we take components and when not :/ im so confused..
  • DLS
my teacher taught me though,im still messed up with everything.
"components of a force" you are talking about magnitude. do not include that cos... it'd show up automatically with the vectors.
when you take vectors like they are, all the things come with it like a package.
  • DLS
okay so dealing with only mag,im correct till here?
yes. but if you take cos 45, you'd have to take unit vectors.. this increases your work. my advice, do not take that cos45 and simply find the vectors
  • DLS
yes i eliminated that ,rest is correct right? and just that a root 2 thing
yes. rest seems fine.
take hintf from the previous picture
I gotta go now.
  • DLS
please tell me how to do that XZ thing :|
  • DLS
okay,later.
  • DLS
what XZ thingy?
  • DLS
how do i find the component of XZ on that vertice?
still not done? :O
  • DLS
no:/
which one? cricle it
or name it
  • DLS
|dw:1364843986230:dw|