DLS
  • DLS
Find the force on any vertice on a cube because of the rest.All edges have same charge i.e q.
Physics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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DLS
  • DLS
|dw:1364748658135:dw|
goformit100
  • goformit100
Use coulombs law to find the Required Answer...
DLS
  • DLS
thanks

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anonymous
  • anonymous
|dw:1364749278793:dw|
anonymous
  • anonymous
and ON the surface of cube?
DLS
  • DLS
please leave it till here,i want to try it.
anonymous
  • anonymous
knock your self out :)
DLS
  • DLS
yeah!
anonymous
  • anonymous
but, your question says, "find the charge"?
DLS
  • DLS
that place u assumed as 0,0,0 we have to find there due to rest 7 charges
anonymous
  • anonymous
find what? force?
DLS
  • DLS
yeah!
anonymous
  • anonymous
no. force at any point on the cube. so, a point P on the cube.
DLS
  • DLS
on the edge
anonymous
  • anonymous
well, off you go then.
anonymous
  • anonymous
answer mila kya?! :P
DLS
  • DLS
will try in smtym
anonymous
  • anonymous
and i didn't get the question.. is it to find the electric field at any point on the cube?!
DLS
  • DLS
on any edge
DLS
  • DLS
all edges have a charge
anonymous
  • anonymous
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
  • DLS
yeah
anonymous
  • anonymous
its 3d vector analysis man :-/
DLS
  • DLS
i suckat it
anonymous
  • anonymous
its not easy.. you ll have to go brutal math here!
anonymous
  • anonymous
yea then its piece of cake.. just use i j k form.. !
DLS
  • DLS
:/
DLS
  • DLS
ata hoga usse yr
yrelhan4
  • yrelhan4
@shubhamsrg Who else? :')
shubhamsrg
  • shubhamsrg
E at any point on the edge ? this can't be the question, please recheck.
DLS
  • DLS
on vertice
shubhamsrg
  • shubhamsrg
|dw:1364751725488:dw| Please circle/mark the position where you want to find the electric field
DLS
  • DLS
|dw:1364751777632:dw|
shubhamsrg
  • shubhamsrg
a charge is kept there? and we have to find E on that? -_-
shubhamsrg
  • shubhamsrg
hmm, so you want to find the net force acting on the charge.
DLS
  • DLS
force
yrelhan4
  • yrelhan4
force nikalni hai? lol
DLS
  • DLS
yes :/
DLS
  • DLS
tume samaj agaya ques :")
DLS
  • DLS
@yrelhan4 ata hai??!!
yrelhan4
  • yrelhan4
Hmm shayad ho jayega. but abhi nhi kr rha mai. biomolecules kr rha hun.
anonymous
  • anonymous
Dls.. you seriously need to get your terminologies right :-/.. !!!!
yrelhan4
  • yrelhan4
^ best response.
anonymous
  • anonymous
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.
anonymous
  • anonymous
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 !!
anonymous
  • anonymous
|dw:1364753502863:dw| and so on
anonymous
  • anonymous
they'd all have a common factor of \[k{q^2\over a^2}\]
anonymous
  • anonymous
just be careful with the signs.. since the forces will be directed along the position vectors but in opposite direction
DLS
  • DLS
@shubhamsrg
DLS
  • DLS
@electrokid
DLS
  • DLS
doubt!
anonymous
  • anonymous
shoot it.
anonymous
  • anonymous
@DLS yes?
DLS
  • 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
  • DLS
in the end i meant Fba=Fda* sorry
DLS
  • DLS
@electrokid !
anonymous
  • anonymous
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
  • DLS
yes i meant a root 2 only sorry!
DLS
  • DLS
I didn't get the direction part,where will the force act?
anonymous
  • anonymous
that cos 45 should not be there!
DLS
  • DLS
why?:|
anonymous
  • anonymous
how did you get it to begin with?
DLS
  • DLS
IDK when we take components and when not :/ im so confused..
DLS
  • DLS
my teacher taught me though,im still messed up with everything.
anonymous
  • anonymous
"components of a force" you are talking about magnitude. do not include that cos... it'd show up automatically with the vectors.
anonymous
  • anonymous
when you take vectors like they are, all the things come with it like a package.
DLS
  • DLS
okay so dealing with only mag,im correct till here?
anonymous
  • anonymous
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
  • DLS
yes i eliminated that ,rest is correct right? and just that a root 2 thing
anonymous
  • anonymous
yes. rest seems fine.
anonymous
  • anonymous
take hintf from the previous picture
anonymous
  • anonymous
I gotta go now.
DLS
  • DLS
please tell me how to do that XZ thing :|
DLS
  • DLS
okay,later.
DLS
  • DLS
@electrokid @Mashy
anonymous
  • anonymous
what XZ thingy?
DLS
  • DLS
how do i find the component of XZ on that vertice?
anonymous
  • anonymous
still not done? :O
DLS
  • DLS
no:/
anonymous
  • anonymous
which one? cricle it
anonymous
  • anonymous
or name it
DLS
  • DLS
|dw:1364843986230:dw|