Here's the question you clicked on:
DLS
Find the force on any vertice on a cube because of the rest.All edges have same charge i.e q.
Use coulombs law to find the Required Answer...
|dw:1364749278793:dw|
and ON the surface of cube?
please leave it till here,i want to try it.
knock your self out :)
but, your question says, "find the charge"?
that place u assumed as 0,0,0 we have to find there due to rest 7 charges
no. force at any point on the cube. so, a point P on the cube.
well, off you go then.
and i didn't get the question.. is it to find the electric field at any point on the cube?!
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?!
its 3d vector analysis man :-/
its not easy.. you ll have to go brutal math here!
yea then its piece of cake.. just use i j k form.. !
@shubhamsrg Who else? :')
E at any point on the edge ? this can't be the question, please recheck.
|dw:1364751725488:dw| Please circle/mark the position where you want to find the electric field
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.
force nikalni hai? lol
Hmm shayad ho jayega. but abhi nhi kr rha mai. biomolecules kr rha hun.
Dls.. you seriously need to get your terminologies right :-/.. !!!!
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
|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.
in the end i meant Fba=Fda* sorry
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
yes i meant a root 2 only sorry!
I didn't get the direction part,where will the force act?
that cos 45 should not be there!
how did you get it to begin with?
IDK when we take components and when not :/ im so confused..
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.
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
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
please tell me how to do that XZ thing :|
how do i find the component of XZ on that vertic