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

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goformit100
 3 years ago
Best ResponseYou've already chosen the best response.0Use coulombs law to find the Required Answer...

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1364749278793:dw

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0and ON the surface of cube?

DLS
 3 years ago
Best ResponseYou've already chosen the best response.1please leave it till here,i want to try it.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0knock your self out :)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0but, your question says, "find the charge"?

DLS
 3 years ago
Best ResponseYou've already chosen the best response.1that place u assumed as 0,0,0 we have to find there due to rest 7 charges

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0no. force at any point on the cube. so, a point P on the cube.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0well, off you go then.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0and i didn't get the question.. is it to find the electric field at any point on the cube?!

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0there 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?!

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0its 3d vector analysis man :/

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0its not easy.. you ll have to go brutal math here!

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0yea then its piece of cake.. just use i j k form.. !

yrelhan4
 3 years ago
Best ResponseYou've already chosen the best response.0@shubhamsrg Who else? :')

shubhamsrg
 3 years ago
Best ResponseYou've already chosen the best response.0E at any point on the edge ? this can't be the question, please recheck.

shubhamsrg
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1364751725488:dw Please circle/mark the position where you want to find the electric field

shubhamsrg
 3 years ago
Best ResponseYou've already chosen the best response.0a charge is kept there? and we have to find E on that? _

shubhamsrg
 3 years ago
Best ResponseYou've already chosen the best response.0hmm, so you want to find the net force acting on the charge.

yrelhan4
 3 years ago
Best ResponseYou've already chosen the best response.0force nikalni hai? lol

yrelhan4
 3 years ago
Best ResponseYou've already chosen the best response.0Hmm shayad ho jayega. but abhi nhi kr rha mai. biomolecules kr rha hun.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Dls.. you seriously need to get your terminologies right :/.. !!!!

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0It 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
 3 years ago
Best ResponseYou've already chosen the best response.0yea... 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
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1364753502863:dw and so on

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0they'd all have a common factor of \[k{q^2\over a^2}\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0just be careful with the signs.. since the forces will be directed along the position vectors but in opposite direction

DLS
 3 years ago
Best ResponseYou've already chosen the best response.1dw: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
 3 years ago
Best ResponseYou've already chosen the best response.1in the end i meant Fba=Fda* sorry

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0if 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
 3 years ago
Best ResponseYou've already chosen the best response.1yes i meant a root 2 only sorry!

DLS
 3 years ago
Best ResponseYou've already chosen the best response.1I didn't get the direction part,where will the force act?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0that cos 45 should not be there!

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0how did you get it to begin with?

DLS
 3 years ago
Best ResponseYou've already chosen the best response.1IDK when we take components and when not :/ im so confused..

DLS
 3 years ago
Best ResponseYou've already chosen the best response.1my teacher taught me though,im still messed up with everything.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0"components of a force" you are talking about magnitude. do not include that cos... it'd show up automatically with the vectors.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0when you take vectors like they are, all the things come with it like a package.

DLS
 3 years ago
Best ResponseYou've already chosen the best response.1okay so dealing with only mag,im correct till here?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0yes. 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
 3 years ago
Best ResponseYou've already chosen the best response.1yes i eliminated that ,rest is correct right? and just that a root 2 thing

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0yes. rest seems fine.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0take hintf from the previous picture

DLS
 3 years ago
Best ResponseYou've already chosen the best response.1please tell me how to do that XZ thing :

DLS
 3 years ago
Best ResponseYou've already chosen the best response.1how do i find the component of XZ on that vertice?