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

- DLS

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

- schrodinger

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- DLS

|dw:1364748658135:dw|

- goformit100

Use coulombs law to find the Required Answer...

- DLS

thanks

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## More answers

- anonymous

|dw:1364749278793:dw|

- anonymous

and ON the surface of cube?

- DLS

please leave it till here,i want to try it.

- anonymous

knock your self out :)

- DLS

yeah!

- anonymous

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

- anonymous

find what? force?

- DLS

yeah!

- anonymous

no. force at any point on the cube. so, a point P on the cube.

- DLS

on the edge

- anonymous

well, off you go then.

- anonymous

answer mila kya?! :P

- DLS

will try in smtym

- anonymous

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

- 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

yeah

- anonymous

its 3d vector analysis man :-/

- DLS

i suckat it

- anonymous

its not easy.. you ll have to go brutal math here!

- anonymous

yea then its piece of cake.. just use i j k form.. !

- DLS

:/

- DLS

ata hoga usse yr

- yrelhan4

@shubhamsrg Who else? :')

- shubhamsrg

E at any point on the edge ? this can't be the question, please recheck.

- DLS

on vertice

- shubhamsrg

|dw:1364751725488:dw|
Please circle/mark the position where you want to find the electric field

- DLS

|dw:1364751777632:dw|

- shubhamsrg

a charge is kept there? and we have to find E on that? -_-

- shubhamsrg

hmm, so you want to find the net force acting on the charge.

- DLS

force

- yrelhan4

force nikalni hai? lol

- DLS

yes :/

- DLS

tume samaj agaya ques :")

- DLS

@yrelhan4 ata hai??!!

- yrelhan4

Hmm shayad ho jayega. but abhi nhi kr rha mai. biomolecules kr rha hun.

- anonymous

Dls.. you seriously need to get your terminologies right :-/.. !!!!

- yrelhan4

^ best response.

- 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

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

|dw:1364753502863:dw|
and so on

- anonymous

they'd all have a common factor of \[k{q^2\over a^2}\]

- anonymous

just be careful with the signs.. since the forces will be directed along the position vectors but in opposite direction

- DLS

@shubhamsrg

- DLS

@electrokid

- DLS

doubt!

- anonymous

shoot it.

- anonymous

@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

@electrokid !

- 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

yes i meant a root 2 only sorry!

- DLS

I didn't get the direction part,where will the force act?

- anonymous

that cos 45 should not be there!

- DLS

why?:|

- anonymous

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.

- anonymous

"components of a force"
you are talking about magnitude. do not include that cos... it'd show up automatically with the vectors.

- anonymous

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?

- 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

yes i eliminated that ,rest is correct right? and just that a root 2 thing

- anonymous

yes. rest seems fine.

- anonymous

take hintf from the previous picture

- anonymous

I gotta go now.

- DLS

please tell me how to do that XZ thing :|

- DLS

okay,later.

- DLS

@electrokid @Mashy

- anonymous

what XZ thingy?

- DLS

how do i find the component of XZ on that vertice?

- anonymous

still not done? :O

- DLS

no:/

- anonymous

which one? cricle it

- anonymous

or name it

- DLS

|dw:1364843986230:dw|