8-point charges on a vertice of a cube

[seanster1324]

Homework Statement

I attached a picture to make it easier...

Homework Equations

Coulomb's Law: F=k(q1*q2)/(r)^2

a^2+b^2=c^2

Charge properties

The Attempt at a Solution

I uploaded a picture of one of the cleaner sheets of work I used so far...Basically, I suppose I'm at a loss at figuring out where to start, and where to finish. I thought I was on the right track by calculating the magnitude of each charge affecting the vertex A, but apparently it hasn't got me very far.

 

[PeterO]

Homework Statement

I attached a picture to make it easier...

Homework Equations

Coulomb's Law: F=k(q1*q2)/(r)^2

a^2+b^2=c^2

Charge properties

The Attempt at a Solution

I uploaded a picture of one of the cleaner sheets of work I used so far...Basically, I suppose I'm at a loss at figuring out where to start, and where to finish. I thought I was on the right track by calculating the magnitude of each charge affecting the vertex A, but apparently it hasn't got me very far.

I would start by calculating the force between two adjacent charge - for example the one directly below A. All the other forces and components will be fractions of that - the size of the fraction determined by the different distances involved. [and angles when looking at components].
That will keep the arithmetic simple and you may be able to keep track of the forces.

 

[seanster1324]

Okay, I understand finding the force between the two adjacent ones. No problem. And I get that the other components are fractions of them, but how would I go about calculating them simply?

 

[PeterO]

Okay, I understand finding the force between the two adjacent ones. No problem. And I get that the other components are fractions of them, but how would I go about calculating them simply?

For "diagonal" charges, the separation is up by a factor of sqrt(2), so the force is down by a factor of 2 courtesy of the inverse square law [there is a factor of R² in the denominator of the formula.
The body diagonal distance is up by a factor of sqrt(3)

 

[seanster1324]

For "diagonal" charges, the separation is up by a factor of sqrt(2), so the force is down by a factor of 2 courtesy of the inverse square law [there is a factor of R² in the denominator of the formula.
The body diagonal distance is up by a factor of sqrt(3)

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