What is the Tension in a Conducting Wire Connecting Two Charged Spheres?

[feelau]

Homework Statement

Two identical conducting spheres each having a radius of .5 cm are connected by a light 2.3 m long conducting wire. A charge of 53 uC is placed on one of the conductors. Assume that the surface distribution of charge on each sphere is uniform. Determine the tension in the wire.

Homework Equations

F(electricity)=(k*q1*q2)/r^2
E=Q/epsilon
E=F/Q
F(electric)-T=0

3. Attempt
Since it's connected by a wire(I'm not sure how the wire plays a roll in this other than tension), I said that the charge instantaneously distributes to the other spheres so that both of them will have equal amount of charge(I'm not sure if the assumption is correct) I then solved for F(electric) which is equal to T, but I'm missing something because answer is not right. Can someone please help?

 

[marcusl]

How much charge did you put on each sphere?
What distance did you use for the separation?

 

[feelau]

well the the thing is I wasn't sure what the length of wire is for, now I think I know it's for how far apart they are. For the F(electric) and the question just says 53uC of charge is put into one sphere so I assumed there were no charges in them beforehand. So then, I assumed that each sphere will have 26.5 uC of charge so F(electric) would just be (k*q^2)/(length of wire)^2? But it seems like since they're spheres and not point charge(unless that's what we're suppose to assume) I need to include another equation?

 

[mukundpa]

the uniform charge distribution hints that the effective points are the centers of the spheres

 

[feelau]

so that means we can take into account that they're just like pt charges then?
So the distance between them is the length of wire and they both have same charge correct? So I just use F(electric)=k(q^2)/distance of wire^2 then i putinto force equation?

 

[marcusl]

so that means we can take into account that they're just like pt charges then?

Exactly

So the distance between them is the length of wire and they both have same charge correct?

Not quite. You need the distance between the centers of the spheres.

 

[feelau]

ah, my TA talked about this and he said that since r is so small, it's negligible

 

[marcusl]

Why wouldn't you include it, however? It's no harder to enter the correct value (2.31) than the wrong one (2.3)...