Electricity and magnetism

1. Jan 24, 2007

feelau

1. The problem statement, all variables and given/known data
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.

2. Relevant 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 plz help?

2. Jan 24, 2007

marcusl

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

3. Jan 24, 2007

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?

4. Jan 24, 2007

mukundpa

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

5. Jan 24, 2007

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?

6. Jan 25, 2007

marcusl

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

7. Jan 25, 2007

feelau

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

8. Jan 25, 2007

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)...