# Help! 3 well separted conductin spheres

1. Mar 29, 2004

### belleamie

Help! 3 well separted conductin spheres....

three well separated conducting spheres of radii 2.5,4cm, and 6.5cm are charged to .04uC, 0.55 uC and 0.3 uC respectively.
a) Calculate the voltage at the surface of each sphere(assume a zero voltage at infinity) hint: outside the surface, charged spheres behave like pt charges.
b)now assume all three spheres are connected by wires. Calculate the new charges on each sphere
hint: continuous metallic surface are equipotential surfaces.

Thats the problem I have to solve
my soultion so far
a)v=kq/r
for sphere 1: V=(8.99x10*9)(0.4uC)/(.025m)=1.43x10*11
For Sphere 2: V=(8.99x10*9)(.55uC)/(.04m)=1.23x10*11
For Sphere 2: V=(8.99x10*9)(.3uC)/(.065m)=4.15x10*9

b)for all use the formula q/r?

2. Mar 30, 2004

1) Yes, now add units and you're done.

2) They want you to calculate the charge on each. Since the spheres are conducting, then each is will have the same potential at its surface. Since all the potentials are equal, you have three equations (each pair of potentials) and three variables (the charge on each sphere). Solve the system of equations.

3. Mar 30, 2004

### belleamie

hmm lol, you confused me?

4. Mar 30, 2004

### Chen

The potential on the surface of each sphere is:
$$V = K\frac{q}{r}$$
As the monster said, the spheres are conducting and therefore each will have the same potential at its surface. The first set of equations is:
$$\frac{q_1}{r_1} = \frac{q_2}{r_2} = \frac{q_3}{r_3}$$
You have three uknowns there, since the charges change but the radii don't. But you only have 2 equations up there, so you can't solve the system yet. You need one more equation, which is the preservation of charge:
$$Q_T = q_1 + q_2 + q_3$$
The total charge of the system doesn't change, since the charges only move between spheres and don't disappear. Now you have 3 equations with 3 unknowns and you can solve the system.