# Multiple-choice question, Electric field and potential

A positively-charged metal sphere A of reaius a is jointed by a long conducting wire to another positively charged metal sphere B of radius b. Assume that B is far away from A. If the charges on A and B are respectively q1 and q2, what is the ratio q1:q2?

Relevant equations
V=q/(4∏εr)

a) a^2:b^2
b) a:b
c) b^2:a^2
d) b:a

Solution:
The conducting wire ensures that the potentials of the two spheres are the same. Since they are far apart, the charge of any sphere will not influence the potential of one another.
V(sphere a)=V(sphere b)
q1/(4∏εa) = q2/(4∏εb)
∴ q1/q2=a/b
This is the solution from the book.
Why the potentials of the two sphere are the same when they joined by a conducting wire?

Related Introductory Physics Homework Help News on Phys.org
D H
Staff Emeritus
Suppose you

- Cut the wire somewhere between the spheres,
- Put a charge on the spheres such that two spheres have different potentials, and
- Reattach the cut ends of the wire.

What flows down the wire?

The charges in the sphere?