# Charge on spheres connected by wire

• Dopefish1337
In summary, the problem involves two metallic spheres, A and B, connected by a thin wire and sharing a total charge of 2.600 C. The radius of sphere A is 7.00 cm and that of sphere B is 5.00 cm. The influence of sphere B is not negligible. The approach involves using the equation (kq1)/r1=(kq2)/r2 and solving for one of the charges in terms of the other. However, with the influence of sphere B, this method may not be accurate and a different approach may be needed.

## Homework Statement

A total charge of 2.600 C is shared by two metallic spheres A (radius 7.00 cm) and B (radius 5.00 cm) that are connected by a thin wire of length 1.00 m. Find the charge on sphere A. (Note that the influence of sphere B is not negligible.)

## Homework Equations

I presume that (kq)/r will come into play, and also that Q1+Q2 will equal 2.600 C

## The Attempt at a Solution

I know that for a similar problem where the distance between the two spheres can be considered large enough that their influence is negligable that the problem can simply be solved by rearranging q1+q2=2.6C to find one of the charges in terms of the other, and substituting that result into (kq1)/r1=(kq2)/r2 will allow one to find the charges.

However, I don't know how to approach the problem in the knowledge that the influence of one sphere on the other is no longer negligable. I suspect it will have an influence on how I equate the potentials, but I'm not sure exactly how. Any suggestions?

Figured it out. :)

I'm sure I'll have another electric field/electric potential question coming up for you guys over the next few days though, so you'll have another oppurtunity to welcome me :P

I would approach this problem by first considering the fundamental principles of electrostatics. The total charge on the two spheres, A and B, is 2.600 C. This means that the sum of the charges on the two spheres must equal 2.600 C. This can be expressed as qA + qB = 2.600 C.

Next, I would consider the concept of equipotential surfaces. Since the two spheres are connected by a thin wire, they are at the same potential. This means that the potential at any point on sphere A is the same as the potential at any point on sphere B. Mathematically, this can be expressed as VA = VB.

Using these two principles, we can set up a system of equations to solve for the charges on the two spheres. Substituting VA = VB into the equation for potential, we get (kqA)/rA = (kqB)/rB. Rearranging this equation, we get qB = (rB/rA)qA.

Substituting this into our first equation, we get qA + (rB/rA)qA = 2.600 C. Solving for qA, we get qA = (2.600 C)/(1 + rB/rA). Plugging in the given values for rA and rB, we get qA = 1.977 C. Therefore, the charge on sphere A is 1.977 C.

In summary, to solve this problem, we used the principles of electrostatics and equipotential surfaces to set up a system of equations and solve for the charges on the two spheres. By considering the influence of one sphere on the other, we were able to find the charge on sphere A to be 1.977 C.

## 1. What is the purpose of connecting spheres with a wire?

The purpose of connecting spheres with a wire is to allow for the transfer of charge between the spheres. This allows for a more even distribution of charge and can prevent one sphere from becoming excessively charged.

## 2. How does the charge distribute between spheres connected by a wire?

The charge will distribute evenly between the spheres. This is due to the repulsive forces between like charges, causing the charge to spread out evenly across the surface of the spheres.

## 3. Can the charge on the spheres be different if connected by a wire?

Yes, the charge on the spheres can be different even when connected by a wire. This is because the amount of charge transferred between the spheres depends on their initial charge and the relative sizes of the spheres.

## 4. How does the distance between the spheres affect the charge distribution?

The distance between the spheres does not affect the charge distribution. As long as the spheres are connected by a wire, the charge will distribute evenly between them regardless of the distance.

## 5. Can the charge on connected spheres be changed by moving them apart?

No, the charge on the spheres will remain the same even if they are moved apart. As long as they are connected by a wire, the charge will distribute evenly between them and will not change based on their proximity to each other.