Three spheres attached by conducting wire

[valarking]

Homework Statement

Three conducting spheres of radii a, b and c, respectively, are connected by negligibly
thin conducting wires as shown in figure 4. Distances between spheres are much larger
than their sizes. The electric field on the surface of the sphere of radius a is measured to
be equal to $$E_a$$. What is the total charge Q that this system of three spheres holds? How
much work do we have to do to bring a very small charge q from infinity to the sphere of
radius b?

http://www.vkgfx.com/physics/fig4.jpg

The attempt at a solution

This one I wasn't so sure about. In order to find the total charge, I first applied Gauss's Law to find the charge of one sphere, A, given the electric field on its surface. So $$\oint {\vec{E} \cdot d\vec{A}} = \frac{Q_{enclosed}}{\epsilon_0}$$ is used to get:
$$4\pi{a^2}E_a = \frac{q_a}{\epsilon_0}$$, or $$q_a = 4\pi{a^2}\epsilon_0{E_a}$$.

To get the relationship between one sphere and the total charge in a system of conducting connected spheres, I used this equation, which I'm not so sure about.
$$q_A = \frac{Qa}{a+b+c}$$
I'm not sure if the radii should be squared.

Solving for Q gives:
$$Q = \frac{{a}E_a}{k_e}(a+b+c)$$

Now for the second part, again I wasn't sure of this. But you could say the work going from infinitely far to right at sphere B would be the difference in potential energies.
So:
$$k_e\frac{q{q_b}}{b} - k_e\frac{q{q_b}}{\infty}$$
where the infinity cancels out the second fraction and leaves:
$$k_e\frac{q{q_b}}{b}$$
as the answer.
Plugging in b's charge gives:
$$\frac{k_e{q}{Q}}{a+b+c}$$

I believe that any effect a or c would have on charge q's potential would be small enough not to matter. I am correct in this assumption?

 

[Delphi51]

The spheres are connected by conductors - so they have the same potential.
Figure out the potential on sphere a, then use that to find the charges on the others.

 

[nlightner]

I would like to provide some clarification and possible corrections to your solution. First, your use of Gauss's Law to find the charge on one sphere is correct. However, the equation q_A = \frac{Qa}{a+b+c} is not applicable in this case. This equation is used to find the charge on one sphere in a system of charged spheres that are not connected by conductive wires. In this case, since the spheres are connected by conductive wires, they will have the same potential and thus the same charge distribution. Therefore, the total charge Q can simply be found by summing the charges on each sphere, which is Q = q_a + q_b + q_c.

Secondly, for the second part of the question, the work required to bring a small charge q from infinity to the surface of sphere b would be given by the equation:
W = k_e\frac{q{q_b}}{b} - k_e\frac{q{\infty}}{\infty} = k_e\frac{q{q_b}}{b}
since the potential at infinity is taken to be zero. Therefore, the work required would simply be k_e\frac{q{q_b}}{b}.

In regards to your assumption about the effect of spheres a and c on the potential of sphere b, it is correct that their effect would be negligible since the distances between the spheres are much larger than their sizes. This can also be seen from the fact that the work required to bring a charge from infinity to the surface of sphere b is only dependent on the charge and the distance b, and not on the charges or distances of the other spheres.

I hope this helps clarify any confusion and provides a more accurate solution to the problem.