# Three spheres attached by conducting wire

1. Mar 1, 2009

### valarking

The problem statement, all variables and given/known data

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 [Broken]

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?

Last edited by a moderator: May 4, 2017
2. Mar 2, 2009

### 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.

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