Electrostatic energy of system of two uniformly charged spheres Suppose I have two different spheres uniformly charged. One of them positively charged and the other one negatively charged at a distance $L$ from each other. I know that the electrostatic energy of each sphere is positive and I know that the energy of interaction between them is negative.
But as I see from this equation:
$$W=\frac{\varepsilon_0}{2}\int E^2d\tau$$
Does the total energy of the system have to be positive?
 A: Mathematically: $$W\sim \frac{\mathbf{E}^2}{2}=\frac{\mathbf{(E_1+E_2)}^2}{2}=\frac{\mathbf{E_1}^2}{2}+\frac{\mathbf{E_2}^2}{2}+\mathbf{E_1 E_2}=W_1+W_2+W_{12}$$
where $W_{1,2}=\frac{\mathbf{E}_{1,2}^2}{2}$ are self energies, $W_{12}=\mathbf{E_1 E_2}$ is mutual or interaction energy. Observing that $\frac{\mathbf{(E_1-E_2)}^2}{2}=\frac{\mathbf{E_1}^2}{2}+\frac{\mathbf{E_2}^2}{2}-\mathbf{E_1 E_2}>0$ we conclude that $$W_1+W_2>W_{12}$$
Physically: the positive "self energy", corresponding to the charge blowing itself apart is always larger than the energy of interaction with other charges elsewhere, because the charge is  closer to itself than to the others.
A: You can get a good estimate of the total energy if you have two conducting spheres with enough separation that you can assume that the charges are uniformly distributed on the surface of each.  Start with both spheres uncharged.  Then find the work  required to move (positive) charge from one to the other (leaving negative behind).  The work to move a small segment dq would be dW = ΔV dq ={[ kq/R – kq/(R+D)] – [kq/(R+D) – kq/R]}dq  where D is the surface to surface separation, and q is the previously moved charge.  This assumes that each charge redistributes after each small move and can be treated as a point charge. (You do get a positive result.)
