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## Homework Statement

We have a spehere of radius ##r_1## and on of ##r_2## far away from each other. The first sphere has a charge ##Q##. What is the change in electro static energy when they are connected together?

## Homework Equations

Potential of a charged sphere: ## V = \frac{Q}{4\pi\epsilon_0 r}##

Elctrostatic energy: ##W = q \Delta V ##

## The Attempt at a Solution

Initially, the potential of the system is: ## V = \frac{Q}{4\pi\epsilon_0 r_1}##

When the spheres are connected together, they become equipotential and will have charges:

## Q_1 = Q \frac{r_1}{r_1+r_2} ## and ## Q_2 = Q \frac{r_2}{r_1+r_2} ##. Hence the change in voltage is: ## \Delta V = \frac{Q}{4\pi\epsilon_0 r_1} - \frac{Q_1}{4\pi\epsilon_0 r_1} = \frac{Q}{4\pi\epsilon_0} (\frac{r_2}{r_1(r_1+r_2)}) ##. So the change in potential energy is: ## W=\frac{Q^2}{4\pi\epsilon_0} (\frac{r_2}{r_1(r_1+r_2)})##.

However, this is wrong. The answer is supposed to be: ## W=\frac{Q^2}{8\pi\epsilon_0} (\frac{r_2}{r_1(r_1+r_2)})##. Where do I loose the factor of 1/2?