Question

The four 1.0 g spheres shown in FIGURE P25.42 are released simultaneously and allowed to move away from each other. What is the speed of each sphere when they are very far apart?

Short answer

The value of the speed of each sphere are 0.49m/s

Step-by-step solution

Theory of the speed of each sphere 

The potential energy of each point

$$\begin{gathered} U_{ij}=1/\left(4\pi {\epsilon }_0\right)\left(\frac{q_1{q}_j}{r_{ij}}\right) \\ ij=1,2....... \\ {q}_{1}to{q}_4=q \\ {r}_{13}=\sqrt{2}r \\ U=4\left(1/(4\pi {\epsilon }_0\frac{\left(q\right)\left(q\right)}{r}\right)+2\left(\mathbf{1}\mathbf{/}\mathbf{(}\mathbf{4}\mathit{\pi }{\mathit{\epsilon }}_{\mathbf{0}}\frac{\left(q\right)\left(q\right)}{\sqrt{\mathbf{2}}\mathbf{r}}\right) \\ q={10}^{-18}C \end{gathered}$$

The equation of U is applicable for all position of particle with their different distance .

By separation of each charges

$$\begin{gathered} r_{12}={10}^{-2}cm \\ {r}_{13}={10}^{-2}\sqrt{2}m \end{gathered}$$

applying all data

$$\begin{gathered} U=4\left(1/(4\pi {\epsilon }_0\frac{\left(q\right)\left(q\right)}{r}\right)+2\left(\mathbf{1}\mathbf{/}\mathbf{(}\mathbf{4}\mathit{\pi }{\mathit{\epsilon }}_{\mathbf{0}}\frac{\mathbf{\left(}\mathbf{q}\mathbf{\right)}\mathbf{\left(}\mathbf{q}\mathbf{\right)}}{\sqrt{\mathbf{2}}\mathbf{r}}\right) \\ U=9.5\times {10}^{-5}(5.1414)J \end{gathered}$$

Definition of the speed of each sphere

Now the kinetic energy for the four charge is

$$\begin{gathered} k=4\left(\frac{1}{2}\times {10}^{-3}kg\right)\left(v^2\right) \\ v=0.49m/s \end{gathered}$$

The value of the speed of each sphere are 0.49m/s