A 1 μC point charge is held at the origin of a cartesian coordinate system. If a second point charge of 10 μC is moved from (0, 10, 0) to (5, 5, 5) and subsequently to (5, 0, 0), then the total work done is _________ mJ. (Round off to 2 decimal places). Take $\frac{1}{4\pi {\epsilon }_o}=9\times {10}^9$ in SI units. All coordinates are in meters.

The correct answer is 9.

1. Understand the Concepts:
The electrostatic force is a conservative force, meaning that the work done in moving a charge between two points in an electrostatic field depends only on the initial and final positions of the charge, not on the path taken. Therefore, the intermediate position (5, 5, 5) does not affect the calculation of the total work done.
The total work done $W$ by an external agent in moving a point charge $q_2$ in the field of a static point charge $q_1$ is equal to the change in electrostatic potential energy of the system:
$W=\Delta U=U_{\text{final}}-U_{\text{initial}}$
where the potential energy $U$ at a distance $r$ is given by:
$U=\frac{1}{4\pi {\epsilon }_o}\frac{q_1{q}_2}{r}$

2. Calculate the Distances:
The charge $q_1=1\text{μC}$ is fixed at the origin (0, 0, 0).
The initial position of the second charge $q_2=\mathrm{10<mtext>\; \mu C</mtext>}$ is (0, 10, 0). Thus, the initial distance $r_1$ is:
$r_1=\sqrt{0^2+{10}^2+0^2}=10\text{m}$
The final position of the charge $q_2$ is (5, 0, 0). Thus, the final distance $r_2$ is:
$r_2=\sqrt{5^2+0^2+0^2}=5\text{m}$

3. Compute the Total Work Done:
Substitute the given values into the work-energy relation:
$q_1=1\times {10}^{-6}\text{C}$
$q_2=10\times {10}^{-6}\text{C}={10}^{-5}\text{C}$
$\frac{1}{4\pi {\epsilon }_o}=9\times {10}^9\text{N}{\text{m}}^2\text{/}{\text{C}}^2$
Now, compute $W$:
$W=(9\times {10}^9)\times \left({10}^{-6}\right)\times \left({10}^{-5}\right)\times (\frac{1}{5}-\frac{1}{10})$
$W=(9\times {10}^{-2})\times (0.2-0.1)$
$W=9\times {10}^{-2}\times 0.1=9\times {10}^{-3}\text{J}$
Converting the work done to millijoules ($\text{mJ}$):
$W=9\text{mJ}$