A uniform spherical volume charge distribution of radius $2\text{ m}$ , centered at the origin, has a strength of $\frac{3}{\pi }\times {10}^{-6}\text{ C/m}{3}^{}$ . A point charge of strength $\pi \times 8.854\times {10}^{-12}\text{ C}$ is moved from $(-3,0,-4)$ to $(0,0,4)$ . The work done is $\_\_\_\_\_\_$ $\text{ μJ}$  (Round off to two decimal places)

The correct answer is 0.40.

Step-by-step Derivation and Explanation:

1. Find the total charge of the sphere
The volume charge distribution has a radius $R=2\text{m}$ and a uniform charge density:
$$\rho =\frac{3}{\pi }\times {10}^{-6}{\text{C/m}}^3$$
The total charge $Q$ of the sphere is calculated by multiplying the charge density by the volume of the sphere:
$$Q=\rho \times \left(\frac{4}{3}\pi R^3\right)$$
Substituting the given values:
$$Q=\left(\frac{3}{\pi }\times {10}^{-6}\right)\times \frac{4}{3}\pi \times 2^3$$
$$Q=32\times {10}^{-6}\text{C}$$

2. Determine the initial and final positions relative to the sphere's center
Since the sphere is centered at the origin $(0,0,0)$, the distance $r$ from the origin to any point $(x,y,z)$ is given by:
$$r=\sqrt{x^2+y^2+z^2}$$
For the initial position $A(-3,0,-4)$:
$$r_i=\sqrt{{(-3)}^2+0^2+{(-4)}^2}=\sqrt{9+0+16}=5\text{m}$$
For the final position $B(0,0,4)$:
$$r_f=\sqrt{0^2+0^2+4^2}=4\text{m}$$
Since both distances $r_i=5\text{m}$ and $r_f=4\text{m}$ are greater than the radius of the sphere ($R=2\text{m}$), both points lie outside the sphere.

3. Calculate the potential difference
For any point outside a uniformly charged sphere, the electric potential is identical to that of a point charge located at the origin:
$$V\left(r\right)=\frac{Q}{4\pi {\epsilon }_{0}r}$$
The potential difference $\Delta V=V_f-V_i$ is:
$$\Delta V=\frac{Q}{4\pi {\epsilon }_0}\left(\frac{1}{r_f}-\frac{1}{r_i}\right)$$

4. Calculate the work done on the point charge
The work done by an external agent in moving a point charge $q$ is:
$$W=q\Delta V=\frac{qQ}{4\pi {\epsilon }_0}\left(\frac{1}{r_f}-\frac{1}{r_i}\right)$$
Given the permittivity of free space ${\epsilon }_0\approx 8.854\times {10}^{-12}\text{F/m}$, the point charge is:
$$q=\pi \times 8.854\times {10}^{-12}\text{C}=\pi {\epsilon }_0$$
Substituting $q=\pi {\epsilon }_0$ into the work formula:
$$W=\frac{\left(\pi {\epsilon }_0\right)Q}{4\pi {\epsilon }_0}\left(\frac{1}{4}-\frac{1}{5}\right)$$
Simplifying the constants:
$$W=\frac{Q}{4}\left(\frac{1}{20}\right)=\frac{Q}{80}$$
Substitute the total charge $Q=32\times {10}^{-6}\text{C}$:
$$W=\frac{32\times {10}^{-6}\text{J}}{80}=0.40\times {10}^{-6}\text{J}=0.40\text{μJ}$$