Two point charges, 4 µC and-3 µC (with no external field) are placed at (-6 cm, 0, 0) and (6 cm, 0, 0), respectively. The amount of work required to separate the two charges infinitely away from each other will be

The correct option is 0.9 J.

To understand why this is the correct answer, we must calculate the work required to separate the two point charges infinitely far apart.

First, let us identify the given values:
First charge, $q_1=4\times {10}^{-6}\text{C}$ placed at position $(-6\text{cm},0,0)$.
Second charge, $q_2=-3\times {10}^{-6}\text{C}$ placed at position $(6\text{cm},0,0)$.

The distance $r$ between the two charges lies along the x-axis and is given by:
$r=6\text{cm}-(-6\text{cm})=12\text{cm}=0.12\text{m}$

The initial electrostatic potential energy ($U_{\text{initial}}$) of the system of these two charges is given by the formula:

$U_{\text{initial}}=\frac{1}{4\pi {\epsilon }_0}·\frac{q_1{q}_2}{r}$

where the constant $\frac{1}{4\pi {\epsilon }_0}\approx 9\times {10}^9{\text{N·m}}^2{\text{/C}}^2$.

Substituting the given values into the equation:

$U_{\text{initial}}=9\times {10}^9\times \frac{(4\times {10}^{-6})\times (-3\times {10}^{-6})}{0.12}$

Let's calculate the numerator first:
$9\times {10}^9\times (-12\times {10}^{-12})=-108\times {10}^{-3}=-0.108$

Now, dividing by the distance $0.12$:

$U_{\text{initial}}=\frac{-0.108}{0.12}=-0.9\text{J}$

When the charges are separated infinitely far away from each other, the final distance between them becomes infinity ($\infty$). Therefore, the final potential energy ($U_{\text{final}}$) is:
$U_{\text{final}}=0\text{J}$

The work required ($W$) by an external agent to separate the charges is equal to the change in the potential energy of the system:

$W=U_{\text{final}}-U_{\text{initial}}$

$W=0-(-0.9\text{J})=0.9\text{J}$

Thus, the work required to separate the two charges infinitely away from each other is 0.9 J.