Two charges $$7\mu C$$ and  $$-2\mu C$$ are placed at (-9,0,0)cm and (9,0,0)cm respectively in an external field $$E=\frac{A}{r^{2}}\overline{r}$$, where $$A=9\times 10^{5}N/C.m^{2}.$$ Considering the potential at infinity is 0, the electrostatic energy of the configuration is ______J.

We need to find the total electrostatic energy of a system of two charges in an external electric field.

First, the total electrostatic energy of a system of charges in an external field has three contributions: $$U_{\text{total}} = U_1^{\text{ext}} + U_2^{\text{ext}} + U_{12}^{\text{mutual}}$$ where $$U_i^{\text{ext}} = q_i V_{\text{ext}}(\vec{r}_i)$$ is the energy of charge $$q_i$$ in the external potential, and $$U_{12}^{\text{mutual}} = \frac{kq_1 q_2}{r_{12}}$$ is the mutual interaction energy.

Next, we find the external potential associated with the radial field $$\vec{E} = \frac{A}{r^2}\hat{r}$$ where $$A = 9 \times 10^5 \, \text{N/C·m}^2$$. By integrating from infinity to $$r$$, we obtain $$V(r) = -\int_\infty^r \vec{E} \cdot d\vec{r} = -\int_\infty^r \frac{A}{r'^2}\,dr' = \frac{A}{r}$$ with the reference $$V(\infty) = 0$$.

Then we calculate the positions and distances of the charges. The first charge is $$q_1 = 7 \, \mu\text{C} = 7 \times 10^{-6} \, \text{C}$$ located at $$(-9,0,0)\text{ cm}$$, so $$r_1 = 9 \, \text{cm} = 0.09 \, \text{m}$$. The second charge is $$q_2 = -2 \, \mu\text{C} = -2 \times 10^{-6} \, \text{C}$$ at $$(9,0,0)\text{ cm}$$, giving $$r_2 = 9 \, \text{cm} = 0.09 \, \text{m}$$. The distance between the charges is $$r_{12} = 18 \, \text{cm} = 0.18 \, \text{m}$$.

Substituting these into the expression for the potential, the external potential at each charge location is $$V(r_1) = \frac{A}{r_1} = \frac{9 \times 10^5}{0.09} = 10^7 \, \text{V}$$ and similarly $$V(r_2) = \frac{A}{r_2} = \frac{9 \times 10^5}{0.09} = 10^7 \, \text{V}$$.

Consequently, the energy of each charge in the external field becomes $$U_1^{\text{ext}} = q_1 V(r_1) = 7 \times 10^{-6} \times 10^7 = 70 \, \text{J}$$ and $$U_2^{\text{ext}} = q_2 V(r_2) = -2 \times 10^{-6} \times 10^7 = -20 \, \text{J}$$.

Meanwhile, the mutual interaction energy between the two charges is $$U_{12} = \frac{kq_1q_2}{r_{12}} = \frac{9 \times 10^9 \times 7 \times 10^{-6} \times (-2 \times 10^{-6})}{0.18} = \frac{9 \times 10^9 \times (-14 \times 10^{-12})}{0.18} = \frac{-126 \times 10^{-3}}{0.18} = -0.7 \, \text{J}$$.

Finally, summing all contributions yields $$U_{\text{total}} = 70 + (-20) + (-0.7) = 49.3 \, \text{J}$$.

The correct answer is Option (3): 49.3 J.