Electrostatics Part 3

Concept:Total electrostatic energy of a system in an external field equals the sum of each charge's self-energy in that field plus the interaction energy between the charges.Explanation:Given external field $E = \frac{A}{r^2}\hat{r}$ with $A = 9\times10^5\ \mathrm{N\,m^2/C}$.Potential from this field: $V(r) = -\int E\,dr = \frac{A}{r}$ (zero at infinity).Self‑energy of a charge $q$ at distance $r$: $U = qV = \frac{qA}{r}$.Charge $q_1 = 7\ \mu\mathrm{C} = 7\times10^{-6}\ \mathrm{C}$ at $r_1 = 9\ \mathrm{cm} = 0.09\ \mathrm{m}$: $U_1 = \frac{(7\times10^{-6})(9\times10^{5})}{0.09} = \frac{6.3}{0.09} = 70\ \mathrm{J}$.Charge $q_2 = -2\ \mu\mathrm{C} = -2\times10^{-6}\ \mathrm{C}$ at $r_2 = 9\ \mathrm{cm} = 0.09\ \mathrm{m}$: $U_2 = \frac{(-2\times10^{-6})(9\times10^{5})}{0.09} = -\frac{1.8}{0.09} = -20\ \mathrm{J}$.Interaction energy between charges: $U_{\mathrm{int}} = k\frac{q_1q_2}{d}$, with $k = 9\times10^9\ \mathrm{N\,m^2/C^2}$ and separation $d = 18\ \mathrm{cm} = 0.18\ \mathrm{m}$.$U_{\mathrm{int}} = \frac{(9\times10^9)(7\times10^{-6})(-2\times10^{-6})}{0.18} = -\frac{0.126}{0.18} = -0.7\ \mathrm{J}$.Total energy: $U_{\mathrm{total}} = U_1 + U_2 + U_{\mathrm{int}} = 70 + (-20) + (-0.7) = 49.3\ \mathrm{J}$.Answer:$49.3\ \mathrm{J}$, which corresponds to option A.