Electrostatics Part 3

Concept:For charges placed in a fixed external field, the total electrostatic energy is $U=\sum_i q_i V_{\text{ext}}(r_i)+U_{\text{interaction}}$.The external-field contribution is not multiplied by $\frac{1}{2}$ because the external field is fixed; the interaction between the two charges is counted once.Explanation:For $E=\frac{A}{r^2}\hat{r}$ and $V(\infty)=0$, $V(r)=-\int_{\infty}^{r}E\,dr=\frac{A}{r}$.Given $A=9\times10^5\ \mathrm{N\,m^2/C}$.Charge $q_1=7\times10^{-6}\ \mathrm{C}$ is at $r_1=9\ \mathrm{cm}=0.09\ \mathrm{m}$.Its external potential energy is $U_1=q_1V(r_1)=\frac{(7\times10^{-6})(9\times10^5)}{0.09}=70\ \mathrm{J}$.Charge $q_2=-2\times10^{-6}\ \mathrm{C}$ is at $r_2=9\ \mathrm{cm}=0.09\ \mathrm{m}$.Its external potential energy is $U_2=q_2V(r_2)=\frac{(-2\times10^{-6})(9\times10^5)}{0.09}=-20\ \mathrm{J}$.The separation between the charges is $d=18\ \mathrm{cm}=0.18\ \mathrm{m}$.So $U_{\mathrm{int}}=\frac{kq_1q_2}{d}=\frac{(9\times10^9)(7\times10^{-6})(-2\times10^{-6})}{0.18}=-0.7\ \mathrm{J}$.Therefore $U_{\mathrm{total}}=70-20-0.7=49.3\ \mathrm{J}$.Answer:$49.3\ \mathrm{J}$, matching option option A.