Potential Energy of a System of Two Charges in an External Electric Field

Worked out. (a) Charges $q_1=-2\,\mu C$ at $(-8\text{ cm},0,0)$ and $q_2=+4\,\mu C$ at $(+8\text{ cm},0,0)$, so $r=16\text{ cm}=0.16$ m apart. Their mutual PE alone is $\dfrac{1}{4\pi\epsilon_0}\dfrac{q_1q_2}{r}=9\times10^9\times\dfrac{(-2\times10^{-6})(4\times10^{-6})}{0.16}=-0.45$ J. (b) The SAME pair is now placed in an external field $E=A/r^2$, $A=8\times10^5\,\text{cm}^{-2}$; from $E=-dV/dr$, integrating gives $V(r)=A/r$ (taking $V\to0$ as $r\to\infty$), so each charge's own individual PE in this external field is $q_1V(r_1)=q_1A/r_1$ and $q_2V(r_2)=q_2A/r_2$ (with $r_1=r_2=0.08$ m). Adding the mutual PE, external-field PE of $q_1$, and external-field PE of $q_2$: total $=-0.45+(-20)+(40)=19.55$ J -- the mutual PE from part (a) plus the two individual external-field terms from section 8.6.3, added exact …