The reciprocity theorem for electrostatics
Consider a set of $n$ conductors of arbitrary sizes, shapes, and positions. Conductors $1,2,3, \ldots$ carry charges $Q_{1}, Q_{2}, Q_{3}, \ldots$, and their voltages are $V_{1}, V_{2}, V_{3}, \ldots$. Without disturbing the conductors, you change the charges to $Q_{1}^{\prime}, Q_{2}^{\prime}, Q_{3}^{\prime}$.
According to the reciprocity theorem for electrostatics,
$$
Q_{1} V_{1}^{\prime}+Q_{2} V_{2}^{\prime}+Q_{3} V_{3}^{\prime}+\cdots=Q_{1}^{\prime} V_{1}+Q_{2}^{\prime} V_{2}+Q_{3}^{\prime} V_{3}+\cdots
$$
or
$$
\sum Q V^{\prime}=\sum Q^{\prime} V
$$
We shall find analogous reciprocity theorems in Chaps. 8 and 27 .
You can prove this theorem by calculating the energy required to change the charges from $Q$ to $Q^{\prime}$ and equating this energy to $\Sigma Q^{\prime} V^{\prime} / 2-\Sigma Q V / 2$. To do this, set the charge and voltage on conductor 1 equal to $(1-x) Q_{1}+$ $x Q_{1}^{\prime}$ and $(1-x) V_{1}+x V_{1}^{\prime}$. Then you can go from one state to the other by letting $x$ go from zero to unity.