(a) Situation (1): actual. Situation (2): right plate at V_{0}, left plate at V = 0, no charge at x.
\int V_{1} \rho_{2} d \tau=V_{l_{1}} Q_{l_{2}}+V_{x_{1}} Q_{x_{2}}+V_{r_{1}} Q_{r_{2}} .
But V_{l_{1}}=V_{r_{1}}=0 \text { and } Q_{x_{2}}=0, \text { so } \int V_{1} \rho_{2} d \tau=0 .
\int V_{2} \rho_{1} d \tau=V_{l_{2}} Q_{l_{1}}+V_{x_{2}} Q_{x_{1}}+V_{r_{2}} Q_{r_{1}} .
But V_{l_{2}}=0 Q_{x_{1}}=q, V_{r_{2}}=V_{0}, Q_{r_{1}}=Q_{2}, \text { and } V_{x_{2}}=V_{0}(x / d) . \text { So } 0=V_{0}(x / d) q+V_{0} Q_{2} , and hence
Q_{2}=-q x / d .
Situation (1): actual. Situation (2): left plate at V_{0}, right plate at V = 0, no charge at x.
\int V_{1} \rho_{2} d \tau=0=\int V_{2} \rho_{1} d \tau=V_{l_{2}} Q_{l_{1}}+V_{x_{2}} Q_{x_{1}}+V_{r_{2}} Q_{r_{1}}=V_{0} Q_{1}+q V_{x_{2}}+0 .
But V_{x_{2}}=V_{0}\left(1-\frac{x}{d}\right) , so
Q_{1}=-q(1-x / d) .
(b) Situation (1): actual. Situation (2): inner sphere at V_{0}, outer sphere at zero, no charge at r.
\int V_{1} \rho_{2} d \tau=V_{a_{1}} Q_{a_{2}}+V_{r_{1}} Q_{r_{2}}+V_{b_{1}} Q_{b_{2}} .
But V_{a_{1}}=V_{b_{1}}=0, Q_{r_{2}}=0 . \text { So } \int V_{1} \rho_{2} d \tau=0 .
\int V_{2} \rho_{1} d \tau=V_{a_{2}} Q_{a_{1}}+V_{r_{2}} Q_{r_{1}}+V_{b_{2}} Q_{b_{1}}=Q_{a} V_{0}+q V_{r_{2}}+0 .
But V_{r_{2}} is the potential at r in configuration 2: V(r)=A+B / r, \text { with } V(a)=V_{0} \Rightarrow A+B / a=V_{0} , or a A+B=a V_{0} \text {, and } V(b)=0 \Rightarrow A+B / b=0, \text { or } b A+B=0 . \text { Subtract: }(b-a) A=-a V_{0} \Rightarrow A=-a V_{0} /(b-a) ; B\left(\frac{1}{a}-\frac{1}{b}\right)=V_{0}=B \frac{(b-a)}{a b} \Rightarrow B=a b V_{0} /(b-a) . \text { So } V(r)=\frac{a V_{0}}{(b-a)}\left(\frac{b}{r}-1\right) . Therefore
Q_{a} V_{0}+q \frac{a V_{0}}{(b-a)}\left(\frac{b}{r}-1\right)=0 ; \quad Q_{a}=-\frac{q a}{(b-a)}\left(\frac{b}{r}-1\right) .
Now let Situation (2) be: inner sphere at zero, outer at V_{0}, no charge at r.
\int V_{1} \rho_{2} d \tau=0=\int V_{2} \rho_{1} d \tau=V_{a_{2}} Q_{a_{1}}+V_{r_{2}} Q_{r_{1}}+V_{b_{2}} Q_{b_{1}}=0+q V_{r_{2}}+Q_{b} V_{0} .
This time V(r)=A+\frac{B}{r} \text { with } V(a)=0 \Rightarrow A+B / a=0 ; V(b)=V_{0} \Rightarrow A+B / b=V_{0} , so
V(r)=\frac{b V_{0}}{(b-a)}\left(1-\frac{a}{r}\right) . \text { Therefore } q \frac{b V_{0}}{(b-a)}\left(1-\frac{a}{r}\right)+Q_{b} V_{0}=0 ; \quad Q_{b}=-\frac{q b}{(b-a)}\left(1-\frac{a}{r}\right) .
