91Ó°ÊÓ

The introduced point charge at a distance x in between the two grounded plates of a parallel-plate capacitor of plate separation d is q.

The introduced point charge at a radius r in between the two grounded spherical shells of radii a and b is q.

The Green's reciprocity theorem can be written as

$\mathbf{∫}{\mathbf{V}}_{\mathbf{1}}{\mathbf{ÒÏ}}_{\mathbf{2}}\mathbf{»åÏ„}\mathbf{=}\mathbf{∫}{\mathbf{V}}_{\mathbf{2}}{\mathbf{ÒÏ}}_{\mathbf{1}}\mathbf{»åÏ„}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{\left(}\mathbf{1}\mathbf{\right)}$

$V_1\mathit{}\mathrm{and}\mathit{}{V}_{\mathit{2}}$ are the potentials.

Let $Q_1,Q_2$and $Q_x$ be the charge on the two plates and the position of charge q . Similarly, $V_1,V_2$and $V_x$ is the potential on the two plates and the position of charge q.

Situation 1 can be summarized as

$Q_1=Q_{1,ind},V_1=0,Q_x=q,V_x=V,Q_2=Q_{2,ind},V_2=0$

In situation 2 remove the charge, keep the first plate at zero potential and keep the second plate at potential $V_0$.

This can be summarized as

$Q_1=Q_{12},V_1=0,Q_x=0,V_x=\frac{V_{0}x}{d},Q_2=Q_{22}{V}_2=0$

Apply Green's reciprocity defined in equation (1) to the first and second equation and get

$$\begin{gathered} Q_{1,ind}×0+q×\frac{V_{0}x}{d}+Q_{2,ind}×V_0=Q_{12}×0+0×V+Q_{22}×0 \\ \frac{V_{0}x}{d}+Q_{2,ind}{V}_0=0 \\ {Q}_{2,ind}=-\frac{qx}{d} \end{gathered}$$

In situation 3 remove the charge, keep the first plate at potential $V_0$and keep the second plate at zero potential.

This can be summarized as

$Q_1=Q_{13},V_1=0,Q_x=0,V_x=V_0\left(1-\frac{x}{d}\right),Q_2=Q_{23}{V}_2=0$

Apply Green's reciprocity defined in equation (1) to the first and third situation and get

$$\begin{gathered} Q_{1,ind}×V_0+q×V_0\left(1-\frac{x}{d}\right)+Q_{2,ind}×0=Q_{12}×0+0×V+Q_{22}×0 \\ {Q}_{1,ind}{V}_0+q{V}_0\left(1-\frac{x}{d}\right)=0 \\ {Q}_{1,ind}=-q\left(1-\frac{x}{d}\right) \\ {Q}_{1,ind}=-q\left(\frac{x}{d}-1\right) \end{gathered}$$

Thus, the charge induced in the first plate is $q\left(\frac{x}{d}-1\right)$and the charge induced in the second plate is $-\frac{qx}{d}$.

Situation 1 can be summarized as

$Q_a=Q_{a,ind},V_a=0,Q_r=q,V_r=V,Q_b=Q_{b,ind,}{V}_a=0$

In situation 2 remove the second charge, keep the inner shell at zero potential and keep the outer shell at potential $V_0$. The potential at any point can be written as

$V\left(r\right)=A+\frac{B}{r}$

The boundary conditions are

$V\left(a\right)=A+\frac{B}{a}=0$

and

$V\left(b\right)=A+\frac{B}{b}=V_0$

Solve the above two equations to get the following expressions:

$B=\frac{V_0}{\left({\displaystyle \frac{1}{b}}-{\displaystyle \frac{1}{a}}\right)}$

and

$A=\frac{b{V}_0}{b-a}$

Thus, the potential function becomes,

$$\begin{gathered} V\left(r\right)=\frac{b{V}_0}{b-a}+\frac{V_0}{\left({\displaystyle \frac{1}{b}}-{\displaystyle \frac{1}{a}}\right)}\frac{1}{r} \\ =\frac{b{V}_0}{b-a}\left(1-\frac{a}{r}\right) \end{gathered}$$

Thus, the second situation can be summarized as,

$Q_a=Q_{a2},V_a=0,Q_r=0,V_r=\frac{b{V}_0}{b-a}\left(1-\frac{a}{r}\right),Q_b=Q_{b2,}{V}_b=0$

Apply Green's reciprocity defined in equation (1) to the first and second equation.

$$\begin{gathered} Q_{a,ind}×0+q×\frac{b{V}_0}{b-a}\left(1-\frac{a}{r}\right)+Q_{b,ind}×V_0=Q_{a2}×0+0×V+Q_{b2}×0 \\ \frac{qb{V}_0}{b-a}\left(1-\frac{a}{r}\right)+Q_{b,ind}{V}_0=0 \\ {Q}_{b,ind}=-\frac{qb}{b-a}\left(1-\frac{a}{r}\right) \end{gathered}$$

In situation 3 remove the second charge, keep the inner shell at potential $V_0$and keep the outer shell at zero potential. The potential at any point can be written as

$V\left(r\right)=A+\frac{B}{r}$

The bound conditions are

$V\left(a\right)=A+\frac{B}{a}=V_0$

and

$V\left(b\right)=A+\frac{B}{b}=0$

Solve the above two equations to get the following expressions.

$B=\frac{V_0}{\left({\displaystyle \frac{1}{a}}-{\displaystyle \frac{1}{b}}\right)}$

and

$A=\frac{a{V}_0}{a-b}$

Thus, the potential function becomes,

$$\begin{gathered} V\left(r\right)=\frac{a{V}_0}{a-b}+\frac{V_0}{\left({\displaystyle \frac{1}{a}-\frac{1}{b}}\right)}\frac{1}{r} \\ =\frac{a{V}_0}{a-b}\left(1-\frac{b}{r}\right) \end{gathered}$$

Thus, the second situation can be summarized as,

$Q_a=Q_{a2},V_a=0,Q_r=0,V_r=\frac{b{V}_0}{b-a}\left(1-\frac{a}{r}\right),Q_b=Q_{b2,}{V}_b=0$

Apply Green's reciprocity defined in equation (1) to the first and second equation.

localid="1658561439076" $$\begin{gathered} Q_{a,ind}×V_0+q×\frac{q{V}_0}{a-b}\left(1-\frac{b}{r}\right)+Q_{b,ind}×0=Q_{a2}×0+0×V+Q_{b2}×0 \\ \frac{qb{V}_0}{a-b}\left(1-\frac{b}{r}\right)+Q_{a,ind}{V}_0=0 \\ {Q}_{a,ind}=-\frac{qa}{a-b}\left(1-\frac{b}{r}\right) \end{gathered}$$

Thus, the charge induced in the inner shell is $-\frac{qa}{a-b}\left(1-\frac{b}{r}\right)$and the charge induced in the outer shell is $-\frac{qb}{b-a}\left(1-\frac{a}{r}\right)$.