Question 6.10: Conducting spherical shells with radii a = 10 cm and b = 30 ...

Consider the spherical shells shown in Figure 6.18 and assume that V depends only on r. Hence Laplace’s equation becomes

\nabla^2 V=\frac{1}{r^{2}}\frac{d}{dr}\left[r^{2}\frac{dV}{dr}\right]=0

Since r\neq 0 in the region of interest, we multiply through by r^{2} to obtain

\frac{d}{dr}\left[r^{2}\frac{dV}{dr}\right]=0

Integrating once gives

r^{2}\frac{dV}{dr}=A

or

\frac{dV}{dr}=\frac{A}{r^{2}}

Integrating again gives

V=-\frac{A}{r}+B

As usual, constants A and B are determined from the boundary conditions. When

r=b,  V=0\rightarrow 0=-\frac{A}{b}+B   or   B=\frac{A}{b}

Hence

V=A\left[\frac{1}{b}-\frac{1}{r}\right]

Also when

r=a,  V=V_{o}\rightarrow V_{o}=A\left[\frac{1}{b}-\frac{1}{a}\right]

or

A=\frac{V_{o}}{\frac{1}{b}-\frac{1}{a}}

Thus

V=V_{o}\frac{\left[\frac{1}{r}-\frac{1}{b}\right] }{\frac{1}{a}-\frac{1}{b}}

E=-\nabla V=-\frac{dV}{dr}a_{r}=-\frac{A}{r^{2}}a_{r}=\frac{V_{o}}{r^{2}\left[\frac{1}{a}-\frac{1}{b}\right]}a_{r}

Q=\int_{S}\varepsilon E\cdot dS=\int_{\theta=0}^{\pi}\int_{\phi=0}^{2\pi}\frac{\varepsilon_{o}\varepsilon_{r}V_{o}}{r^{2}\left[\frac{1}{a}-\frac{1}{b}\right]}r^{2}\sin\theta d\phi d\theta=\frac{4\pi\varepsilon_{o}\varepsilon_{r}V_{o}}{\frac{1}{a}-\frac{1}{b}}

Alternatively

\rho_{s}=D_{n}=\varepsilon E_{r},        Q=\int_{s}\rho_{s}dS

The capacitance is easily determined as

C=\frac{Q}{V_{o}}=\frac{4\pi\varepsilon}{\frac{1}{a}-\frac{1}{b}}

which is the same as we obtained in eq (6.32)

C=\frac{Q}{V}=\frac{4\pi\varepsilon}{\frac{1}{a}-\frac{1}{b}}

there in Section 6.5, we assumed Q and found the corresponding V_{o}, but here we assumed V_{o} and found the corresponding Q to determine C. Substituting a=0.1 m, b=0.3 m, V_{o}=100V yields

V=100\frac{\left[\frac{1}{r}-\frac{10}{3}\right]}{10-10/3}=15\left[\frac{1}{r}-\frac{10}{3}\right]V

Check:

\nabla^2 V=0,    V(r=0.3m)=0,    V(r=0.1m)=100

E=\frac{100}{r^{2}[10-10/3]}a_{r}=\frac{15}{r^{2}}a_{r}V/m

Q=\pm4\pi\cdot \frac{10^{-9}}{36\pi}\cdot\frac{(2.5)\cdot(100)}{10-10/3}=\pm4.167nC

The positive charge is induced on the inner shell; the negative charge is induced on the outer shell. Also

C=\frac{|Q|}{V_{o}}=\frac{4.167\times10^{-9}}{100}=41.67pF