Question 6.8: A metal bar of conductivity σ is bent to form a flat 90° sec...

(a) Between the vertical curved ends located at \rho=a and \rho=b, the bar has a nonuniform cross section and hence eq. (5.16)

R=\frac{V}{I}=\frac{\ell}{\sigma S}

does not apply. We have to use eq. (6.16)

R=\frac{V}{I}=\frac{\int_{L}E\cdot dl}{\int_{S}\sigma E\cdot dS}

Let a potential difference V_{o} be maintained between the curved surfaces at \rho=a and \rho=b so that V(\rho=a)=0 and V(\rho=b)=V_{o}. We solve for V in Laplace’s equation \nabla^2 V=0 in cylindrical coordinates. Since V=V(\rho)

\nabla^2 V=\frac{1}{\rho}\frac{d}{d\rho}\left(\rho\frac{dV}{d\rho}\right)=0

As \rho=0 is excluded, upon multiplying by \rho and integrating once, this becomes

\rho\frac{dV}{d\rho}=A

or

\frac{dV}{d\rho}=\frac{A}{\rho}

Integrating once again yields

V=A\ln\rho+B

where A and B are constants of integration to be determined from the boundary conditions.

V(\rho=a)=0\rightarrow 0=A\ln a+B   or   B=-A\ln a

V(\rho=b)=V_{o}\rightarrow V_{o}=A\ln b+B=A\ln b-A\ln a=A\ln\frac{b}{a}   or   A=\frac{V_{o}}{\ln\frac{b}{a}}

Hence

V=A\ln \rho-A\ln a=A\ln\frac{\rho}{a}=\frac{V_{o}}{\ln\frac{b}{a}}\ln\frac{\rho}{a}

E=-\nabla V=-\frac{dV}{d\rho}a_{\rho}=-\frac{A}{\rho}a_{\rho}=-\frac{V_{o}}{\rho\ln\frac{b}{a}}a_{\rho}

J=\sigma E,       dS=-\rho d\phi dza_{\rho}

I=\int_{S}J\cdot dS=\int_{\phi=0}^{\pi/2}\int_{z=0}^{t}\frac{V_{o}\sigma}{\rho\ln\frac{b}{a}}dz\rho d\phi=\frac{\pi}{2}\frac{tV_{o}\sigma}{\ln\frac{b}{a}}

Thus

R=\frac{V_{o}}{I}=\frac{2\ln\frac{b}{a}}{\sigma\pi t}

as required.

(b) Let V_{o} be the potential difference between the two horizontal surfaces so that V(z=0)=0 and V(z=t)=V_{o}.   V=V(z), so Laplace’s equation \nabla^2 V=0 becomes

\frac{d^{2}V}{dz^{2}}=0

Integrating twice gives

V=Az+B

We apply the boundary conditions to determine A and B:

V(z=0)=0\rightarrow 0=0+B   or   B=0

V(z=t)=V_{o}\rightarrow V_{o}=At   or   A=\frac{V_{o}}{t}

Hence

V=\frac{V_{o}}{t}z

E=-\nabla V=-\frac{dV}{dz}a_{z}=-\frac{V_{o}}{t}a_{z}

J=\sigma E=-\frac{\sigma V_{o}}{t}a_{z},       dS=-\rho d\phi d\rho a_{z}

I=\int_{S}J\cdot dS=\int_{\rho=0}^{b}\int_{\phi=0}^{\pi/2}\frac{V_{o}\sigma}{t}\rho d\phi d\rho

=\frac{V_{o}\sigma}{t}\cdot \frac{\pi}{2}\frac{\rho^{2}}{2}|_{a}^{b}=\frac{V_{o}\sigma\pi(b^{2}-a^{2})}{4t}

Thus

R^{'}=\frac{V_{o}}{I}=\frac{4t}{\sigma\pi(b^{2}-a^{2})}

Alternatively, for this case, the cross section of the bar is uniform between the horizontal surfaces at z=0 and z=t and eq. (5.16) holds. Hence

R^{'}=\frac{\ell}{\sigma S}=\frac{t}{\sigma\frac{\pi}{4}(b^{2}-a^{2})}=\frac{4t}{\sigma\pi(b^{2}-a^{2})}

as required.