Two infinitely long wires running parallel to the x axis carry uniform charge densities +λ and −λ (Fig. 2.54).
(a) Find the potential at any point (x, y, z), using the origin as your reference.
(b) Show that the equipotential surfaces are circular cylinders, and locate the axis and radius of the cylinder corresponding to a given potential V_0.
(a) Potential of +\lambda \text { is } V_{+}=-\frac{\lambda}{2 \pi \epsilon_{0}} \ln \left(\frac{s_{+}}{a}\right), \text { where } s_{+} \text {is distance from } \lambda_{+} (Prob. 2.22).
Potential of -\lambda \text { is } V_{-}=+\frac{\lambda}{2 \pi \epsilon_{0}} \ln \left(\frac{s_{-}}{a}\right), \text { where } s_{-} \text {is distance from } \lambda_{-} .
\therefore \operatorname{Total} \quad V=\frac{\lambda}{2 \pi \epsilon_{0}} \ln \left(\frac{s_{-}}{s_{+}}\right) .
\text { Now } s_{+}=\sqrt{(y-a)^{2}+z^{2}} \text {, and } s_{-}=\sqrt{(y+a)^{2}+z^{2}} \text {, so }
V(x, y, z)=\frac{\lambda}{2 \pi \epsilon_{0}} \ln \left(\frac{\sqrt{(y+a)^{2}+z^{2}}}{\sqrt{(y-a)^{2}+z^{2}}}\right)=\frac{\lambda}{4 \pi \epsilon_{0}} \ln \left[\frac{(y+a)^{2}+z^{2}}{(y-a)^{2}+z^{2}}\right] .
(b) Equipotentials are given by \frac{(y+a)^{2}+z^{2}}{(y-a)^{2}+z^{2}}=e^{\left(4 \pi \epsilon_{0} V_{0} / \lambda\right)}=k = constant. That is:
y^{2}+2 a y+a^{2}+z^{2}=k\left(y^{2}-2 a y+a^{2}+z^{2}\right) \Rightarrow y^{2}(k-1)+z^{2}(k-1)+a^{2}(k-1)-2 a y(k+1)=0 , or
y^{2}+z^{2}+a^{2}-2 a y\left(\frac{k+1}{k-1}\right)=0 . \text { The equation for a circle, with center at }\left(y_{0}, 0\right) \text { and radius } R \text {, is }
\left(y-y_{0}\right)^{2}+z^{2}=R^{2}, \text { or } y^{2}+z^{2}+\left(y_{0}^{2}-R^{2}\right)-2 y y_{0}=0 .
Evidently the equipotentials are circles, with y_{0}=a\left(\frac{k+1}{k-1}\right) and
a^{2}=y_{0}^{2}-R^{2} \Rightarrow R^{2}=y_{0}^{2}-a^{2}=a^{2}\left(\frac{k+1}{k-1}\right)^{2}-a^{2}=a^{2} \frac{\left(k^{2}+2 k+1-k^{2}+2 k-1\right)}{(k-1)^{2}}=a^{2} \frac{4 k}{(k-1)^{2}}, \text { or }
R=\frac{2 a \sqrt{k}}{|k-1|} ; \text { or, in terms of } V_{0} :
y_{0}=a \frac{e^{4 \pi \epsilon_{0} V_{0} / \lambda}+1}{e^{4 \pi \epsilon_{0} V_{0} / \lambda}-1}=a \frac{e^{2 \pi \epsilon_{0} V_{0} / \lambda}+e^{-2 \pi \epsilon_{0} V_{0} / \lambda}}{e^{2 \pi \epsilon_{0} V_{0} / \lambda}-e^{-2 \pi \epsilon_{0} V_{0} / \lambda}}= a \operatorname{coth}\left(\frac{2 \pi \epsilon_{0} V_{0}}{\lambda}\right) .
R=2 a \frac{e^{2 \pi \epsilon_{0} V_{0} / \lambda}}{e^{4 \pi \epsilon_{0} V_{0} / \lambda}-1}=a \frac{2}{\left(e^{2 \pi \epsilon_{0} V_{0} / \lambda}-e^{-2 \pi \epsilon_{0} V_{0} / \lambda}\right)}=\frac{a}{\sinh \left(\frac{2 \pi \epsilon_{0} V_{0}}{\lambda}\right)}=a \operatorname{csch}\left(\frac{2 \pi \epsilon_{0} V_{0}}{\lambda}\right)
