- #1

rachmaninoff

I've been going back over Griffiths' E&M this summer, and this question (2.47, electrostatics) is making me stupid:

There are two 'infinite' wires, parallel to each other, with linear charge densities [itex]+ \lambda [/itex] and [itex]- \lambda [/itex];

and the question asks to show that the equipotential surfaces are "circular cylinders".

I don't see how any cylinders could be equipotentials, and in doing the algebra,

(lines parallel to the x axis, + charged one going through (0,d,0), - charged one going through (0,-d,0)

[tex]V(x,y,z)= -\frac{\lambda}{2 \pi \epsilon _0} \left[ \log \sqrt{ (y-d)^2 + z^2 } - \log \sqrt{ (y+d)^2 + z^2 } \right] [/tex]

[tex]\begin{align*}V(r)=V_0 &\Longrightarrow \frac{(y-d)^2+z^2}{(y+d)^2+z^2}=e^{-\frac{4 \pi \epsilon_0}{\lambda} V_0}=C>0 \mbox{ (absolute value signs go away, both the top and bottom are everywhere > 0) } \\

&\Longrightarrow (y-d)^2+z^2 = C\left( (y+d)^2+z^2 \right) \\ &\Longrightarrow y^2+z^2+d^2-2yd=C\left(y^2+z^2+d^2+2yd \right) \\

&\Longrightarrow (1-C)y^2+(1-C)z^2=-(1-C)d^2+4Cyd \end{align}[/tex]

or,

[tex]\begin{align*}r^2&=-d^2+\left( \frac{4C}{1-C}\right) yd, (C > 0) \\

&= d(ky-d), k \in (-\infty,4) \cup (0, + \infty) \end{align}[/tex]

Which seems to make physical sense - the equipotentials are a family of curves which include y=0 (where V=0!) and curves that look sort of like hyperbolas (all extending to infinity). There are no 'cylindrical' solutions.

Where am I going wrong?

There are two 'infinite' wires, parallel to each other, with linear charge densities [itex]+ \lambda [/itex] and [itex]- \lambda [/itex];

and the question asks to show that the equipotential surfaces are "circular cylinders".

I don't see how any cylinders could be equipotentials, and in doing the algebra,

(lines parallel to the x axis, + charged one going through (0,d,0), - charged one going through (0,-d,0)

[tex]V(x,y,z)= -\frac{\lambda}{2 \pi \epsilon _0} \left[ \log \sqrt{ (y-d)^2 + z^2 } - \log \sqrt{ (y+d)^2 + z^2 } \right] [/tex]

[tex]\begin{align*}V(r)=V_0 &\Longrightarrow \frac{(y-d)^2+z^2}{(y+d)^2+z^2}=e^{-\frac{4 \pi \epsilon_0}{\lambda} V_0}=C>0 \mbox{ (absolute value signs go away, both the top and bottom are everywhere > 0) } \\

&\Longrightarrow (y-d)^2+z^2 = C\left( (y+d)^2+z^2 \right) \\ &\Longrightarrow y^2+z^2+d^2-2yd=C\left(y^2+z^2+d^2+2yd \right) \\

&\Longrightarrow (1-C)y^2+(1-C)z^2=-(1-C)d^2+4Cyd \end{align}[/tex]

or,

[tex]\begin{align*}r^2&=-d^2+\left( \frac{4C}{1-C}\right) yd, (C > 0) \\

&= d(ky-d), k \in (-\infty,4) \cup (0, + \infty) \end{align}[/tex]

Which seems to make physical sense - the equipotentials are a family of curves which include y=0 (where V=0!) and curves that look sort of like hyperbolas (all extending to infinity). There are no 'cylindrical' solutions.

Where am I going wrong?

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