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Brian-san
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Homework Statement
Two long cylindrical conductors (wires) of radii a1 and a2 respectively are separated by a distance d >> a1, a2. Find the capacitance per unit length for the system. If d=0.5cm and a1=a2=a, what must be the diameter of the wires to give a capacitance per unit length of 9x10^-3? (CGS units)
Homework Equations
Gauss' Law
Q=CV
[tex]\vec{E}=-\nabla\Phi[/tex]
The Attempt at a Solution
Assuming wire one has linear charge density λ and wire 2 has linear charge density -λ, then by Gauss' Law:
[tex]\vec{E_1}=\frac{2\lambda}{r_1}\vex{\hat{r_1}}, \vec{E_2}=\frac{-2\lambda}{r_2}\vex{\hat{r_2}}[/tex]
Integrating with respect to the radial coordinates gives:
[tex]\Phi_1=2\lambda ln(r_1), \Phi_2=-2\lambda ln(r_2)[/tex]
So at any point, the potential is a superposition of these two,
[tex]\Phi=2\lambda ln(\frac{r_1}{r_2})[/tex]
Going from a1 to d-a2, gives the potential difference (I assume this is safe, since d is much greater than both wire radii)
[tex]V=2\lambda ln(\frac{d-a_2}{a_2})-2\lambda ln(\frac{a_1}{d-a_1})=2\lambda ln(\frac{(d-a_1)(d-a_2)}{a_1a_2})[/tex]
Finally,
[tex]C=\frac{1}{2ln(\frac{(d-a_1)(d-a_2)}{a_1a_2})}[/tex]
For the above case in the problem, I get
[tex]a=\frac{d}{1+e^{\frac{1}{4C}}}[/tex]
and a solution of a=4.318x10^-13 cm
Does everything look about right? I've never dealt with capacitance in this manner before, usually it was stuff like parallel plates, concentric cylindrical/spherical shells, or circuits. Thanks.