- #1
Order
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Homework Statement
Two spherical buoys float half submerged in a calm deep sea. Their radius a is very much smaller than their separation b. Calculate the resistance between them if the resistivity of the water is ρ.
Homework Equations
[tex]R=V/I[/tex]
[tex]V=\int Edl[/tex]
[tex]I=\sigma \int_S \textbf{E} \cdot d \textbf{S}[/tex]
[tex]\rho = \sigma^{-1}[/tex]
The Attempt at a Solution
The field between the bouys is
[tex]E=\frac{Q}{\varepsilon 4 \pi} \Big(\frac{1}{x^2}-\frac{1}{(b-x)^2} \Big) =E_1+E_2[/tex]
I already know how to show that
[tex]V \simeq \frac{Q}{2 \pi \varepsilon_0} \Big( \frac{1}{a}-\frac{1}{b-a} \Big) \simeq \frac{Q}{a 2 \pi \varepsilon_0}[/tex]
Now here comes the tricky part and I am not sure what I am doing but it feels like the current between the two bouys should pass a theoretical Earth before reaching each other. The reason for this is that when the two bouys are a distant apart half of their field lines should still join. Is this reasonable? So reasoning like that
[tex]I= \sigma \int_{S_{between}} \textbf{E} \cdot d \textbf{S}=2 \sigma \int_{earth} \textbf{E}_1 \cdot d \textbf{S} = \sigma \int_{sphere} \textbf{E}_1 \cdot d \textbf{S} = \frac{Q}{ \rho \varepsilon_0} [/tex]
And finally
[tex]R=V/I=\frac{\rho}{2 \pi a}[/tex]
But the correct answer is ##R=\frac{\rho}{\pi a}## so I guess I overestimated the current by a factor 2. (Where did it go wrong? I always tend to miss a factor 2 in electromagnetism.)