A small copper sphere of radius a is half immersed at the surface of a poorly conducting liquid of resistivity $\rho$. The liquid is contained in a spherical copper bath of radius b, where b > a and where the bath is concentric with the sphere. Determine the electrical resistance between the sphere and the walls of the bath.


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My attempt

OK, so this is my thought process in attempting this problem: A current will flow from the copper sphere to the walls of the bath via the conducting liquid, causing a potential difference due to the resistivity of the conducting liquid. The liquid will behave as a dielectric inside a spherical capacitor, so we can use the capacitance to compute the potential difference between the plates, and hence the resistance.

The capacitance will be given by $$C=\epsilon_rC_0,$$ where $C_0$ is the capacitance of a spherical plate capacitor with no dielectric, so we get $$C=\epsilon_r\frac{4\pi\epsilon_0}{\frac{1}{a}-\frac{1}{b}}=\frac{4\pi\epsilon ab}{b-a}.$$ Then the potential difference between the plates will, by definition of capacitance, be $$\Delta V=\frac{Q(b-a)}{4\pi\epsilon ab},$$ where $Q$ is the total charge on the plates. Then the resistance is $$R={V\over I}=\frac{Q(b-a)}{4\pi\epsilon abI}.$$ I have issues with this solution. The first is that I am uncomfortable with having an arbitrary charge and current in the final answer. Secondly, I have not considered the fact that only half of the inner sphere is immersed. Thirdly, I have made no use of the resistivity mentioned in the question. Another thought is maybe that we need to use $$\rho=\frac{RL}{A},$$ where $L=b-a$ if we consider a coordinate system with the origin at the centre of the spheres, and $A$ increases with distance from the smaller sphere. But this then confuses me... Can anyone help constructively? This was a previous exam question on a paper I will be sitting in a few weeks time.


closed as off-topic by John Rennie, CuriousOne, ACuriousMind, Gert, user36790 May 1 '16 at 17:27

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  • $\begingroup$ Why closed? I quite clearly show a good effort to work it out and it is the topic of electrical circuits in physical situations... $\endgroup$ – ODP May 6 '16 at 20:14

Forget anything about a capacitor and just consider the resistance of the conducting liquid.

Think of the liquid as made up of thin $dr$ concentric shells of radius $r$ and find the resistance of a shell in terms of the resistivity, radius and thickness.
Then do the integration to find the resistance of all the liquid.


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