Immersing a copper sphere into a conducting liquid that is itself in a conducting bath [closed]

Question

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.

Visualising

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

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "Homework-like questions should ask about a specific physics concept and show some effort to work through the problem. We want our questions to be useful to the broader community, and to future users. See our meta site for more guidance on how to edit your question to make it better" – John Rennie, CuriousOne, ACuriousMind, Gert, Community
If this question can be reworded to fit the rules in the help center, please edit the question.

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

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.