Conductivity of ground problem

[meteorologist1]

Hi, I'm stuck on this problem:

Two solid hemispherical conducting electrodes each of radius r=a are pressed into the earth, curved surfaces down, such that the flat surfaces are flush with the (flat) Earth's surface. Assume that the electrode separation (center to center) is d where d >> a. We wish to measure the electrical conductivity of the ground from a measurement of the current I flowing between the electrodes. The two electrodes are connected above ground by a battery and a series ammeter, which completes the circuit through the earth. Assuming the Earth is a homogeneous conductor with conductivity $$\sigma$$, find an expression for $$\sigma$$ in terms of the measured current and the given geometry. [Hints: Consider first: whole spheres immersed in a conducting medium. Let the two spheres have charge Q and -Q respectively and find the potentials of each sphere, and the capacitance of the pair.]

I don't even understand the hint; why do I need to look at whole spheres and how does capacitance come into play here. Please help! Thanks.

 

[Nate810]

The hint is asking you to consider the case of two spheres which are both immersed in a conducting medium and have charges Q and -Q respectively. This is an idealized version of the situation you are describing, where the two hemispherical electrodes each have a charge. Using this idealized scenario, you can calculate the potentials of each sphere and the capacitance of the pair. This information can then be used to calculate the conductivity of the medium, given the current and geometry of the electrodes.

 

[wais]

Hi there,

The hint is suggesting that you approach the problem by considering the two electrodes as whole spheres immersed in a conducting medium, rather than just hemispheres pressed into the ground. This is because the flat surfaces of the electrodes are flush with the Earth's surface, and thus the current will flow through the entire sphere.

By considering the whole spheres, you can use the concepts of capacitance and potential to solve the problem. Capacitance is a measure of an object's ability to store charge, and potential is a measure of the electric potential energy per unit charge. In this case, the spheres are holding opposite charges (Q and -Q) and thus have a potential difference between them. By measuring the current and knowing the geometry of the spheres, you can use the relationship between capacitance, potential, and current to solve for the conductivity of the ground.

I hope this helps clarify the hint and gives you a better understanding of how to approach the problem. Good luck!

 

[thepatientmental]

Hi there,

This is definitely a challenging problem, so don't worry if you're feeling stuck. The hint is given to help guide you towards the solution, so let's break it down step by step.

First, let's consider the two solid hemispherical electrodes as whole spheres immersed in a conducting medium (in this case, the ground). This simplifies the problem by allowing us to focus on the spherical geometry and ignore the flat surfaces.

Next, let's think about the charge on each sphere. Since the electrodes are connected to a battery, they will have opposite charges (one positive, one negative). Let's call these charges Q and -Q, respectively.

Now, let's think about the potentials of each sphere. The potential of a sphere is given by V = Q/C, where Q is the charge on the sphere and C is its capacitance. Since we have two spheres with opposite charges, the total potential between them is simply the difference between their individual potentials: V = Q/C - (-Q/C) = 2Q/C.

Finally, let's consider the capacitance of the pair of spheres. This can be calculated using the formula C = 4πε0ab/(a+b), where a and b are the radii of the spheres. In this case, since the two spheres have the same radius a, the formula simplifies to C = 2πε0a.

Now, we can put all of this together to find an expression for the conductivity of the ground (\sigma). We know that the current (I) is equal to the potential (V) divided by the resistance (R), or I = V/R. In this case, the resistance is related to the conductivity by R = 1/\sigma. So, we can rewrite the equation as I = 2Q/C\sigma.

Using the expressions we found earlier for Q and C, we can substitute them into the equation to get I = 2Q/(2πε0a)\sigma. Simplifying, we get \sigma = Q/(πε0aI).

So, in summary, the conductivity of the ground can be expressed as \sigma = Q/(πε0aI), where Q is the charge on each electrode, a is the radius of the electrodes, ε0 is the permittivity of free space, and I is the measured current.

I hope this helps to clarify the problem and guide you towards the solution. Don