Consider Mars to be a spherical conductor of radius 3400 kilometers. What is its capacitance?
Looking at this question, we see it describes Mars as a spherical conductor. That is, the planet is modeled as a solid sphere of conducting material with radius, we’ve called 𝑟, 3.400 times 10 to the sixth meters. This solid sphere is similar to a spherical capacitor of concentric shells.
We can recall that if we have one conducting charge-carrying sphere of radius 𝑟 sub 𝑖, separated by an air gap from another larger spherical shell which conducts and holds charge of radius 𝑟 sub 𝑜, then this object overall forms a spherical capacitor. It separates positive and negative charges one from another on conducting surfaces.
Now the capacitance of this setup takes on a special form. It’s equal to four 𝜋 times the permittivity of free space 𝜖 naught divided by one over the inner radius minus one over the outer radius. Recall that one type of charge rests on the inner radius sphere. And the other type of charge rests on the outer radius sphere. This mathematical relationship is great for a setup like this.
But what about in our case, where all we have is a solid conducting sphere? That is, it’s as though we only have the inner sphere, marked here with a radius 𝑟 sub 𝑖. And since we’ll edit this equation, let’s make a working copy over here. As we said in our particular application, it’s as though we only have the inner solid sphere, labeled 𝑟 sub 𝑖 in our diagram.
Effectively then, as far as this equation goes, 𝑟 sub 𝑜, the radius of the outer sphere, is infinitely large. We know that one divided by ∞ is effectively zero. So our capacitance equation for our solid conducting sphere simplifies to four 𝜋𝜖 naught divided by one over 𝑟 sub 𝑖, or more simply four 𝜋 times 𝜖 naught times 𝑟 sub 𝑖.
Now in our case, 𝑟 sub 𝑖, the radius of our inner conducting sphere, we’ve called simply 𝑟. That’s the radius of this planet Mars we’re considering. Since we know that value, now all we need to do is recall the value for 𝜖 naught, the permittivity of free space, so we can plug in and solve for 𝑐.
This constant is approximately equal to 8.85 times 10 to the negative12th farads per meter. When we plug in this value for 𝜖 naught as well as our radius for Mars, notice that the units of meters cancel out. And we’ll be left with the units of farads, the units of capacitance.
To two significant figures, our answer comes out to be 0.38 millifarads. That’s the capacitance of the planet Mars treating it as a spherical conductor.