Earth can be considered as a spherical capacitor with two plates, where the negative plate is the surface of Earth and the positive plate is the bottom of the ionosphere, which is located at an altitude of approximately 70 kilometres. The potential difference between the Earth’s surface and the ionosphere is about 350000 volts. Calculate the capacitance of this system. Find the total charge on this capacitor. Find the energy stored in this system.
In this three-part problem, we want to solve for the capacitance of the system, which we’ll call 𝐶. We want to know also the charge of the capacitor, which we’ll call 𝑄, and also the energy stored in the system, which we’ll call e [𝑢].
The problem statement tells us we’re to consider Earth as a spherical capacitor, where the Earth’s surface is the inner part of the spherical capacitor and the ionosphere located at distance of 70 kilometres above Earth’s surface is the outer layer of the capacitor. So imagine negative charges on the surface of the Earth and positive charges on the ionosphere layer. We’ve labelled the radius of the Earth 𝑟 sub 𝑒 and the radius from the centre of the Earth to the ionosphere 𝑟 sub 𝑖.
Let’s start out by looking up these values and figuring out what these two radii are. If we look up the radius of the Earth in units of metres, it’s approximately 6.37 times 10 to the sixth. So that’s 𝑟 sub 𝑒. 𝑟 sub 𝑖, we’re told is that plus 70 kilometres. If we write both these terms out in units of metres, when we add them together, we find that 𝑟 sub 𝑖 is 6.44 times 10 to the sixth metres. So that’s the outer radius of our spherical capacitor.
In part one, we want to solve for the capacitance of our capacitor 𝐶. For a spherical capacitor, that value is equal to four 𝜋 times ε nought, the permittivity of free space, divided by one over the inner radius of the spherical capacitor minus one over the outer radius. ε nought is a constant. We’ll assume as exactly 8.85 times 10 to the negative 12 farads per metre.
Using our variables, the capacitance of our capacitor is four 𝜋 ε nought over one divided by 𝑟 sub 𝑒 minus one divided by 𝑟 sub 𝑖. Since we know all the values given in this equation, we can plug in now and solve for 𝐶. When we do and then enter these values on our calculator, we find that 𝐶 is equal to 0.065 farads. That’s the capacitance of this Earth-based spherical capacitor.
Next, we want to solve for the charge 𝑄 on this capacitor. To begin, let’s recall the general equation for capacitance 𝐶. Capacitance 𝐶 is equal to charge divided by voltage. And we can recall from the problem statement that the potential difference that exists across both sides of this spherical capacitor 𝑉 is 350000 volts.
𝐶 equals 𝑄 over 𝑉 implies that 𝑄 equals 𝐶 times 𝑉 or in our case it equals 0.065 farads times 350000 volts, which when we calculate this value is approximately equal to 23000 coulombs or 23 kilo coulombs. That’s the charge on the spherical capacitor.
Now, we’re ready to solve for the energy 𝑢 stored in this system. In general, the energy stored in a capacitor is equal to one-half the charge 𝑄 stored in the capacitor squared divided by its capacitance 𝐶. So 𝑢 equals one-half 𝑄 squared over 𝐶. And we’ve solved for both 𝑄 and 𝐶 in the previous part of this problem. So we can plug those in to solve for 𝑢.
When we do, being careful to use units of coulombs for our charge 𝑄, and enter these values on our calculator, we find that 𝑢 to two significant figures equals 4.0 times 10 to the ninth joules or 4.0 gigajoules. That’s how much energy is stored in this spherical capacitor.