Video: Capacitance of a Spherical Capacitor

A capacitor is made from two concentric spheres: one with radius 8.00 cm and the other with radius 7.00 cm, with vacuum between them. What is the capacitance of this set of conductors? If the region between the conductors is filled with a material whose dielectric constant is 5.00, what is the capacitance of the system?

04:30

Video Transcript

A capacitor is made from two concentric spheres: one with radius 8.00 centimeters and the other with radius 7.00 centimeters, with vacuum between them. What is the capacitance of this set of conductors? If the region between the conductors is filled with a material whose dielectric constant is 5.00, what is the capacitance of the system?

To start out, let’s look at the first of these two questions. What is the capacitance of this set of conductors, where there is vacuum in between them? First, we can sketch out what these two concentric spheres might look like. In this scenario, we’re told that we have two concentric spheres. We’ll call the radius of the inner concentric sphere π‘Ÿ sub 𝑖. And the radius of the outer concentric sphere we’ll call π‘Ÿ sub π‘œ. And in units of meters, π‘Ÿ sub 𝑖 is 0.07. And π‘Ÿ sub π‘œ is 0.08 meters.

Along with that, we’re told that these two spheres are separated by vacuum. So that means that all the space between π‘Ÿ sub 𝑖 and π‘Ÿ sub π‘œ, this spherical shell of space, is vacuum. There’s nothing there.

We’re asked to solve for the capacitance of these two conductors. And to do that, we can recall the relationship for the capacitance of a spherical set of conductors. That capacitance is equal to four πœ‹ times πœ– naught, the permittivity of free space divided by one over the inner radius minus one over the outer radius. Here, the inner radius is the radius of the smaller conducting sphere. And the outer radius is the radius of the larger one.

In this expression though, notice that we’ve used the permittivity of free space, meaning we’re assuming that vacuum separates the conductors. In this particular instance, that’s correct. And so we’ll use this exact equation to solve for the capacitance of these conductors. We already know π‘Ÿ sub 𝑖 and π‘Ÿ sub π‘œ. And πœ– naught is a constant which, when we look up its value, we find it to be 8.85 times 10 to the negative 12th farads per meter.

So we’re now ready to apply this equation for capacitance 𝐢. With all the numbers plugged in, we check and make sure that all these values are in standard units, that is, meters for length and so on. We find they are. So we’re ready to enter this expression on our calculator. We find that, to three significant figures, the capacitance of this set of conductors is 62.3 picofarads or 62.3 times 10 to the negative 12th farads. That answers the first question. Now let’s turn to the second one.

In part two, something changes about our setup. We’re no longer calculating the capacitance between a set of conductors separated by vacuum. Now in this part, we filled that previously empty space with a material called a dielectric material. The purpose of a dielectric material inserted between the two sides of a capacitor is to enable the capacitor to store more charge on it, that is, overall to increase the capacitance of the capacitor. The particular dielectric material inserted here has a dielectric constant that we’ll refer to as πœ…. And it’s given as 5.00, no units.

Now πœ…, the dielectric constant, interacts with the permittivity of free space πœ– naught to give rise to a new permittivity of this particular material. This permittivity known as a relative permittivity symbolised πœ– sub π‘Ÿ is equal to the product of πœ…, the dielectric constant, times πœ– naught, the permittivity of free space, that is, for a vacuum. And this relative permittivity, πœ– sub π‘Ÿ, feeds back into our equation for capacitance. It replaces πœ– naught when the plates of our capacitor are not separated by vacuum.

So let’s do this. Let’s call the capacitance of this set of conductors which now have a dielectric in between them. We’ll name this 𝐢 sub 𝑑. And 𝐢 sub 𝑑 is equal to four πœ‹ times πœ… times πœ– naught, divided by one over π‘Ÿ sub 𝑖 minus one over π‘Ÿ sub π‘œ. So to calculate this capacitance, 𝐢 sub 𝑑, we’ll plug in the same numbers as before except with the addition of our value for πœ…, 5.00.

When we look at this expression, we see it’s identical to our previous expression for capacitance except for πœ…. So we expect an answer to be five times greater than our previous answer. And indeed, when we calculate 𝐢 sub 𝑑, we find a result which rounds to 311 picofarads. So by inserting this dielectric to replace vacuum between the set of conductors, we’ve increased the capacitance of our capacitor.

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