# Video: Determining the Capacitances Produced by Using Different Dielectrics

An air-filled capacitor is made from two flat parallel plates located 1.0 mm apart. The area of each plate is 8.0 cm². What is the capacitance of the plates? What is the capacitance of the plates if the region between the plates is filled with a material whose dielectric constant is 6.0?

03:31

### Video Transcript

An air-filled capacitor is made from two flat parallel plates located 1.0 millimeters apart. The area of each plate is 8.0 centimeters squared. What is the capacitance of the plates? What is the capacitance of the plates if the region between the plates is filled with a material whose dielectric constant is 6.0?

We can call the distance separating the two plates, 1.0 millimeters, 𝑑 and the area of each one of the plates, 8.0 centimeters squared, 𝐴. In part one, we want to solve for the capacitance of the plates, what we’ll call 𝐶. And in part two, we want to again solve for the capacitance, but this time there is a dielectric that fills the area in between the plates. So we’ll call this capacitance 𝐶 sub 𝑑. And we know that that dielectric constant is 6.0, what we’ll call 𝐾.

Let’s start our solution by drawing a sketch of these plates. We have two parallel plates, each of area 𝐴 separated by a distance 𝑑. We want to solve for the capacitance of these plates. We recall that for parallel plates, the capacitance 𝐶 is equal to the permittivity of free space, 𝜀 naught, times the area of the plates divided by the distance separating them. The constant 𝜀 naught, we’ll treat as exactly 8.85 times 10 to the negative 12th farads per meter. In this equation for capacitance, we’ve been given 𝐴 and 𝑑. And 𝜀 naught is a known constant. So we can now plug in to solve for 𝐶.

When we do plug these values in, we make sure to use units of meters for our distance and meters squared for our area. Entering these values on our calculator, to two significant figures, we find that 𝐶 is 7.1 picofarads. That’s the capacitance of these plates with no dielectric between them.

Now we move on to solving for 𝐶 sub 𝑑, the capacitance of these plates, when a dielectric of dielectric constant 𝜅 is inserted between them. Looking back at our equation for a parallel plate capacitor, 𝜀 naught, the permittivity of free space, assumes that there is no dielectric material in between our two plates, that it’s air. But when we do have a dielectric material separating the two plates, we insert the dielectric constant for that value right before 𝜀 naught in this equation. So this now is the general equation for the capacitance of a parallel plate capacitor. If no material separates the two plates, then 𝐾 is equal to one. And the equation reduces to what we saw earlier. But in this case, 𝐾 is equal to 6.0. And so we want to account for that term in this equation.

When we write out this equation for 𝐶 sub 𝑑, we see that it’s equal to the dielectric constant times 𝐶, the value we solved for previously. Knowing both 𝐾 and 𝐶, we can plug in and solve for 𝐶 sub 𝑑 now. When we calculate this value, we find that, to two significant figures, 𝐶 sub 𝑑 is 42 picofarads. That’s the capacitance of these plates with the dielectric inserted between them.