# Video: Analyzing the Properties of a Parallel Plate Capacitor Using Different Dielectrics

A parallel plate capacitor has charge of magnitude 9.00 𝜇C on each plate and capacitance 3.00 𝜇F when there is air between the plates. The plates are separated by 2.00 mm. With the charge on the plates kept constant, a dielectric with 𝜅 = 5.00 is inserted between the plates, completely filling the volume between the plates. What is the potential difference between the plates of the capacitor before the dielectric has been inserted? What is the potential difference between the plates of the capacitor after the dielectric has been inserted? Before the dielectric is inserted, what is the electric field magnitude at the point midway between the plates? After the dielectric is inserted, what is the electric field magnitude at the point midway between the plates?

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### Video Transcript

A parallel plate capacitor has charge of magnitude 9.00 microcoulombs on each plate and capacitance 3.00 microfarads when there is air between the plates. The plates are separated by 2.00 millimeters. With the charge on the plates kept constant, a dielectric with 𝜅 equals 5.00 is inserted between the plates, completely filling the volume between the plates. What is the potential difference between the plates of the capacitor before the dielectric has been inserted? What is the potential difference between the plates of the capacitor after the dielectric has been inserted? Before the dielectric is inserted, what is the electric field magnitude at the point midway between the plates? After the dielectric is inserted, what is the electric field magnitude at the point midway between the plates?

In this four-part problem, we have a dielectric that’s being inserted in between the plates of a parallel plate capacitor. We want to know the potential difference across the plates before and after insertion, what we’ll call 𝑉 sub 𝑏 and 𝑉 sub 𝑎. And we also want to know the electric field magnitude at the point midway between the plates, again both before and after dielectric insertion. We’ll call these values 𝐸 sub 𝑏 and 𝐸 sub 𝑎.

We can start our solution by sketching out the capacitor. In our parallel plate capacitor setup, we’re told that one side has a charge, we’ll call it 𝑄, of positive 9.00 microcoulombs. And that the opposite plate has the opposite charge, negative 𝑄. These equal and opposite charges separated by a gap create an electric field, that we’ll call 𝐸, in between the plates. Along with the charge on each plate, we’re also told the capacitance 𝐶 of this capacitor, 3.00 microfarads. We’re told further that the plates are separated by a distance 𝑑 of 2.00 millimeters. And also that at some time, a dielectric with a constant 𝜅 equals 5.00 is inserted to completely fill the space between the plates.

We want to start off by solving for the potential difference across the plates before the dielectric is inserted. To do that, we can recall that mathematically, capacitance 𝐶 is equal to charge divided by potential difference 𝑉. So 𝐶 equals 𝑄 over 𝑉 sub 𝑏. Rearranging, we see that 𝑉 sub 𝑏 equals 𝑄 divided by 𝐶. And both 𝑄 and 𝐶 are given to us in the problem statement. So we can plug in for those values now.

When we do, being careful to use units of coulombs and farads, we see that this fraction 𝑉 sub 𝑏 is equal to 3.00 volts. That’s the potential difference across the plates before the dielectric is inserted.

Now we move on to solving for the potential difference across the plates 𝑉 sub 𝑎 after the dielectric has been inserted. Recall that the dielectric constant of the dielectric is 5.00. The equation we’ve identified already is a general capacitance equation. There is another equation for capacitance we can recall that speaks specifically to a parallel plate capacitor, like the one we have here. The capacitance of a parallel plate capacitor 𝐶 is equal to 𝜅, the dielectric constant which is one when no material is inserted between the plates, multiplied by 𝜀 naught, the permittivity of free space, times 𝐴, the area of one of the plates, divided by the distance separating the two plates.

With these two equations for capacitance, we can set them equal to one another. Specifically, 𝑄 divided by 𝑉 sub 𝑎 is equal to 𝜅𝜀 naught 𝐴 over 𝑑. May if we rearrange this equation to solve for 𝑉 sub 𝑎, we see it’s equal to 𝑄 times 𝑑 over 𝜅𝜀 naught 𝐴. As we look at this equation, we notice that 𝑄 times 𝑑 over 𝜀 naught 𝐴 is equal to something we already solved for. This expression is equal to 𝑉 sub 𝑏, the potential difference we solved for earlier. So 𝑉 sub 𝑎 equals 𝑉 sub 𝑏 divided by 𝜅. Since we’ve solved for 𝑉 sub 𝑏 and we’re given 𝜅, we can now plug in to solve for 𝑉 sub 𝑎.

When we do and calculate this fraction, we find that, to three significant figures, it’s 0.600 volts. That’s the potential difference across the capacitor plates after the dielectric is inserted.

Now we move on to solving for the electric field midway between the plates before the dielectric is inserted. To do that, we can recall that the electric field within a parallel plate capacitor 𝐸 is equal to the potential difference across the capacitor divided by the distance separating the plates. So 𝐸 sub 𝑏 equals 𝑉 sub 𝑏 over 𝑑, where we solved for 𝑉 sub 𝑏 in part one and we’re given 𝑑 in the problem statement.

When we plug in for these values, being careful to use units of meters in our distance 𝑑, and then calculate 𝐸 sub 𝑏, we find it’s equal to 1500 volts per meter. That’s the electric field magnitude midway between the plates before the dielectric is inserted.

Now we want to solve for the electric field midway between the plates after the dielectric, with dielectric constant 𝜅, is inserted. To do this, we only need to change out 𝑉 sub 𝑏 for 𝑉 sub 𝑎 in our equation which would change 𝐸 sub 𝑏 to 𝐸 sub 𝑎. With both 𝑉 sub 𝑎 and 𝑑 known, we can plug in and solve for 𝐸 sub 𝑎.

When we calculate this fraction, we find that 𝐸 sub 𝑎 equals 300 volts per meter. That’s the electric field midway between the capacitor plates after the dielectric has been inserted.