### 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 π
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 thatβs 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 weβll call π sub π and π sub π. And we also want to know the electric field magnitude at the point midway between the plates, again both before and after dielectric insertion. Weβll call these values πΈ sub π and πΈ sub π.

We can start our solution by sketching out the capacitor. In our parallel plate capacitor setup, weβre told that one side has a charge, weβll call it π, of positive 9.00 microcoulombs. And that the opposite plate has the opposite charge, negative π. These equal and opposite charges separated by a gap create an electric field, that weβll call πΈ, in between the plates. Along with the charge on each plate, weβre also told the capacitance πΆ of this capacitor, 3.00 microfarads. Weβre told further that the plates are separated by a distance π of 2.00 millimeters. And also that at some time, a dielectric with a constant π
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 πΆ is equal to charge divided by potential difference π. So πΆ equals π over π sub π. Rearranging, we see that π sub π equals π divided by πΆ. And both π and πΆ 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 π sub π is equal to 3.00 volts. Thatβs the potential difference across the plates before the dielectric is inserted.

Now we move on to solving for the potential difference across the plates π sub π after the dielectric has been inserted. Recall that the dielectric constant of the dielectric is 5.00. The equation weβve 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 πΆ is equal to π
, the dielectric constant which is one when no material is inserted between the plates, multiplied by π naught, the permittivity of free space, times π΄, 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, π divided by π sub π is equal to π
π naught π΄ over π. May if we rearrange this equation to solve for π sub π, we see itβs equal to π times π over π
π naught π΄. As we look at this equation, we notice that π times π over π naught π΄ is equal to something we already solved for. This expression is equal to π sub π, the potential difference we solved for earlier. So π sub π equals π sub π divided by π
. Since weβve solved for π sub π and weβre given π
, we can now plug in to solve for π sub π.

When we do and calculate this fraction, we find that, to three significant figures, itβs 0.600 volts. Thatβs 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 πΈ is equal to the potential difference across the capacitor divided by the distance separating the plates. So πΈ sub π equals π sub π over π, where we solved for π sub π in part one and weβre given π in the problem statement.

When we plug in for these values, being careful to use units of meters in our distance π, and then calculate πΈ sub π, we find itβs equal to 1500 volts per meter. Thatβs 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 π
, is inserted. To do this, we only need to change out π sub π for π sub π in our equation which would change πΈ sub π to πΈ sub π. With both π sub π and π known, we can plug in and solve for πΈ sub π.

When we calculate this fraction, we find that πΈ sub π equals 300 volts per meter. Thatβs the electric field midway between the capacitor plates after the dielectric has been inserted.