A capacitor is constructed from two square, metallic plates...

A capacitor is constructed from two square, metallic plates of sides l and separation d. Charges +Q and -Q are placed on the plates, and the power supply is then removed. A material of dielectric constant ? is inserted a distance x into the capacitor as shown in the figure below. Assume d is much smaller than x. Suggestion: The system can be considered as two capacitors connected in parallel. (Use the following as necessary: ?0, ?, l, Q, d, and x.) (a) Find the equivalent capacitance of the device. C_eq = (b) Calculate the energy stored in the capacitor. U = (c) Find the direction and magnitude of the force exerted by the plates on the dielectric. magnitude: direction: (d) Obtain a numerical value for the force when x = l/2, assuming l = 5.20 cm, d = 1.90 mm, the dielectric is glass (? = 4.50), and the capacitor was charged to 2.20 × 10^3 V before the dielectric was inserted. magnitude: (?N) direction:

Transcript

00:01 All right, so we have two parallel plate capacitors in the shape of a square, and they have a side length of l, and they're separated by a distance d, and a dielectric is inserted a distance x into the capacitors like this.

00:19 Let me write this as l.

00:21 And we want to know a couple of things.

00:23 What is the equivalent capacitance of this? what is the energy stored in the capacitor, and then what is the magnitude and direction? and of course, one other thing is, you know, they're charged up to an amount plus q and minus q on the plates like this.

00:39 And so we can treat these as basically two capacitors in parallel.

00:44 One is sort of the capacitor without the dielectric, and then one is the capacitor with the dielectric partially wedged in between.

00:51 So the formula for capacitance is like epsilon knot a over d for a parallel plate capacitor.

00:59 So if we have two of these in parallel, we just add them together.

01:03 So for the first, for the capacitor without the dielectric, we have epsilon knot over d, and the area of that whole plate is just l squared.

01:17 So that's sort of our first capacitor.

01:18 And we're going to add to that the capacitance of the other capacitor.

01:21 So i'm actually going to include another term here.

01:24 So if we wedge this dielectric in a distance x, then basically the area of that, section is x times l and it has a dielectric constant of kappa so you know the dielectric of like a or i mean the capacitance of a dielectric uh capacitors like kappa minus one times the area so this is basically what we'll have um and so that's our formula and then the energy stored in the capacitor so there are you know multiple ways of doing this um but you could write this as one -half c delta v squared and delta v if you remember it's you know the voltage is the charge divided by the capacitance so we could write this as one half q squared over c and so if we use our formula we should have something like one half q squared times d over epsilon not and then we'll have l squared plus x times l times kappa minus one so this is our energy.

02:32 And then the force between the plates we can just write as like the negative derivative of the potential energy with respect to x in this case.

02:40 Or really like you can write it as the with respect to, well, yeah, let's do it with respect to x because that's the sort of parameter that we're changing...

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