A constant potential difference of 12 V is maintained between the terminals of a 0.25-μF , paral

A constant potential difference of 12 V is maintained...

A constant potential difference of 12 $\mathrm{V}$ is maintained between the terminals of a $0.25-\mu \mathrm{F}$ , parallel-plate, air capacitor. (a) A sheet of Mylar is inserted between the plates of the capacitor, completely filling the space between the plates. When this is done, how much additional charge flows onto the positive plate of the capacitor (see Table 18.1$) ?$ (b) What is the total induced charge on either face of the Mylar sheet? (c) What effect does the Mylar sheet have on the electric field between the plates? Explain how you can reconcile this with the increase in charge on the plates, which acts to increase the electric field.

Key Concepts

Dielectric Materials

Dielectric materials are insulating substances that, when placed between the plates of a capacitor, increase its capacitance. This increase is due to the material’s ability to become polarized in an electric field, which effectively reduces the electric field strength in the capacitor for a given charge, allowing more charge to be stored without increasing the voltage.

Dielectric Insertion Effect

When a dielectric is inserted into a capacitor that maintains a constant voltage, the overall capacitance increases. This causes additional charge to flow onto the plates from the voltage source in order to balance the new charge storage capacity of the capacitor, even though the voltage remains constant.

Polarization and Induced Charge

Polarization occurs when a dielectric material is subjected to an electric field, causing its molecular dipoles to align with the field. This alignment leads to the formation of induced surface charges on the faces of the dielectric, which oppose the applied field. These induced charges are responsible for modifying the effective electric field within the capacitor and play a key role in the dielectric’s influence on the overall behavior of the capacitor.

Electric Field in a Capacitor

The electric field between the plates of a capacitor is determined by the surface charge density on the plates and the properties of the intervening medium. Although inserting a dielectric under a constant voltage condition increases the charge on the plates, the polarization of the dielectric reduces the effective electric field within the material. This apparent contradiction is resolved by noting that the dielectric's polarization partially cancels the external field, allowing the increased charge to coexist with a net reduced field in the dielectric.

Capacitance

Capacitance is the measure of a capacitor's ability to store charge per unit voltage. It depends on the geometric factors of the capacitor, such as the area of the plates and the distance between them, as well as the properties of the medium between the plates. An increase in capacitance allows the capacitor to store more charge at the same voltage.

Transcript

00:01 So on this problem, we're discussing concerning a piece of mylar in a capacitor and determining how it changes the charge on the capacitor and sort of modeling that effect.

00:15 And so the first thing i want to do as i get set up is to look up the dielectric constant of mylar, which is 3 .1.

00:24 So i'll go ahead and write that kappa is 3 .1.

00:31 And what else are we given in the problem that the voltage is 12 volts? and we are also given that the capacitance is 0 .25 microfarads.

00:55 And with that, we want to figure out how much additional charge flows onto the positive plate of the capacitor.

01:05 So let's figure out how much was on there originally.

01:11 So originally, so c is q over v, so q is equal to cv.

01:16 So q originally is just 0 .25 times 10 to the minus 6.

01:22 Oh, well, maybe i'll go back.

01:25 I got excited about the numbers because i don't have to use a calculator to calculate them.

01:31 So it's cv and then just multiplying those, that's just taking 12 divided by 4.

01:38 So that's 3 times 10 to the minus 6 quolones.

01:43 And then if you insert kappa, you've got a new charge, which is q prime, and that's going to be kappa times the original c, v.

01:55 So we just want to take this whole thing and multiply it by 3 .1.

01:59 Oh, i can even do that in my head, too.

02:01 9 .3.

02:04 Wonderful.

02:05 Excellent.

02:05 And so now we can find the delta q is equal to q prime minus q.

02:16 So that's going to be 6 .3 times 10 to the minus 6 kuloms...

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