The plates of a parallel-plate capacitor are 2.50 mm apart, and each carries a charge of magnitu

The plates of a parallel-plate capacitor are 2.50 mm apart,...

The plates of a parallel-plate capacitor are 2.50 $\mathrm{mm}$ apart, and each carries a charge of magnitude 80.0 $\mathrm{nC}$ . The plates are in vacuum. The electric field between the plates has a magnitude of $4.00 \times 10^{6} \mathrm{V} / \mathrm{m}$ . (a) What is the potential difference between the plates? (b) What is the area of each plate? (c) What is the capacitance?

Key Concepts

Relationship between Electric Field and Potential Difference

In a parallel-plate configuration, the electric field (E) is uniform and directly related to the potential difference (V) between the plates by the equation V = E × d, where d is the separation between the plates. This concept is central to understanding how a constant field produces a voltage difference over a particular distance.

Parallel-Plate Capacitor Capacitance

The capacitance (C) of a parallel-plate capacitor is determined by its geometric factors and the properties of the medium between the plates. The formula C = ??A/d shows that capacitance increases with the plate area (A) and decreases with the separation (d) between the plates, with ?? being the permittivity of free space. This concept highlights how physical dimensions affect a capacitor’s ability to store electrical charge.

Charge-Voltage Relationship in Capacitors

The relationship Q = CV connects the stored charge (Q) to the potential difference (V) across the capacitor and its capacitance (C). This fundamental principle in capacitors explains how the ability to hold charge is influenced by both the voltage applied and the inherent capacitance of the system.

Permittivity of Free Space

The permittivity of free space (??) is a constant that quantifies the ability of a vacuum to permit electric field lines. It is a critical parameter in the equations that describe the behavior of electric fields and capacitance in a vacuum, emphasizing its role in governing electromagnetic interactions.

Transcript

00:01 In this question, we have another parallel plate capacitor, and the capacitor has a distance d equals 2 .5 millimeter apart, and then each of them carry a charge of 80 .0 .cuncule.

00:21 This is negative 80 .0 .9cool.

00:24 So the plates are in vacuum, and we know the electric field between the two plates has a magnitude of e equals 4 .00 times 10 to the third folds per meter.

00:45 And these are all the numbers we know.

00:49 We want to know a, what is the potential difference between the plates? so in order to find the potential difference, we want to know a potential difference can be created by an electric field across a distance.

01:02 So we have v across a surface equals a strength of the electric field between the two points times the distance between them.

01:10 So we have e equals 4 .00 times 10 to the 6, 6 volts per meter and times 2 .5 times 10 to the negative 3 meter.

01:33 And this gives us a total of 1 .0 times 10 to the 4 volts.

01:42 Okay, so there is that.

01:45 And part b asks us, what is the area of each play? so how do we do that? in order to do that, we need to utilize the equation that is a book right before equation 18 .16.

01:59 So in the book, it tells us in chapter 17, that is in the previous chapter, we learned through goss's law.

02:05 We know that the electric field equals the charge density divided by epsilon.

02:12 Not...

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