A parallel-plate air capacitor has a capacitance of 500.0 pF and a charge of magnitude 0.200μC o

A parallel-plate air capacitor has a capacitance of 500.0 pF...

Key Concepts

Capacitance

Capacitance is a measure of a capacitor’s ability to store electrical energy per unit potential difference between its conductors. It depends on the geometry of the conductors and the dielectric material between them, and is defined by the relation Q = C × V, where Q is the stored charge and V is the potential difference.

Parallel-Plate Capacitor

A parallel-plate capacitor consists of two conductive plates separated by a small distance. Its capacitance is determined by the area of the plates, the separation distance, and the permittivity of the dielectric medium between them. This configuration is idealized for calculating the electric field, potential difference, and stored charge.

Relationship between Charge and Potential Difference

This concept involves the formula V = Q / C, which directly relates the stored charge in the capacitor to the potential difference across its plates. It is a fundamental principle used to determine one quantity when the other two (charge and capacitance) are known.

Capacitance and Geometric Factors

The capacitance of a parallel-plate capacitor is calculated using the formula C = ??A/d, where A is the area of one of the plates, d is the separation between the plates, and ?? is the permittivity of free space. This relationship shows how the physical dimensions of the capacitor directly affect its ability to store charge.

Electric Field in a Capacitor

The electric field in a parallel-plate capacitor is uniform and is related to the potential difference and plate separation by the equation E = V/d. This represents the strength of the field between the plates and is essential for understanding the behavior of charges in the capacitor.

Surface Charge Density

Surface charge density, denoted as ?, is the amount of charge per unit area on a conductor. In the context of a capacitor, it is calculated using the formula ? = Q/A. It is a key concept in relating the charge on the capacitor plates to the resulting electric field between them.

Transcript

00:01 So in this problem, we're told that we have a parallel plate capacitor, and then it has a capacitance of 500 .0 pico -farrad, so that's times 10 to the minus 12 ferad.

00:20 And the charge is 0 .20.

00:25 Zero.

00:29 What was it? microculomes.

00:33 Microculeomes and their distance is 0 .6 millimeters apart.

00:53 So i'm going to convert that to meters by multiplying by 10 of the minus 3.

00:57 And then our goal is to get for a, the potential difference.

01:01 So v equals, so c is q over v, so v is q over c.

01:11 So i'm going to divide this 500 by, or excuse me, this point.

01:18 I'm just going to emphasize if that's a point.

01:21 And then put a zero here maybe to remind myself.

01:25 So 0 .2 times 10 to the minus 6 divided by the 500 times 10 to the minus 12.

01:39 So i got 400 when i did that.

01:43 So i guess i could have almost done that one in my head.

01:47 400.

01:48 I don't know what happened when i wrote.

01:52 I tried to write my v.

01:53 So 4.

01:54 Hundred volts for b what's the area of each plate so c is equal to epsilon not a over oops not times d but divided by d therefore the area is is c d divided by epsilon not so so if i plug c times d divided by epsilon not into a calculator, so that's 500.

02:44 And the distance is 0 .6 times 10 to the minus 3.

02:53 And then i'm dividing that by 8 .85 times 10 to the minus 12.

02:59 I get 0 .339 for the area.

03:06 Was it 2 zeros or 1? 1 .30 -339 meters squared.

03:21 And cool...

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