ILW Two long, charged, thin-walled, concentric cylindrical...

ILW Two long, charged, thin-walled, concentric cylindrical shells have radii of 3.0 and 6.0 $\mathrm{cm} .$ The charge per unit length is $5.0 \times 10^{-6} \mathrm{Clm}$ on the inner shell and $-7.0 \times 10^{-6} \mathrm{Clm}$ on the outer shell. What are the (a) magnitude $E$ and (b) direction (radially inward or outward) of the electric field at radial distance $r=4.0 \mathrm{cm} ?$ What are $(\mathrm{c}) E$ and (d) the direction at $r=8.0 \mathrm{cm} ?$

ILW Two long, charged,
thin-walled, concentric cylindrical shells have radii of 3.0 and
6.0 $\mathrm{cm} .$ The charge per unit length is $5.0 \times 10^{-6} \mathrm{Clm}$ on the inner
shell and $-7.0 \times 10^{-6} \mathrm{Clm}$ on the outer shell. What are the (a)
magnitude $E$ and (b) direction (radially inward or outward) of the
electric field at radial distance $r=4.0 \mathrm{cm} ?$ What are $(\mathrm{c}) E$ and
(d) the direction at $r=8.0 \mathrm{cm} ?$

Key Concepts

Electric Field Direction

The direction of the electric field is influenced by the sign of the enclosed charge. A positive charge produces an electric field that radiates outward, while a negative charge creates an inwardly directed field. Understanding this principle is important for correctly interpreting the vector nature of the electric field in symmetric charge distributions.

Enclosed Charge

The concept of enclosed charge is critical when applying Gauss's law as it is the net amount of charge contained within the Gaussian surface. Accurately determining the enclosed charge is necessary for computing the resulting electric field, especially when dealing with multiple charge distributions with different signs.

Cylindrical Symmetry

Cylindrical symmetry implies that the system's properties are invariant under rotations about a central axis and translations along that axis. This symmetry enables the assumption that the electric field depends solely on the radial distance from the axis, which simplifies the calculations and understanding of the electric field distribution.

Gaussian Surface

A Gaussian surface is an imaginary closed surface chosen to match the symmetry of a charge distribution, thereby simplifying the electric flux integral. In cylindrical symmetry, a coaxial cylindrical surface is typically used so that the magnitude of the electric field remains constant over the surface, making it easier to relate the enclosed charge to the field using Gauss’s law.

Gauss's Law

Gauss's law is a fundamental principle in electromagnetism stating that the net electric flux through a closed surface is directly proportional to the total electric charge enclosed within that surface. In problems with high symmetry, such as cylindrical distributions, Gauss's law provides an efficient method to calculate the electric field by simplifying the integration required over the field's distribution.

Transcript

00:02 Okay, so in this problem you're given, basically there's two shells, an inner shell and an outer shell.

00:12 And then they have radiuses r1 and r2, and then you want to find the electric field here, somewhere in between, four centimeters, and then on the outside.

00:28 And you're also given the linear charge density.

00:31 Oops, not the total charge.

00:38 There's like a delay in the eraser.

00:40 I guess i just try to make less mistakes.

00:45 We'll see.

00:47 Oh, not sigma.

00:49 Why did i think it was sigma? thinking about making mistakes can make you make mistakes.

00:57 Okay.

00:58 So we're actually given lambda inner and lambda outer.

01:06 Great.

01:07 So basically to find the electric field within here, it only depends on what's going on inside that comes out from gauss's law.

01:17 I explained it in more detail for problem 29 in this chapter if you wanted to look at that video.

01:24 But i'm going to just shorten up and say we're going to completely ignore the outer surface and only focus on the inner surface.

01:31 And again, you can make a gauss's law argument for that.

01:33 And i'm sure it's also written up nicely somewhere.

01:38 So, yeah, so it'd be, if we solve it through gauss's law, making our gaussian surface through this radius.

01:49 So the first one we're considering is, i think, i believe it's four centimeters.

01:57 Yeah, so if we do that, then we can kind of draw this gaussian surface here.

02:05 And then it really becomes the same physics problem of finding the electric field due to an infinitely long wire.

02:13 And then i'll just quote the result from that, which is the linear charge density of the inner, divided by 2 pi r, the position that you're at.

02:23 So it doesn't depend actually on the radius at all of r1.

02:28 It doesn't depend on r1.

02:31 Divided by epsilon knot.

02:33 Now is a good time to write down the given.

02:35 So lambda inner is 5 times 10 to the minus 6 coulomes per meter.

02:45 Lambda outer is minus 7 .0.

02:52 Oh yeah, that was 0...

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