A long coaxial cable (Fig. 2.26 ) carries a uniform volume charge density ρon the inner cylinder

A long coaxial cable (Fig. 2.26 ) carries a uniform volume...

A long coaxial cable (Fig. 2.26 ) carries a uniform volume charge density $\rho$ on the inner cylinder (radius $a$ ), and a uniform surface charge density on the outer cylindrical shell (radius $b$ ). This surface charge is negative and of just the right magnitude so that the cable as a whole is electrically neutral. Find the electric field in each of the three regions: (i) inside the inner cylinder $(s < a),$ (ii) between the cylinders $(a < s < b),$ (iii) outside the cable $(s > b) .$ Plot $|\mathbf{E}|$ as a function of $s$.

Key Concepts

Cylindrical Symmetry

Many problems in electrostatics, such as those involving long coaxial cables, exhibit cylindrical symmetry. This symmetry implies that the electric field depends only on the radial distance from the axis, greatly simplifying the mathematical analysis of the field and the choice of Gaussian surfaces.

Charge Distributions

Understanding the difference between volume charge density and surface charge density is fundamental in electrostatics. Volume charge density describes how charge is distributed throughout a material's volume, while surface charge density describes charge confined to a surface. This distinction is crucial when calculating the total charge in regions with different types of charge distributions.

Gauss's Law

Gauss's Law relates the net electric flux through any closed surface to the charge enclosed by that surface. This law is instrumental in electrostatics because it simplifies the calculation of electric fields for systems with high symmetry by allowing the use of appropriately chosen Gaussian surfaces.

Region-wise Analysis of Electric Fields

In problems where different regions (for instance, inside the inner conductor, between the conductors, and outside the system) have different charge configurations, it is essential to analyze the electric field in each region separately. This approach involves applying Gauss's Law to Gaussian surfaces that lie entirely within each region, ensuring that only the appropriate charges contribute to the net flux.

Transcript

00:01 Hi there, so for this problem, we have a long coaxial cable that is shown in this figure that carries an uniform volume charge density row on the inner cylinder of radius a and an uniform charge density on the outer cylindrical shelf of radius b.

00:20 Now, we are told that this surface charge is negative and of the surface.

00:30 Just the right magnitude so that the cable as a hold is electrical neutral.

00:36 So we need to find the electric field in each of the three regions.

00:41 Now the first region, region one, is inside the inner cylinder.

00:47 This is when adds, the radius where we are measuring the electric field, is less than a.

00:55 So in this case, we have this situation.

01:01 We have this one in here.

01:03 So because we have a cylindrical wire, we just simply choose a cylindrical surface, a cylindrical gaussian surface.

01:14 So with that, we just simply apply gauss law.

01:23 So in this case, it's going to be the close integral of the part between the electric field and the differential area.

01:32 So because the electric field is a constant, we can take it out of the integral, and then the integral is going to be the the integral of the differential area that we know is just simply the area.

01:42 So we will have the electric field times the area of the gaussian surface.

01:49 That because it is a cylinder, we will have two times pi times x times the length of the gaussian surface.

01:58 So this is equal by the gauss law as the enclosed charge divided by epsilon sub -zero.

02:07 Now, in this case, the enclosed charge is the density, the charge density row times pi times s squared times l, because that is the volume of, in this case.

02:27 So with that said, if we solve for the electric field, we will obtain that that is the charge density row times s divided by 2.

02:39 Times epsilon sub -0.

02:42 And this in the radial direction s.

02:47 Now for part two of this problem, we are told now that the, we need to find the electrode field between the cylinders a and b.

03:07 So the region in this case is when x is between a and b.

03:12 Now in this case, we choose the following gaussian surface, and that is this one right here...

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