A hollow spherical shell carries charge density ρ=(k)/(r^2) in the region a ≤r ≤b (Fig. 2.25 ).

A hollow spherical shell carries charge density ρ=(k)/(r^2)...

Key Concepts

Graphical Representation of Electric Field

Plotting the magnitude of the electric field as a function of distance from the center helps visualize how the field varies in different spatial regions. The graph typically reflects transitions in the behavior of the field, such as having a zero field in regions with no enclosed charge and distinct variations within and outside the charged region, thus providing insight into the underlying physics of the system.

Volume Charge Density Integration

When dealing with continuous charge distributions, the total charge enclosed within a given volume is found by integrating the charge density over that volume. In spherical coordinates, this involves integrating the charge density function with respect to the radius, which is critical to applying Gauss's Law in regions where the charge density is non-zero.

Piecewise Analysis of Regions

In problems with non-uniform charge distributions, the behavior of the electric field can vary in different regions of space. A piecewise analysis involves dividing the space into regions based on whether they lie inside the charged region, within the charge distribution, or outside it. This approach is essential for applying Gauss's Law correctly in each region by accurately determining the enclosed charge.

Gauss's Law

Gauss's Law is a fundamental principle in electromagnetism that relates the electric flux through a closed surface to the charge enclosed by that surface. It is particularly useful for calculating electric fields when the charge distribution has symmetry, as it simplifies the integration process and allows one to relate the field directly to the total enclosed charge.

Spherical Symmetry

Spherical symmetry in a charge distribution implies that the physical properties of the system remain identical in all directions from a central point. This symmetry simplifies the analysis because the electric field at any point depends only on the radial distance, allowing the use of spherical Gaussian surfaces to easily relate the field to the total enclosed charge.

Transcript

00:01 Hi there, so for this problem, we are told that a hollow spherical shell carries charge density row of k divided by the radius to the square in the region between a and b, as is shown in this figure.

00:26 So we need to find the electric field in the three regions.

00:33 Region in this case is when the radius is less than the area so we are inside the spherical shield now in that case because we are inside the sphere we know that inside we know that inside we don't have we are not enclosing any charge we with our gaussian surface.

01:07 So because we know that enclosed charge is equal to zero, then with this we can conclude that the electric field is equal to zero at that point.

01:21 Now for part two, we are told that the radius, where we want to measure the electric field is between a and b.

01:37 So with that, set, we just used gauss law and that states that the close integral between the product of the electric field and the differential area is equal to the magnitude of the electric field times the integral because we can take out this from the integral because the electric field is constant and we will have the integral of the differential in area which is just simply total area, which is 4 times pi times the radius s -square.

02:15 And by gauss law, we know that this should be equal to the enclosed charge divided by epsilon sub -zero.

02:24 And the enclosed charge in this case, we need to use the integral of the charge density times a differential that we need to see what the frame sure are going to use, because we know that this is an spherical shell.

02:42 It is convenient to use spherical coordinates.

02:46 So we're going to have one over epsilon sub -zero times the integral of k divided by the radius square.

02:57 And this times the radius square.

03:00 Let's call it prime to differentiate of the radius art.

03:04 And this sign of theta, the differential in the radius, the differential in theta and the differential in fee.

03:14 Now we know that in here nothing depends, well, and also these councils.

03:19 In here, nothing depends on theta or fee, just the sign of theta.

03:32 So from there we obtain the integral of a solid angle that we know is 4 times pi.

03:38 So we can take out everything that is a constant or out of the integral.

03:43 And this is the integral from a to art of the differential in the radius.

03:51 So with this, we can conclude that at that point, we obtain 4 times pi times the constant k divided by epsilon sub 0 times the radius r minus a.

04:05 So with this we can conclude that the electric field is equal to the constant k divided by epsilon sub zero times the radius r minus a divided by the radius square.

04:22 And this in a radial direction...

Please give Ace some feedback