A spherical shell has inner radius Rin and outer radius Rout...

Key Concepts

Gauss's Law

Gauss's law relates the net electric flux through any closed surface to the charge enclosed within that surface. In problems with high symmetry, such as spherical symmetry, it simplifies the calculation of the electric field by allowing the use of a Gaussian surface on which the field has a constant magnitude, leading to straightforward integration.

Spherical Symmetry

Spherical symmetry implies that a system’s properties are invariant under any rotation about its center. This symmetry means that the electric field depends only on the radial distance from the center, greatly simplifying the analysis of electric fields in spherically distributed charge systems.

Uniform Charge Distribution

A uniform charge distribution indicates that the charge is spread evenly throughout the region, resulting in a constant charge density. This concept is essential when determining the amount of charge enclosed within a specific volume and is a key factor in calculating the electric field in problems involving spread-out charge regions.

Calculation of Enclosed Charge

In symmetric charge distributions, calculating the enclosed charge by a given Gaussian surface can often be done by integrating the constant charge density over the volume enclosed or by relating the volume of the charge distribution to the total charge. This process is crucial when applying Gauss's law to find the electric field in different regions.

Electrostatic Boundary Conditions

Electrostatic boundary conditions govern the behavior of the electric field at interfaces where the material or the charge distribution changes. These conditions ensure that the field solutions are physically consistent, showing continuity or appropriate discontinuities at boundaries, which is critical when matching piecewise solutions in different regions.

Graphical Analysis of Fields

Graphical analysis of the electric field as a function of distance offers a visual representation of how the field varies across distinct regions of space. Such plots help in understanding the transitions in field behavior, especially near boundaries where charge distributions change, thus aiding in the qualitative assessment of the system.

Transcript

00:01 So here we are just going to draw the cross section of the shell.

00:08 So for the first case, this is your r.

00:15 So electric field over here.

00:17 First of all, using the gaussian formula, e .orgia equals flux, equals q, nclosed over epsilon.

00:27 So e comes out to be 1 over epsilon not a over q enclosed.

00:35 And here because the gaussian surface will be spherical at a distance of r it will be 1 over 4 pi epsilon r squared times q and closed so first we are trying to find e for r r greater than uppercase r so uh this much is r out so this much is r out and this much is r out and this much is r in.

01:17 So for this condition we will have 1 over 4 pi epsi epsomr square times the entire charge is inside the cylinder so we can just have q and close equals q for e so let's do it in blue so if we take a gaussian surface over here that is e for r less than r in and that has got no charge enclosed so it is zero now for the third case we have to find the volume charge density so volume charge density equals total charge divided by total volume which is four third pi r out cube minus r in cube and e for r for r four third for r greater than r in but less than r out will be one over four pi epsilon not r square but then the chart enclosed will be volume charge density so let's just write it volume charge density so cube over four third pi r not cube minus r i cube times the volume so basically this is this is the volume that we have to take care of now which will be four third by r cube minus r in simplifying this will come out as q over four by epsilon not r square times r cube minus r cube equals r0 minus r i cube now let's figure out if they match at the boundary condition so from here if we choose e for r equals r in then we will see that r equals r in makes the numerator to be zero so that equals zero which matches with this case now e for r equals r equals r out will give us put r equals r out and then this term becomes one so we have got q equals 4 pi epsilon not are out 4 pi yeah so this r squared now will become r out square and then this will match with this where if r equals r out then put r out over here and then they will match up now e versus r graph will look like as follows.

05:16 Let's do it as a function of r over r out...

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