A nonconducting sphere of radius r2 contains a concentric...

Key Concepts

Boundary Conditions and Continuity of Potential

The continuity of the electric potential across boundaries between different regions is a key requirement dictated by the physics of electrostatics. Even if the electric field might experience discontinuities due to surface charges or abrupt changes in charge density, the potential itself must remain continuous to ensure physical realizability. This concept is applied to verify that the potential function is smooth at the boundaries of the cavity and the outer edge of the sphere, which is critical for a correct and meaningful solution.

Electric Potential and Its Relation to the Electric Field

The electric potential is related to the electric field via the gradient, specifically the potential difference is obtained by integrating the electric field. In this problem, determining the electric potential involves integrating the electric field obtained from Gauss's Law over a proper distance, while ensuring the correct sign and limits. This concept is crucial when calculating the potential in regions separated by different charge distributions, as it provides a method to recover the scalar potential function from a known vector field.

Integration of Charge Distributions

When dealing with a material that carries a volume charge density, the total charge enclosed within a Gaussian surface is found by integrating the charge density over the volume. This integration is essential for calculating the electric field within and outside the charged region. In problems involving cavities or regions with missing charge distributions, the integration must be performed carefully over the relevant regions, ensuring that the absence of charge is correctly taken into account.

Gauss's Law

Gauss's Law is a fundamental principle relating the electric flux through a closed surface to the charge enclosed by that surface. It is particularly useful in situations with high symmetry, such as spherical, cylindrical, or planar geometries. In this context, Gauss's Law is used to determine the electric field in different regions of a sphere with a cavity by taking advantage of spherical symmetry, simplifying the calculation of the field produced by a uniform charge distribution.

Spherical Symmetry

Spherical symmetry refers to the property of a system where physical quantities depend only on the radial distance from a central point and not on the direction. This symmetry allows one to simplify the mathematical analysis by reducing complex three-dimensional integrals to one-dimensional integrals with respect to the radial coordinate. For this problem, spherical symmetry ensures that the electric potential and field are functions solely of the distance from the center, making the application of Gauss's Law straightforward.

Transcript

00:01 Okay, so we're in chapter 23, problem 81.

00:03 So it says a non -conducting sphere of radius r2 contains a concentric spherical cavity of radius r1.

00:11 The material between r1 and r2 contains a uniform charge density row d.

00:19 Determine the electric potential v relative to v of r equals infinity equaling zero as a function of distance r from the center.

00:31 So for a, we want r greater than r2.

00:36 Ok, so since we know that the potential is zero at infinity, the potential outside the spell, or outside of the shell, follows that as a point charge.

00:56 So we already know that this is given by 1 over 4 pi epsilon not, total charge over r.

01:08 Ok, but we don't know in terms of the total charge.

01:12 So we got to write out this in terms of row so this is now 1 over 4 pi epsilon not times r and now the total charge is given by row e times the big volume minus the carved out small volume.

01:40 Cool and if we simplify that this comes out to being row e over 3 epsilon knot r2 cubed minus r1 cubed all over r.

01:52 Cool so this is for r greater than r2.

01:57 Moving on to part b it asks now for r1 less than r less than r2.

02:04 So for this we need to use gauces law where we have the surface integral of e.

02:09 D .a and if we draw a gaussian surface just outside of the shell, then we should see that this equals the charge enclosed over epsilon not.

02:24 And we know this is given by since the spherical symmetry of our problem, e times the surface area of a sphere, which is 4 pi r squared.

02:37 So now we can just write this as 1 over 4 pi epsilon not charge enclosed over r squared.

02:50 But what is the charge enclosed now? we've got to figure that out.

02:54 So we have 1 over 4 pi epsilon not r squared.

02:58 And now the charge enclosed is again similarly to what we have here, but instead of being the big radius r2 outside, it's just as a function of r.

03:09 So this is row e times the volume enclosed from r minus the carved out no charge volume.

03:21 Cool.

03:21 So this looks very similar, i'm guessing.

03:24 And you'll see that it comes out to being units.

03:27 So we get the electric.

03:30 Field to be row e over 3 epsilon not r cubed minus r1 cubed all over r squared.

03:43 So this is what the electric field is like, but we're still trying to find the potential.

03:48 So if we're trying to find the potential, we can take the potential at r and subtract off the potential at the surface...

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