An insulating hollow sphere has inner radius a and outer...

An insulating hollow sphere has inner radius $a$ and outer radius $b$. Within the insulating material the volume charge density is given by $\rho(r)=\alpha / r,$ where $\alpha$ is a positive constant. (a) In terms of $\alpha$ and $a$, what is the magnitude of the electric field at a distance $r$ from the center of the shell, where $a<r<b ?$ (b) A point charge $q$ is placed at the center of the hollow space, at $r=0$. In terms of $\alpha$ and $a$, what value must $q$ have (sign and magnitude) in order for the electric field to be constant in the region $a<r<b$, and what then is the value of the constant field in this region?

Key Concepts

Gauss's Law

This law relates the electric flux through a closed surface to the charge enclosed by that surface. It is especially useful for problems with high symmetry, allowing one to simplify the integration process and directly relate the electric field to the total charge present within a Gaussian surface.

Spherical Symmetry

Spherical symmetry implies that the physical properties of the system, such as the charge distribution and the resulting electric field, depend only on the radial distance from the center. This symmetry allows for the selection of spherical Gaussian surfaces that simplify the calculation of the electric field, as the field’s magnitude remains constant over these surfaces.

Volume Charge Density Integration

When dealing with continuous charge distributions, it is necessary to compute the total charge by integrating the volume charge density over the region of interest. In cases where the density varies with position (for example, inversely with the radius), careful integration over the volume is required to obtain the enclosed charge for use in Gauss’s Law.

Superposition Principle

This principle states that the resulting electric field from multiple charge distributions is the vector sum of the individual fields produced by each distribution. It is especially important when additional charges, such as a central point charge, are introduced to modify the overall electric field in a desired way.

Electric Field Uniformity Conditions

In some configurations, it is desirable to achieve a constant electric field in a specific region. By appropriately adjusting additional charges—like a point charge in the center of a charged shell—the position-dependent variations of the field due to non-uniform charge distributions can be counteracted, leading to a uniform electric field across that region.

Transcript

00:01 So we have a holo sphere with inner radius a, outer radius b, okay, as it shows the diagram.

00:10 And we know the volume charge density, okay? it's the function given here bar row.

00:16 And we want to find the magnitude of the field e, the electric field, when we are like in this, in a point here, that is greater than a and smaller than b okay so any point in the isolating and how do we do this okay so first notice that we do have the volume charge density okay we also know that dv equals we know the volume of the sphere so if we take the derivative, we can find the this is 4 pi r squared, and this is the radius of the sphere, okay, the r, okay? in terms, so we have like the differentiation of the volume, in terms of the differential of the radio.

01:21 That's just derivating the function of the volume.

01:24 Okay, you can do it.

01:26 And this is actually the area, no? but we'll talk about it later, okay? we know that the charge, okay, will be equal for the integral, okay, from a to r of the density, okay, the charge density, times the div, the differentiation of the volume, okay? and you can actually see we have these two values, okay? now i use here our prime, okay, because i, i, i'm calling r at the point at which i am here.

02:06 So r prime is just the r i am coming through here.

02:12 So when you substitute the values, okay, from i to r, okay, the way substitute the charge density is alpha divided by r, times the differential of the volume, that it's this value, for pi r squared d r, r, okay.

02:31 I can eliminate 1r and then i find that this is the integral from a to r.

02:39 I can leave the constants out of the integral.

02:44 Okay, and so this is alpha 4 pi.

02:48 And i have the r, okay, r prime, okay? this will be r prime...

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