A solid, insulating sphere of radius a has a uniform charge...

A solid, insulating sphere of radius $a$ has a uniform charge density throughout its volume and a total charge $Q$ . Concentric with this sphere is an uncharged, conducting, hollow sphere whose inner and outer radii are $b$ and $c$ as shown in Figure P 24.51. We wish to understand completely the charges and electric fields at all locations. (a) Find the charge contained within a sphere of radius $r< a$ (b) From this value, find the magnitude of the electric field for $r< a$ (c) What charge is contained within a sphere of radius $r$ when $a< r < b ?$ (d) From this value, find the magnitude of the electric field for $r$ when $a< r < b$ . (e) Now consider $r$ when $b < r< $ c. What is the magnitude of the electric field for this range of values of $r ?(\mathrm{f})$ From this value, what must be the charge on the inner surface of the hollow sphere? (g) From part (f), what must be the charge on the outer surface of the hollow sphere? (h) Consider the three spherical surfaces of radii $a, b,$ and $c .$ Which of these surfaces has the largest magnitude of surface charge density?

Key Concepts

Surface Charge Density

Surface charge density is defined as the amount of charge per unit area on a surface. It is a critical concept for understanding how charges distribute themselves on the surfaces of conductors and insulators, influencing the local electric field and potential. This parameter is used to compare different surfaces and understand the intensity of the electric field generated by surface charges.

Gauss's Law

Gauss's Law is a fundamental principle in electromagnetism that relates the net electric flux through a closed surface to the total charge enclosed by that surface. In problems with high symmetry, such as spherical distributions, it simplifies the calculation of electric fields by allowing the selection of a Gaussian surface that makes the integral manageable.

Uniform Charge Distribution

A uniform charge distribution means that the charge density is constant throughout the volume of the object. This concept is important for calculating the enclosed charge within any spherical volume by scaling the total charge proportionally to the volume considered, which is integral to determining the electric field through Gauss's Law.

Spherical Symmetry in Electrostatics

Spherical symmetry allows the simplification of complex electrostatic problems by reducing the spatial variables to one radial coordinate. This symmetry means that the electric field at any point depends only on the distance from the center, facilitating the use of Gauss's Law to find the electric field in regions both inside and outside the charge distribution.

Behavior of Conductors in Electrostatic Equilibrium

In electrostatic equilibrium, the electric field within a conductor is zero. This principle leads to the redistribution of charges on a conductor's surfaces when exposed to external electric fields or nearby charges, ensuring that the field inside remains null. Understanding this behavior is crucial when analyzing systems involving conductors with internal or external charge distributions.

Charge Induction on Conducting Surfaces

When a conductor is exposed to an external or nearby charge distribution, charges within the conductor rearrange themselves to cancel internal electric fields. This induced charge appears on the conductor's surface, often splitting between inner and outer surfaces, and is key to maintaining electrostatic balance. The concept explains how initially uncharged conductors can develop surface charges in response to enclosed charges.

Transcript

00:01 Okay, so in this problem, we have the reference figure and this figure represents insulated sphere.

00:10 So let's put here in green the insulated sphere.

00:16 That means that the charge of the sphere is distributed uniformly and we have this sphere surrounded by a concentric hollow sphere.

00:28 So we have here the first radius and the the second radius, okay? which means the radius of the insulated sphere is a, this is b, and this is c.

00:47 And this problem, the first part of this problem, part a, b, c and d, and e, we have to calculate the charge of this configuration in all of this.

01:07 Regions in every region that we have in this problem so in order to begin this problem let's begin with part a in part a we have to discover the charge when the radius of the gaussian surface is smaller than the radius of the insulated sphere so the radius of the insulated sphere is just a to begin with this you must remember of the gauss law and we know that according to the gauss law, actually we don't need the gauss law.

01:48 We just want the charge.

01:49 We don't want the electrical field.

01:51 So what is charge? charge by definition when we do not have the entire charge of the system.

02:01 This is going to be just the density multiplied by the gaussian surface, the volume of the gaussian surface.

02:10 So, first of all, we know that the density, can be described as the total charge divided by the total volume of the insulated sphere.

02:21 And that's multiplied by the volume of the gaussian surface.

02:26 So if you want to simplify this, we can say that the charge q, oops, the charge q is going to be equal, let's put here, this is the charge inside, the charge inside, so the charge the charge inside the gaucian surface it's better if we do not mix our kills in here so the charge inside the gaucian surface it's going to be the entire charge kill now it's better this way it's better to understand so the entire charge kill they multiplies let me see four divided by three pi are cubic divided by 4 pi 3 a cubic we can cross 4 pi 3 4 pi 3 and finally we have that the answer to the part a the charge inside the gaussian surface when the radius is smaller than the radius of the insulated sphere is going to be the entire charge there multiplies are divided by a cubic.

03:59 So that's our first answer.

04:01 Now let's move on to the second part.

04:04 Part b we have to calculate.

04:08 Let's see.

04:11 In part b we have to find the magnitude of the electric field.

04:17 So in part b we have to find the electric field when the radius is smaller than the radius of the insulated sphere.

04:27 And this time we have to use the gausslo.

04:30 So according to the gauss law, we know that, let's see here, we know that the electrical field, electric field that multiplies the surface area of the gaussian surface is going to be equal the charge inside divided by epsilon zero.

04:51 The charge inside we already calculated.

04:54 That means that the electrical field in this part is just going to be, then multiplies r divided by a cubic divided by epsilon 0 divided by 1 divided by 4 pi r square therefore to simplify the answer for part b is going to be q r divided by 4 pi let me see 4 pi a cubic epsilon zero so that's the answer for part b now for part c let's see what we have to calculate in part c in part c we are asked to find the electric field actually no in part c we are asked to find the charge in close and a sphere so let's put here we have to find the charge inside when the radius goes from a to b.

06:20 That's it.

06:21 That's it.

06:26 In this region, as we can see here, the region for a to b is this region here.

06:34 Okay? so in this region we can say that the entire charge q has enclosure, which means that it is the answer is the charge inside is just the entire charge cube.

06:50 Nice...

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