A small conducting spherical shell with inner radius a and...

A small conducting spherical shell with inner radius $a$ and outer radius $b$ is concentric with a larger conducting spherical shell with inner radius $c$ and outer radius $d$ (Fig. 22.38). The inner shell has total charge $+2 q$, and the outer shell has charge $+4 q$. (a) Calculate the electric field (magnitude and direction) in terms of $q$ and the distance $r$ from the common centre of the two shells for (i) $r<a$; (ii) $a<r<b$; (iii) $b<r<c$; (iv) $c<r<d$; (v) $r>d$. Show your results in a graph of the radial component of $\vec{E}$ as a function of $r$.(b) What is the total charge on the (i) inner surface of the small shell; (ii) outer surface of the small shell; (iii) inner surface of the large shell; (iv) outer surface of the large shell? (FIGURE CANT COPY)

Key Concepts

Graphical Analysis of Electric Fields

Graphical analysis is used to depict the variation of the electric field as a function of distance from the center of the spherical shells. By plotting the electric field versus radial distance, one can visually interpret the regions where the field is constant, null, or changes abruptly due to the presence of charge on the surfaces. This helps in understanding and communicating the behavior of the electric field across different regions surrounding the charged conductors.

Electric Field in and around Conductors

In electrostatics, the electric field inside a conductor is zero when in equilibrium. This property, combined with the principles of charge redistribution on conductors, ensures that any excess charge resides on the surfaces. For problems involving spherical shells, this means that the field inside the material of the conductor is zero, while the field in nearby regions is determined by the charge on the surfaces, as dictated by Gauss's Law.

Charge Distribution on Conducting Shells

Conducting shells in electrostatic equilibrium exhibit a specific pattern of charge distribution: charges redistribute themselves on the inner and outer surfaces to ensure that the field within the conducting material remains zero. The induced charges are determined by the need to balance the external field and any charges present within the cavity. This concept is critical when solving for the electric fields in multiple-shell systems, as the net charge on each surface must be determined by applying Gauss's Law to appropriate regions.

Gauss's Law

Gauss's Law is a fundamental principle in electromagnetism that relates the electric flux through a closed surface to the net charge enclosed by that surface. This law is especially useful in analyzing problems with high symmetry, as it allows the electric field to be calculated by choosing an appropriate Gaussian surface, greatly simplifying the computations of fields in regions with spherical, cylindrical, or planar symmetry.

Spherical Symmetry

Spherical symmetry is the property of a system where physical quantities depend only on the radial distance from a central point and not on the direction. In the context of charged spherical shells, this symmetry allows us to assume that the electric field has solely a radial component. This simplification significantly reduces the complexity of using Gauss's Law, as the flux through a spherical Gaussian surface becomes a straightforward calculation involving the surface area of a sphere.

Transcript

00:01 Okay, so for this problem, we have an inner shell and an outer shell, and the inner shell has a charge of plus 2q.

00:09 The outer shell has a charge of plus 4q.

00:12 The inner shell is bound by the radii a and b.

00:16 The outer shell are bound by the radii c and d.

00:19 We're looking for the electric field everywhere and see if we can find where the charges are going and what's happening.

00:27 So let's go ahead and look for the electric field everywhere.

00:33 So, in order to find the electric field, we're going to write down galses law, which says that the closed line integral of e .da equals the total charge that is enclosed in our region divided by epsilon 0.

00:54 So in general, this is our galses law, and this is our area element.

01:01 This is our total charge enclosed in our gaussian surface.

01:04 This is epsilon zero, just a constant.

01:06 And e is what we're looking for.

01:08 So let's look for our first region, which is when we have a radius less than a.

01:16 So four or less than a, well, we don't have any charge there.

01:21 So that means that q equals zero, q equals zero, e equals zero.

01:34 So e equals zero, this region.

01:39 All right, what about if we are in between a and b? so now we're in this shell right here.

01:44 Well, if it is a conducting shell, then by definition, it must have zero electric field.

01:57 So we have the same thing.

01:59 E must equal zero here.

02:08 Now what about this cavity between b and c? so for b less than r less than c, well now we do have some charge and we have the total charge on this show plus 2q.

02:25 So, just in general, we know that the formula for the electric field of a point charge, e equals k, which is 1 over 4 pi epsilon, or 99, kq, divided by r squared, we can use this for our region between b and c, and we see that the electric field, e, equals k, q, which is 2 q.

03:08 Q right here is the total charge divided by r squared.

03:11 R squared is just this.

03:17 We can write this more explicitly.

03:20 1 divided by 4 pi epsilon times 2 q over r squared.

03:32 Okay.

03:34 Now what about the region c and d? so now we are in this outer ring.

03:41 Well, when we are in this outer ring, remember that the electric field must equal zero because it is a conductor.

03:49 So for the region between c, less than r less than d, e equals zero.

04:03 To run inside that conductor.

04:06 Okay.

04:07 And then finally, what about if we are outside of this object? so what happens if r is greater than d? well, when r is greater than d, then we have the total charge and close.

04:21 Which is 4q plus 2q, 6q.

04:26 So we just can use this formula right here for our regular point charge.

04:32 And we get our electric field, e, equals 1 divided by 4 pi, epsilon 0, times 6q over r squared.

04:56 And is a 6.

04:59 So there we go.

05:04 And those are all of our regions.

05:05 Now we can go ahead and graph this...