A metal sphere with radius ra is supported on an insulating...

A metal sphere with radius $r_{a}$ is supported on an insulating stand at the center of a hollow, metal, spherical shell with radius $\eta_{b}$. There is charge $+q$ on the inner sphere and charge $-q$ on the outer spherical shell. (a) Calculate the potential $V(r)$ for (i) $r<r_{a}:$ (ii) $r_{a}<r<n_{b}$ (iii) $r>n$. (Hint: The net potential is the sum of the potentials due to the individual spheres.) Take $V$ to be zero when $r$ is infinite. (b) Show that the potential of the inner sphere with respect to the outer is $$ V_{a b}=\frac{q}{4 \pi \epsilon_{0}}\left(\frac{1}{r_{a}}-\frac{1}{n_{b}}\right) $$ (c) Use Eq. (23.23) and the result from part (a) to show that the electric field at any point between the spheres has magnitude $E(r)=\frac{V_{a b}}{\left(1 / r_{a}-1 / r_{b}\right)} \frac{1}{r^{2}}$ (d) Use Eq. (23.23) and the result from part (a) to find the electric field at a point outside the larger sphere at a distance $r$ from the center, where $r>r_{b} .$ (e) Suppose the charge on the outer sphere is not $-q$ but a negative charge of different magnitude, say $-Q .$ Show that the answers for parts (b) and (c) are the same as before but the answer for part (d) is different

Key Concepts

Boundary Conditions for Conductors

In electrostatics, conductors have the property that the electric potential is constant over their surfaces, and any excess charge resides on the exterior. This boundary condition simplifies the analysis of problems involving spherical conductors or shells since the potential inside a conductor is uniform and the electric field inside is zero. These properties are essential when calculating potentials and fields in systems where inner and outer conductors interact.

Relationship Between Electric Field and Potential

The electric field is related to the electric potential by the spatial derivative, typically given by E = -dV/dr in radially symmetric situations. This relationship is exploited when one calculates the potential differences between two points (such as between a charged sphere and a surrounding shell) and then uses those differences to determine the electric field at any point in the system. Understanding this connection is crucial for transitioning between the concepts of potential energy and the force experienced by a test charge.

Gauss's Law and Spherical Symmetry

Gauss's law relates the electric flux through a closed surface to the charge enclosed by that surface. In the context of spherical charge distributions, it is particularly useful because the symmetry allows one to easily derive the electric field in different regions. This law is instrumental in establishing relationships between the field and the charge, and by integrating the electric field, one can determine the potential differences between regions.

Superposition Principle in Electrostatics

The principle of superposition states that the total electric potential created by multiple charges is equal to the algebraic sum of the potentials due to each individual charge. This fundamental concept simplifies the calculation of the overall potential in multi-charge systems, including configurations involving a sphere inside a concentric spherical shell, where potentials due to each are added together to obtain the net potential.

Electric Potential from Distributed Charges

Electric potential is a scalar quantity that represents the work done per unit charge in bringing a test charge from infinity to a given point in an electric field. In problems with spherical symmetry, such as charged spheres or shells, the potential at any point can be calculated using the formula V = kq/r (with appropriate modifications), where the contributions from different charge distributions are summed. This concept is essential because the potential is continuous and provides a basis for determining the energy differences between points in the field.

Transcript

00:01 So in this problem, we're looking at the case of two spheres with the same center.

00:07 So we have sphere a with an inner radius ra and a charge of positive q that is concentric with sphere b, radius rb, which has a negative q charge.

00:21 So same magnitude, but obsolete charged.

00:24 So what we know about solid metal spheres or spherical shells is that the potential outside the shell, is going to act like that of a point charge.

00:35 So we'll have v equal to k times q over r, and where r is some distance from the center.

00:45 And when we're inside a shell, we know that the potential is going to look pretty similar, but rather than using the distance from the center, we're just going to use the radius, so denoted by big r here.

00:59 So this is a general expression, finding the potential inside of a size, spherical shell.

01:04 So at any point inside of a shell, the potential will be the same.

01:10 And we also know that potential due to multiple charges can be treated as a scalar sum.

01:20 So this means that we can find the potential at any one point due to multiple charges, and we can just add up those potentials to get what the total potential would be.

01:32 Another thing to right down here is that the electric field when its radial can be expressed in terms of the potential as a derivative.

01:42 So it'll be negative delta v over delta r.

01:47 So that change of potential divided by the change of distance in terms of r.

01:54 So now that we have those facts, let's get started and look at problem a, part a of this problem.

02:03 And what we're asked to do is we want to calculate the potential v in terms of r for three different scenarios.

02:10 So when we're inside the inner shell, when we're between the two shells, and when we're outside.

02:16 So in the first part, we're looking at when r is less than ra.

02:20 So inside the inner shell.

02:22 So if we're inside both spheres, that means that we're going to be using this, excuse me, this, let me label these here.

02:30 We'll say outside and inside.

02:34 So we're going to be using this equation for both of the spheres.

02:39 So what we're going to do is we're going to have v equal to kq over ra minus kq over rb.

02:53 Remember it's a minus because we have a negative charge on the outer shell.

02:57 So what that'll give us when we simplify is kq times 1 over ra minus 1 over rb.

03:11 So there we have the potential inside of the inner shell.

03:21 And now for the second part, we're looking at the space between ra and rb.

03:26 So we are outside of the inner shell.

03:31 So we can use this equation for the potential due to that inner shell, but we're still inside the outer shell.

03:40 So we're going to use this for rb.

03:43 So what that looks like when we write it out, we have the potential equal to kq divided by r, that whatever distance we are in between ra and rb, minus kq divided by rb.

04:00 And again, we just simplify this.

04:02 We've rearranged a little bit, we'll have kq times one over r minus one over rb.

04:16 And so finally, we want to determine what the potential is when we are outside of the outer shell.

04:26 So we're now outside of both shells where r is greater than rb.

04:31 And this time, since we're outside, we're going to be using this equation for both of the potentials due to both shells.

04:38 So what that looks like is v is equal to kq over r minus kq over r.

04:52 So that is zero.

04:57 The potentials due to each sphere actually cancel themselves out.

05:02 So outside the sphere, the potential is the same as for a point charge as we established here.

05:09 And since the shells are equally and oppositely charged, we will have now.

05:13 No potential actually outside, potential will be zero, outside of both these shells.

05:19 So now moving on to part b...

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