# Concentric conducting spherical shells cut by a horizontal plane

• homer
In summary: Since the integral is over the entire sphere, the limit is when \mathbf{r} = a. For the inner sphere, the limit is when \mathbf{r} = 3a/4.In summary, the net force on the outside surface of the A part of the sphere is given by\begin{align}d\mathbf{F} & = -\frac{Q \, dq}{a^2} \mathbf{n}_{\text{outer}} \\\end{align}
homer

## Homework Statement

A conducting spherical shell of outer radius a and inner radius 3a/4 is cut in two pieces via a horizontal plane a distance a/2 above the center of the spherical shell, as shown in Figure 1. Let us label "A" the upper part of the shell and "B" the lower part of the shell. The shell is initially uncharged and the two pieces that result from the cutting procedure remain in perfect electrical contact. A new conducting sphere of radius a/16 and total charge +Q is inserted in the shell and it is centered on the shell's center as shown in the same figure.

(a) Are there any charge densities on the inner (r=3a/4) and outer (r=a) surfaces of the shell as well as within it? If yes, derive them.
(b) What (if any) is the force per unit area on the inner and outer surfaces of the shell?
(c) From now on we focus only on the "A" part of the shell: set up an integral that will yield the net force acting on the "A" part. What is its direction? Identify which variables you are integrating and what are the limits of integration. (You are not asked to perform the integration!)
(d) Do the same as (c) for the inner shell of the "A" part of the shell.
(e) Do the same as (c) for the "A" part as a whole.

## Homework Equations

Gauss' Law (cgs): $$\nabla \cdot \mathbf{E} = 4\pi \rho$$

Poisson's equation (cgs): $$\nabla^2 \varphi = -4\pi \rho$$

## The Attempt at a Solution

I just don't even know where to start with this problem. I know without the horizontal plane I could apply Gauss' Law within the shell to see that a charge −Q must be distributed on the inside surface (the one at radius r=3a/4), and that it would be uniform by symmetry since the +Q charge is right in the center. But how would I even approach this with the horizontal surface there? Gauss' Law shows there should still be charge −Q along the inside of the shell and on the horizontal plane inside it, but no idea how it's distributed. Since the plane is not assumed to be a conductor the method of images wouldn't seem useful. I have no idea what the field should look like since the plane isn't necessarily an equipotential, and thus I can't assume the field is perpendicular to the plane. Anyone have some ideas as to how to start this problem that is laid out below? It's from problem set 3 of the 8.022 course from Fall 2004 on MIT OCW. Thanks!

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Ooops, bad title for the thread, as it's a conducting sphere inside a conducting spherical shell, the shell of thickness r/4. Not a sphere inside a shell inside another shell.

Nevermind. I see the plane is just something to show how the spheres are cut, not something physically there.

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I have gotten sidetracked from this problem for a couple of days, but back at it now. Can someone tell me if my solution so far is right?

Part (a) By Gauss' Law there is a charge $-Q$ distributed on the inner surface (take a spherical Gaussian surface inside the actual shell, e.g., with $3a/4 < r < a$, where $\mathbf{E} = \mathbf{0}$, so that $\mathbf{E} \cdot d\mathbf{a} = 0$), and it's distributed uniformly by symmetry since it's a spherical conducting surface and the $+Q$ charge is at the center. Thus the charge density on the inner surface is

\sigma_{\text{inside}} = - \frac{Q}{4\pi(\tfrac{3}{4}a)^2} = -\frac{4Q}{9\pi a^2}.

Since the shell is electrically neutral, there is also a charge $+Q$ on its outer surface. It is uniform since the outer surface is a sphere, so the charge density on the outer surface is

\sigma_{\text{outside}} = \frac{Q}{4\pi a^2}.

Part (b) There is no net force on the inner nor outer surface of the entire sphere, due to symmetry.

Part (c) I'm going to assume the question is asking for the net force on the outside surface of the A part of the shell. Let's use spherical coordinate with the origin at the center of the shell, so that the shell is parametrized as

\mathbf{r} = a\sin{\theta}\cos{\phi}\,\hat{\mathbf{x}} + a\sin{\theta}\sin{\phi}\,\hat{\mathbf{y}} + a\cos{\theta}\,\hat{\mathbf{z}}

for $0 \leq \theta \leq \pi$ and $0 \leq \phi \leq 2\pi$, where $\theta$ is the colatitude and $\phi$ is the longitude. For the outer surface the unit normal is

\mathbf{n}_{\text{outer}} = \sin{\theta}\cos{\phi}\,\hat{\mathbf{x}} + \sin{\theta}\sin{\phi}\,\hat{\mathbf{y}} + \cos{\theta}\,\hat{\mathbf{z}}

while for the inner it is

\mathbf{n}_{\text{inner}} = -\mathbf{n}_{\text{outer}}.

A differential area element of the sphere is given by

dS = a^2\sin{\theta}\,d\phi\,d\theta,

and carries a charge
\begin{align}
dq & = \sigma_{\text{outside}}\,dS \\
& = \frac{Q}{4\pi a^2}\,a^2\sin{\theta}\,d\phi\,d\theta \\
& = \frac{Q}{4\pi}\,\sin{\theta}\,d\phi\,d\theta.
\end{align}

For the outer sphere, the differential force on the area element is
\begin{align}
d\mathbf{F} & = \frac{Q \, dq}{a^2} \mathbf{n}_{\text{outer}} \\
& = \frac{Q^2}{4\pi a^2}\sin{\theta}\,d\phi\,d\theta\,\mathbf{n}_{\text{outer}} \\
& = \frac{Q^2}{4\pi a^2}\,(\sin^2{\theta}\cos{\phi}\,\hat{\mathbf{x}} + \sin^2{\theta}\sin{\phi}\,\hat{\mathbf{y}} + \sin{\theta}\cos{\theta}\,\hat{\mathbf{z}})\,d\phi\,d\theta.
\end{align}

For the entire A section, the colatitude varies from $\theta = 0$ to $\theta = \cos^{-1}(1/2) = \pi/3$ (since $\cos{\theta} = (a/2)/a = 1/2$ at the maximum $\theta$) and the longitude from $\phi = 0$ to $\phi = 2\pi$, so the net force acting on the outer surface of the A part of the sphere should be
\begin{align}
\mathbf{F} & = \int\!\!\!\! \int d\mathbf{F} \\
& = \int_{0}^{\pi/3}\!\!\!\!\int_{0}^{2\pi} \frac{Q^2}{4\pi a^2}\,(\sin^2{\theta}\cos{\phi}\,\hat{\mathbf{x}} + \sin^2{\theta}\sin{\phi}\,\hat{\mathbf{y}} + \sin{\theta}\cos{\theta}\,\hat{\mathbf{z}})\,d\phi\,d\theta.
\end{align}

Does this look correct? The calculation for the force acting on the inner surface would be similar.

The integral is pretty trivial to do, so I'll spare the details, but I get

\mathbf{F} = \frac{3Q^2}{16a^2}\,\hat{\mathbf{z}}

for the force on the outside of the A part of the shell, which seems reasonable since it has positive charge on it that should repel the positive charge in the center and since the force components in the $x, y$ directions should cancel out due to symmetry.

Then I'd just compute the force on the inner surface similarly, but a little messier since the distance from the center of the $+Q$ charge is $3a/4$ instead of $a$. Then add that to the force on the outer and that gives the force on the entire A part, since there is no force due to the interior of the shell since there is no charge in it, as its a conductor; all the charge is on the surface.

Is that correct, or am I screwing something up somewhere? Thanks.

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Oops, I'd have to compute the force on the annulus that forms the boundary between the A and B parts of the sphere also. But is what I have for the outer part correct?

I guess that annulus of boundary between A and B parts should have no net charge on it though, since A and B are hypothesized to be in perfect electrical contact.

@homer I think that for part b), there is a net force/unit area on the inner surface due to the charge in the center. A for part c), I don't see why there should be any force on the outer surface unless we consider ONLY the corresponding inner surface and nothing else (no charge at center or the rest of the sphere considered). Did you ever resolve this?

## 1. What is the purpose of cutting a concentric conducting spherical shell with a horizontal plane?

The purpose of this is to create a simplified model for studying the electric field and potential inside the shell. It allows for easier calculations and analysis.

## 2. How does cutting the shell affect the electric field and potential?

Cutting the shell creates two separate regions with different electric fields and potentials. Inside the shell, the electric field is zero and the potential is constant. Outside the shell, the electric field and potential follow the same laws as a point charge.

## 3. How do the radii of the shells affect the electric field and potential?

The ratio of the inner and outer radii of the shells determines the strength of the electric field and the potential difference between the two regions. A smaller inner radius and larger outer radius will result in a stronger electric field and larger potential difference.

## 4. Can the electric field and potential also be studied for non-concentric shells?

Yes, the same principles apply for non-concentric shells. However, the calculations become more complicated as the distance between the centers of the shells must be taken into account.

## 5. How is this concept used in real-life applications?

This concept is used in various technological and scientific applications, such as in the design of capacitors and electromagnetic shielding. It also helps understand the behavior of charged particles in a spherical environment, which is important in fields like astronomy and particle physics.

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