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PS4 Solutions

# PS4 Solutions - Physics 217 Problem Set 4 Solutions Yao 1-5...

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Unformatted text preview: Physics 217: Problem Set 4 Solutions Yao 1-5 Jolyon 6-9 Due, 2010 1 Purcell 3.01 Conductors with cavities have a common feature, even if the surfaces are not spherical. Since E inside a conductor (outside the cavity) is zero, we can draw any Gaussian surface in the “meat” of the conductor, and have zero charge enclosed! For one thing, no charge can reside in the meat, and must be on the boundary. If the Gaussian surface encloses a cavity, the total charge (charge inside the cavity + those induced on the inner surface) has to be zero. [Picture: Spherical conductor A has two spherical cavities, one with a charge q b at its center, and the other with q c at its center. Outside is a charge q d far apart.] For this problem, the cavity with q b has an amount of- q b somehow distributed over the inner sphere. Similarly- q c is induced on the other cavity. Since the conductor A is neutral, there has to be q b + q c on the outer surface. To see what the distribution of charge is, consider a Poisson’s equation inside the b sphere, subject to the boundary condition that ϕ on the surface is a constant ϕ . I can give one solution, and that has to be the solution by the uniqueness theorem. That solution is simply the ϕ produced by a single charge q b , which is at the center of the sphere. That means the induced charge is uniformly distributed. If you visualize the field lines, you should expect q b feel no force, no matter what goes on outside the cavity. We say q b is shielded. Also, to the outside, the shell of uniform charge- q b is the same as if all that charge is concentrated at the center, which would then cancel the q b there. Therefore, to the outside, the cavity as a whole has zero effect. That is also true if q b is not placed at the center, or there are multiple charges in the cavity, or the cavity is not spherical at all: there would just have to be complicated distribution on the inner surface to cancel whatever is in there. The q b now does feel a force from that surface charge distribution. But whatever it is, it does not depend on what goes on outside, since we are solving the same Poisson’s equation in the cavity with the same boundary condition, and the uniqueness theorem dictates the same solution. The distribution of the outer surface is more complicated, but remarkably it is known exactly. Anyways, if r is large, relative to the size of A , then your intuition tells you that the distribution should also be uniform. Now to figure out what force is on q d , we can think of the collective effect of all the known charges. The cavity b gives zero effect, as does cavity c . The uniform charge q b + q c on the outer sphere is as if all that charge is concentrated at the center of the sphere. Therefore the force on q d is F ≈ ( q b + q c ) q d r 2 The force on A is equal and opposite by Newton’s third law. Curiously, even though q b and q c are shielded from outside, we can detect their presence. However, all we can infer is that there’s a total amountfrom outside, we can detect their presence....
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PS4 Solutions - Physics 217 Problem Set 4 Solutions Yao 1-5...

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