Classic Hemispherical Analyzer

• rbrayana123
In summary, the conversation discusses solving a problem involving two concentric cylinders with equal but opposite charge. The potential difference between the cylinders can be calculated using Gaussian surfaces and is equal to 2kQ/L * ln(b/a). The conversation also mentions the use of the equations for uniform circular motion. The individual's solution involved checking the potential difference and recognizing the system as a coaxial cable conductor. There is some confusion about symbols and units used in the solution, but it appears that the individual is on the right track.
rbrayana123

Homework Statement

http://oi50.tinypic.com/4i35g.jpg

I remember learning about this in Chemistry but boy I did not expect to actually prove it works haha. I've actually solved the problem but I feel a little uncomfortable with certain aspects.

Homework Equations

E = $\Delta$Vq

For uniform circular motion:
a = v^2/r

For two concentric cylinders with equal but opposite charge, the potential difference can be calculated out via Gaussian surface to be:

2kQ/L * ln(b/a) where L is the length of the cylinder.

The Attempt at a Solution

First, I checked the potential difference and found it to be 2V * ln(b/a) which immediately tells me a lot about the system. Therefore, kQ/L = V (units seem to work out).

Conservation of Energy:

Vq = 0.5mv^2 --> v^2 = 2Vq/m

Using Gaussian surfaces, E(r) = 2kQ/Lr = 2V/r.
F(r) = E(r)q = 2Vq/r
a(r) = 2Vq/mr = v^2/r which is the equation of motion for uniform circular motion of radius r.

What's bothering me is that the reason I was able to solve this problem is because I'm *very* familiar with the coaxial cable conductor and quickly recognized it.

Provided the inner and the outer conductor did not have equal but opposite charges, would the potential still be the same or would it somehow be different? I think I solved this problem operating under the assumption they do but is it necessary to make that assumption?

Last edited:
rbrayana123 said:

Homework Statement

http://oi50.tinypic.com/4i35g.jpg

I remember learning about this in Chemistry but boy I did not expect to actually prove it works haha. I've actually solved the problem but I feel a little uncomfortable with certain aspects.

Homework Equations

E = $\Delta$Vq
This E is energy but later you used E for electric field ... ? Anyway, energy is not relevant here.
For uniform circular motion:
a = v^2/r

For two concentric cylinders with equal but opposite charge, the potential difference can be calculated out via Gaussian surface to be:

2kQ/L * ln(b/a) where L is the length of the cylinder.
That is correct.

The Attempt at a Solution

First, I checked the potential difference and found it to be 2V * ln(b/a) which immediately tells me a lot about the system. Therefore, kQ/L = V (units seem to work out).

What the ...? What is your V? It's not the potential difference between the two shells and it's not V0 either ... I am confused ...
Conservation of Energy:

Vq = 0.5mv^2 --> v^2 = 2Vq/m
This V must be V0.
Using Gaussian surfaces, E(r) = 2kQ/Lr = 2V/r.
Again, I don't know what your V is.

At this point I give up ... you need to get your symbols straightened out.

I think you're on the right track, though. Get the E field as E(Q,r), change that to E(V,r) where V is the pot, diff. betw. the two shells. V is of course -∫abE(r)dr.

Then get E(V,r0), equate qE(V,r0) to mv2/r0, realize (as you kind of did) that qV0 = mv2/2, and solve for V.

Note that the given answer, 2V0*ln(b/r0) - 2V0*ln(a/r0) can be more simply written as 2V0*ln(b/a).

What is a Classic Hemispherical Analyzer?

A Classic Hemispherical Analyzer is a scientific instrument used for studying the electronic and atomic structure of materials. It is typically used in surface science and can analyze the properties of a material's surface, such as its chemical composition and electronic states.

How does a Classic Hemispherical Analyzer work?

The analyzer works by using a beam of electrons to excite the surface of a material. The electrons are then scattered and detected by the analyzer, which measures their kinetic energy and direction. This data is then used to determine the electronic and atomic structure of the material's surface.

What are the advantages of using a Classic Hemispherical Analyzer?

One of the main advantages of using a Classic Hemispherical Analyzer is its ability to provide detailed information about the electronic and atomic structure of a material's surface. It is also a non-destructive technique, meaning that the sample is not altered or damaged during analysis.

What types of materials can be analyzed with a Classic Hemispherical Analyzer?

A Classic Hemispherical Analyzer can be used to analyze a wide range of materials, including metals, semiconductors, and insulators. It is particularly useful for studying thin films and surfaces of materials.

What are some common applications of a Classic Hemispherical Analyzer?

A Classic Hemispherical Analyzer is commonly used in fields such as materials science, surface science, and nanotechnology. It can be used to study the properties of materials for various applications, such as in the development of new electronic devices or in the study of catalytic reactions.

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