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(a) Draw a diagram of the model with the nucleus shifted a distance d from the center, due to the applied field. Add a free body diagram of the nucleus to your drawing, showing the force on the nucleus by the applied field and the force on the nucleus by the field that is due to the negative charges. Note that we are interested in the situation where the nucleus is in equilibrium so the net force on it is zero. (b) You don't need to re-derive any results that we already got in class or that are done in the text for you - you may simply quote the relevant results: What is the field due to the sphere of negative charge at a distance d from the center? Use this and your free body diagram (remember, there

(a) Draw a diagram of the model with the nucleus shifted a distance d from the center, due to the applied field. Add a free body diagram of the nucleus to your drawing, showing the force on the nucleus by the applied field and the force on the nucleus by the field that is due to the negative charges. Note that we are interested in the situation where the nucleus is in equilibrium so the net force on it is zero. (b) You don't need to re-derive any results that we already got in class or that are done in the text for you - you may simply quote the relevant results: What is the field due to the sphere of negative charge at a distance d from the center? Use this and your free body diagram (remember, there

College Physics
College Physics
11th Edition
ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Raymond A. Serway, Chris Vuille
Chapter1: Units, Trigonometry. And Vectors
Section: Chapter Questions
Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...
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capacitors. Remarkably, Gauss's law is all we need to get a theoretical value for the polarizability of atoms that is a surprisingly good match to the experimentally measured value. Let's see how that works: We will model the atom as a uniform sphere, radius a, of negative charge -q, with a positive point charge +q in the middle. We will assume that the external field does not distort the sphere, nor pull the nucleus all the way out of the negatively charged sphere (these are certainly good assumptions for any reasonable field you might be using!) In this problem, we are treating the positive nucleus as a test charge, and the sphere of uniform negative charge as the source charge. PLEASE NOTE THAT THE TEXT WORKS OUT THE GAUSS'S LAW CALCULATION OF E(r) inside the sphere for you; you may use that and just apply the result. Of course, it wouldn't hurt for you to understand HOW the book got the result, but I am NOT requiring you to show all of that here. (a) Draw a diagram of the model with the nucleus shifted a distance d from the center, due to the applied field. Add a free body diagram of the nucleus to your drawing, showing the force on the nucleus by the applied field and the force on the nucleus by the field that is due to the negative charges. Note that we are interested in the situation where the nucleus is in equilibrium so the net force on it is zero. (b) You don't need to re-derive any results that we already got in class or that are done in the text for you - you may simply quote the relevant results: What is the field due to the sphere of negative charge at a distance d from the center? Use this and your free body diagram (remember, there are two equal-magnitude forces) to find the dipole moment p = qd as a function of a and the applied field E. (c) You should have found that the dipole moment is proportional to E. The proportionality =p/E is called the atomic polarizability. Based on these results, are larger atoms constant a easier or harder to polarize than smaller atoms, or harder to polarize? (d) The measured polarizability times k, in units of 10-30 m³, of helium is 0.205 and of carbon is 1.67 (from the Handbook of Chemistry and Physics.) Use this data and the proportionality factor you got using Gauss's law to estimate the radius of each of these atoms. Your calculation (if you did it right) is actually pretty close to the actual value!
Transcribed Image Text:capacitors. Remarkably, Gauss's law is all we need to get a theoretical value for the polarizability of atoms that is a surprisingly good match to the experimentally measured value. Let's see how that works: We will model the atom as a uniform sphere, radius a, of negative charge -q, with a positive point charge +q in the middle. We will assume that the external field does not distort the sphere, nor pull the nucleus all the way out of the negatively charged sphere (these are certainly good assumptions for any reasonable field you might be using!) In this problem, we are treating the positive nucleus as a test charge, and the sphere of uniform negative charge as the source charge. PLEASE NOTE THAT THE TEXT WORKS OUT THE GAUSS'S LAW CALCULATION OF E(r) inside the sphere for you; you may use that and just apply the result. Of course, it wouldn't hurt for you to understand HOW the book got the result, but I am NOT requiring you to show all of that here. (a) Draw a diagram of the model with the nucleus shifted a distance d from the center, due to the applied field. Add a free body diagram of the nucleus to your drawing, showing the force on the nucleus by the applied field and the force on the nucleus by the field that is due to the negative charges. Note that we are interested in the situation where the nucleus is in equilibrium so the net force on it is zero. (b) You don't need to re-derive any results that we already got in class or that are done in the text for you - you may simply quote the relevant results: What is the field due to the sphere of negative charge at a distance d from the center? Use this and your free body diagram (remember, there are two equal-magnitude forces) to find the dipole moment p = qd as a function of a and the applied field E. (c) You should have found that the dipole moment is proportional to E. The proportionality =p/E is called the atomic polarizability. Based on these results, are larger atoms constant a easier or harder to polarize than smaller atoms, or harder to polarize? (d) The measured polarizability times k, in units of 10-30 m³, of helium is 0.205 and of carbon is 1.67 (from the Handbook of Chemistry and Physics.) Use this data and the proportionality factor you got using Gauss's law to estimate the radius of each of these atoms. Your calculation (if you did it right) is actually pretty close to the actual value!
Estimating the induced dipole moment of atoms (how much the opposite charges separate) is an interesting and remarkable - and surprisingly accurate yet simple - application of Gauss's law. When an atom is placed in an external field, the electrons and nucleus move just a tiny bit, so that the nucleus is no longer centered relative to the electrons, until the nucleus is once again in equilibrium. As a result, the atom acquires a dipole moment. For many atoms, the induced dipole moment is proportional to the applied field; these are called linear materials, or dielectrics. This dependence, called the polarizability, can be measured fairly easily in the lab. It is also the underlying mechanism for the dielectric constant of materials, which we will encounter when we study
Transcribed Image Text:Estimating the induced dipole moment of atoms (how much the opposite charges separate) is an interesting and remarkable - and surprisingly accurate yet simple - application of Gauss's law. When an atom is placed in an external field, the electrons and nucleus move just a tiny bit, so that the nucleus is no longer centered relative to the electrons, until the nucleus is once again in equilibrium. As a result, the atom acquires a dipole moment. For many atoms, the induced dipole moment is proportional to the applied field; these are called linear materials, or dielectrics. This dependence, called the polarizability, can be measured fairly easily in the lab. It is also the underlying mechanism for the dielectric constant of materials, which we will encounter when we study
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