Analyzing dipoles (2015)last edited by Joe Redish 7 years, 7 months ago4.2.4.P4Page historyLots of things that are electrically neutral overall have one side that's electrically negative and one side positive (a water molecule, for example). We call such things "electric dipoles," and we can model them as pairsof particles of charge + q and -q (where q is a positive number) separated by a distance d. Usually, d is a very small distance. (For water it would be around 10-11 m, thinking of a few protons worth of charge on oneend - about 6 x 10-19 C -- and a few electrons worth at the other.) Furthermore, because of the magic of quantum mechanics, in many molecules it behaves more like a rigid rod than like a soft spring. So we can treatd as a fixed distance.*d-9Suppose you have a dipole that's free to move in any way (including rotate - imagine it floating in space). And there's an object with charge Q a distance r away. That distance r would be much larger than d, thedistance between the charges of the dipole, so we draw the dipole small.a) Consider the forces between the charge Q and the dipole. If the dipole is free to rotate:•Would the dipole end up attracted to the charge, repelled, or neither?Would the charge Q be attracted, repelled, or neither to the dipole?r• Do your answers depend on the sign of Q?Think of the dipole as lined up the way it's drawn below, so the -q end is a distance r from Q, and the + q end is a distance r + d.rdb) Write a formula for the total force on the dipole by the charge Q.c) If you were to double the distance r between Q and the dipole, what would happen to the magnitude of the force? Would it double? Cut in half? Something else?You could work on this in a couple of ways; pick whichever one you prefer. One way would be to try numbers. You could use the water molecule if you like, q = 6 x 10-19 C and d = 10-11 m. A reasonable amount ofcharge for Q might be 10-6 Coulombs, and r could be 1/2 cm = 5 x 10-3 m. Find F, and then double r. Or you could make up your own numbers - you could pick easier numbers - just be sure d is much smaller than r.The other, more advanced way is to simplify the formula you found in part b). A trick to make that easier is to ignore the term d ² you'll get along the way. Remember that d is a very small number compared to r, andso d² is a very very small number compared to r². When you're adding numbers, and you know one is tiny compared to the others, you can often ignore it. (As in 1,000,000 + 2 is about equal to 1,000,000.) So - atone point you'll have (r² + 2rd + d²), and you can ignore the d² - just cross it out. (Don't ignore the 2rd term, though! Why not?)* Water doesn't actually look like a pair of charges separated by a distance. It's more like a pair of "Mickey Mouse ears" with the oxygen atom being the head and the hydrogen atoms being the ears. The hydrogen atoms have excess positive charge andthe oxygen has excess negative charge. It turns out that the difference between a "two-ball-and-stick" dipole and "Mickey Mouse ears" becomes irrelevant to the electric field or forces calculated once the distance to the molecule is more than a fewmolecular diameters. Coulomb's law describes the force between two charges. We have written it as follows:FEQ-9=kcqQroa2(1)In various textbooks you will see this written in other ways: for exampleF=kF1 on 222.k992=(2)(3)A. For the expression (1) explain what each of the terms in the equation means and what they tell you about the electric force between two charges.B. Expressions (2) and (3) differ in various ways from expression (1). Explain some of these differences (at least one for each) saying what information is representeddifferently (or not represented) compared to (1). What do the vertical bars mean in expression (3)?C. Electric forces satisfy Newton's third law. Explain how this information is coded into expression (1).D. In the problem Analyzing dipoles, we discovered that the force between a single charge and a dipole doesn't behave like 1/r², but falls off like 1/r³ at long distances.Does this mean that Coulomb's law doesn't hold? When can we use it and when can we not use it?

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Analyzing dipoles (2015)last edited by Joe Redish 7 years, 7 months ago4.2.4.P4Page historyLots of things that are electrically neutral overall have one side that's electrically negative and one side positive (a water molecule, for example). We call such things &#34;electric dipoles,&#34; and we can model them as pairsof particles of charge + q and -q (where q is a positive number) separated by a distance d. Usually, d is a very small distance. (For water it would be around 10-11 m, thinking of a few protons worth of charge on oneend - about 6 x 10-19 C -- and a few electrons worth at the other.) Furthermore, because of the magic of quantum mechanics, in many molecules it behaves more like a rigid rod than like a soft spring. So we can treatd as a fixed distance.*d-9Suppose you have a dipole that's free to move in any way (including rotate - imagine it floating in space). And there's an object with charge Q a distance r away. That distance r would be much larger than d, thedistance between the charges of the dipole, so we draw the dipole small.a) Consider the forces between the charge Q and the dipole. If the dipole is free to rotate:•Would the dipole end up attracted to the charge, repelled, or neither?Would the charge Q be attracted, repelled, or neither to the dipole?r• Do your answers depend on the sign of Q?Think of the dipole as lined up the way it's drawn below, so the -q end is a distance r from Q, and the + q end is a distance r + d.rdb) Write a formula for the total force on the dipole by the charge Q.c) If you were to double the distance r between Q and the dipole, what would happen to the magnitude of the force? Would it double? Cut in half? Something else?You could work on this in a couple of ways; pick whichever one you prefer. One way would be to try numbers. You could use the water molecule if you like, q = 6 x 10-19 C and d = 10-11 m. A reasonable amount ofcharge for Q might be 10-6 Coulombs, and r could be 1/2 cm = 5 x 10-3 m. Find F, and then double r. Or you could make up your own numbers - you could pick easier numbers - just be sure d is much smaller than r.The other, more advanced way is to simplify the formula you found in part b). A trick to make that easier is to ignore the term d ² you'll get along the way. Remember that d is a very small number compared to r, andso d² is a very very small number compared to r². When you're adding numbers, and you know one is tiny compared to the others, you can often ignore it. (As in 1,000,000 + 2 is about equal to 1,000,000.) So - atone point you'll have (r² + 2rd + d²), and you can ignore the d² - just cross it out. (Don't ignore the 2rd term, though! Why not?)* Water doesn't actually look like a pair of charges separated by a distance. It's more like a pair of &#34;Mickey Mouse ears&#34; with the oxygen atom being the head and the hydrogen atoms being the ears. The hydrogen atoms have excess positive charge andthe oxygen has excess negative charge. It turns out that the difference between a &#34;two-ball-and-stick&#34; dipole and &#34;Mickey Mouse ears&#34; becomes irrelevant to the electric field or forces calculated once the distance to the molecule is more than a fewmolecular diameters. Coulomb's law describes the force between two charges. We have written it as follows:FEQ-9=kcqQroa2(1)In various textbooks you will see this written in other ways: for exampleF=kF1 on 222.k992=(2)(3)A. For the expression (1) explain what each of the terms in the equation means and what they tell you about the electric force between two charges.B. Expressions (2) and (3) differ in various ways from expression (1). Explain some of these differences (at least one for each) saying what information is representeddifferently (or not represented) compared to (1). What do the vertical bars mean in expression (3)?C. Electric forces satisfy Newton's third law. Explain how this information is coded into expression (1).D. In the problem Analyzing dipoles, we discovered that the force between a single charge and a dipole doesn't behave like 1/r², but falls off like 1/r³ at long distances.Does this mean that Coulomb's law doesn't hold? When can we use it and when can we not use it?

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