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anonymous
 one year ago
Calculating the potential due to a defect in cesium chloride crystal structure
anonymous
 one year ago
Calculating the potential due to a defect in cesium chloride crystal structure

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Okay, this is a boring problem :P.. more of geometry... which part are you stuck at?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I took a materials class so I understand what this is saying. But the part that gets me is setting up the reference potential infinitely far away.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Oh.. that is just so that you can use the formula \[V(r) = kQ/r\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0That is the potential due to a point charge at any distance r, from it.. that formula is true provided our reference potential is far away

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Okay, so the other point is going to be at any one of the Cs points, right? And then what is the distance refer to in this scenario?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0You need to calculate the potential at the point of crystal defect (one of the edges of the cube) due to all the eight charges (7 +e and one e charges) So you need to individually calcualte the potential due to each of the eight charges at that point of defect (the task is only to evaluate the distance of each charge from the defect point) and then add them algebraically

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Ah okay, and r is the distance between those points. That makes sense.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I'm assuming I'll have to do the same thing for part b). but use the equation for work? How would I do that given that W=Fd? Maybe use the equation for potential energy, because that would equal the necessary work energy to bring it together?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Work done to bring two charges q1 and q2 together is \[U = \frac{kq_1q_2}{r_{12}}\] if you have three charges then \[U = k (\frac{q_1q_2}{r_{12}}+\frac{q_2q_3}{r_{23}} +\frac{q_3q_1}{r_{31}})\] So if there are totally N charges you would have to do a huge summation taking all possible combinations

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0There would be \[N_{C_2}\] terms in it

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Oh boy.. Couldn't I multiply some of them by a certain factor? Like at a certain corner, there are 3 other corners that have the same charge and are equidistant from that empty corner. And there are also 3 more corners that are equidistant and share the same charge. Then there's that 1 furthest distance as well as the one that's in the center of the cube with the opposite charge.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Oh wait, I guess I could do that for a). but not b). because it's a combination of other charges, not just that point and the empty corner.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0You could do that for b also because some combinations do have the same distances.. but like i said, now everything else is geometry :P

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0@Mashy Okay, thanks! :P Also, I posted again on the thread with the cylinder
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