This homework deals with the subject of electric potential and electric potential energy. It will help you, hopefully, to understand these topics and to distinguish between potential and potential energy, a distinction which mostly has to do with units and meaning. The units of electric potential energy are typically Volt-Coulombs, which are equivalent to Joules. the units of electric potential, on the other hand, are simply Volts. So they differ by a multiplication by a charge of 1 Coulomb.
There is another common and "natural" unit used for electrical potential energy, which is the electron volt (eV). That is what you get when you multiply volts times the charge of an electron. since an electron has a charge of 1.6 x10^-19 coulombs, 1 eV is equal to 1.6 x10^-19 Joules (That is, 1 eV is equal to 1.6 x10^-19 Volt-Coulombs.)
Electron volts are often used to talk about atomic molecular systems, where electron transitions and ionizations typically involve energies of order 2 to 20 eV.
Photons of visible light have energies between about 2 and 4 eV. UV photons have more energy, say 4 to 20 eV. The ability of UV photons to damage cells is associated with their higher energy and the electronic processes that that allows.
Back to the homework, I thought that problem 22.30 was/is an interesting and informative problem, but part C-- the way they express their answer-- seems cryptic. So I'll just tell you: the answer is 180. (You tell me what that means. I mean, it does mean something, once you see it you might know what it means, but it might be difficult to realize in advance that that's what they want.
-------
additional commentary, added October 6:
for problem 54* ( Find the potential 16
I believe they expect you to use an approximate equation for the potential associated with a dipole, which is reasonable (valid) when the distance away from the dipole is much greater than the separation between the two charges of the dipole. (Note how they say: “the dipole separation is much greater than…”.) This equation involves cosine(theta), and is valid even when theta is greater than 90°. I'm not a big fan of using this equation in an intro course, because it engenders a plug and chug mentality, so probably on tests you would be given the dipole separation and the magnitude of the individual charges and asked to find the potential, or field, at particular points in space or, as a function of x or y along a symmetry axis.
However, I realize that later in this class we will need to understand cosine and sine functions, including their behavior beyond the intuitive 0 to 90° range. so let's review that here.
For theta in the range from 0 to 90°, the cosine can be thought of in terms of adjacent/ hypotenuse ratio with reference to the standard picture of a triangle and an angle. ( sketch that now) cosine starts at one, at theta = 0, and reaches zero at theta = 90°. Beyond that it goes negative, it returns to zero at theta = 270°, and returns to one at 360°. Cosine has a minimum value of -1 at theta = 180°. (graph that now...). The sine function is similar, and also different. It starts at zero and grows to a maximum of one at 90°. It is one when the cosine is zero, and zero when the cosine is one. In that sense they are complementary. Well, I guess that's enough about sine and cosine for now. Please feel free to ask any questions here.
Hi Professor Schlesinger,
ReplyDeleteI'm currently stuck on problem 22.39.
I'm just confused on how to get units to work out.
I started out with the 20 Nm/c and then thought to multiply by
1.6x10^-19c for the charge of electrons, which then gives me Nm. I believe I can convert these to Joules which gives units of Kgm^2/s^2. So then at this point I thought you could probably divide by the mass of an electron which would leave units of just m^2/s^2, but then at this point I didn't know how to get rid of the squares to get m/s. Could I divide by something like speed of light or something? It would give me m/s but I didn't see why I would be allowed to do this. Thanks for your help
Gregory: your method seems totally sound. Let me try to explain what's behind your method and then it will be clear what to do.
ReplyDeleteYou are using conservation of energy. Essentially, one wants to set the increase in kinetic energy equal to the loss, or diminishment, of electrical potential energy.
Thus one obtains: m v^2/2 = q V .
What you did with the units looks perfect to me. What you need to do next is take a square root. Does that make sense? Do you see why?
yeah I checked that kinetic energy is represented by (mv^2)/2 so then I set up my equation as ((9.11*10^-31kg)*(v^2))/2=(1.6*10^-19c)*20Nm/c. When I rearranged my equation I got
ReplyDeletev=sqrt(2qV/m) Is this right? I tried it on mastering physics and it's telling me it's wrong. Is there a unit conversion I need to do?
I believe if you are using kilograms, meters, Volts and Coulombs, that should work out fine. A V-C is a Joule=kg m^2/s^2
ReplyDeleteThe potential difference given in the problem is in kV so you would have to convert it to v.
ReplyDeleteThe book says that the potential energy difference is negative work (delta U= -W). And since delta V= delta U/q, wouldn't delta U= -W/q? I got a couple problems wrong because my answer was negative...I learned, but I don't understand. Is it because they only want the magnitude (the unstated rule)?
ReplyDeleteFor problem 22.39, I also get two different answers from using the unit method and the KE=loss of electrical potential energy equation. The difference is in the dividing by 2 (KE=mv^2/2). What does this mean? And which one is right?
ReplyDeleteWhen problem 22.54 tells us the point is 12cm from the dipole moment, does that mean 12cm from in between the two charges?
ReplyDelete"does that mean 12cm from in between the two charges?"
ReplyDeleteYes, I think so.
ps. I will think about your other questions.
Emily: good point.
ReplyDeleteThis comment has been removed by the author.
ReplyDeletetry getting rid of the micros. they are each 10^-6 factors. when multiplied together the make 10^-12...
ReplyDelete