Recent content by Chetlin

Hi everyone, Would anyone be able to give me a few examples of places where units with non-integer exponents come up? I know of a couple right now: noise voltage density (?) in electronics has units \text{V}/\text{Hz}^\frac{1}{2} and the statcoulomb has equivalent units...

Thanks for the link! So it all just comes from Snell's Law? That page doesn't have any example of a case where the interface is convex but the image is virtual, however this must happen at least sometimes, since the thin-lens formula is derived from the first interface giving a virtual image...

Homework Statement If you have a spherical interface between two different "media" (like air and water), and an object is placed in the one with the lower index of refraction, with the interface being convex toward the object, how can you tell if the image will be real or virtual? Here's a...

Yeah, this was an old exam that they gave us, so I was pretending I only had the resources I'll be provided, which doesn't seem like it includes a periodic table. We are given the mass of a proton though, so I'll just memorize the four most common that we seem to get (1 proton mass for hydrogen...

Homework Statement On a planet not entirely unlike earth, the ratio of the partial pressure of N2 to that of O2 equals 1 at an altitude of 1 km: \frac{p_{\text{N}_2}}{p_{\text{O}_2}} = 1. Assuming that T = 200 K, and the gravitational constant is 5 m/s2, what is the ratio...

Thanks, that helped a lot! :D I got the right answer. I set the voltages of the two capacitors equal to each other: V3 = V4 so \frac{Q_3}{C_3} = \frac{Q_4}{C_4}. Then, I also saw that the initial charge on C3 was 400 μC and C4 was uncharged. This would distribute over the two capacitors, so...

Homework Statement This is the circuit diagram I am working with: There is actually a switch under C3 that I didn't draw. Initially, the switch was set so that the red dashed line was part of the circuit and the green dashed line was not (so C1, C2, and C3 charged but C4 did not). Then...

All right yeah, I did have an issue with using a volume integral, since it seems like I still have to use all coordinates. Also, instead of considering the charge dQ at (0,0,z), I thought you had to use (0,0,z′) or some other variable that was not the same as the axis. But since you're...

Homework Statement Charge is distributed uniformly along an infinite straight line with density λ. Develop the expression for E at the general point P. Homework Equations The electric field at a point P, caused by N point charges Qi, each a distance ri from P, is given by \mathbf{E}...

It's a unit vector, so it's magnitude has to be 1, so you have to normalize it too by dividing each component by its magnitude. You'd get \hat{r} = \left\langle \frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}}\right\rangle. The purpose of a unit vector is to provide the direction of something, so you...

Yep, that unit vector is just the vector r2,1 divided by its magnitude ( \hat{r}_{2,1} = \frac{1}{r_{2,1}}\mathbf{r}_{2,1}, where r_{2,1} = ||\mathbf{r}_{2,1}|| = \sqrt{{r_{{2,1}_x}}^2 + {r_{{2,1}_y}}^2 + {r_{{2,1}_z}}^2}). Sorry for the really weird notation inside the square root sign...

Well, I have my question answered, but to be complete, here is the text of the problem. It's from an old Schaum's Outlines book from the early 1980's. Point charge Q1 = 300 µC, located at (1, −1, −3) m, experiences a force F2,1 = (8, −8, 4) N due to point charge Q2 at (3, −3, −2) m...

That's true, but \hat{r}_{2,1} is a vector, and I would want that to be on the other side as well. I just realized what I tried to do is impossible, because it would result in a scalar on one side of the = sign, and whatever you would (imaginatively) get if you divided two vectors, on the other...

Homework Statement I have the equation: \mathbf{F}_{2,1} = \frac{Q_1 Q_2}{4 \pi \varepsilon_0 {r_{2,1}}^2}\hat{r}_{2,1} (standard electric force equation for 2 charges) I know the value of everything except Q2 and have to find it. The vectors each have 3 components. Normally in an...

Thanks! I understand this a lot more now. That's something I considered too, setting c = 2 or something. It seems that if you do that, you can't express p and m with the same units now. The units for p become eV/2 and those for m become eV/4. The only way I can see to work with these...