Pranav-Arora and TSny,
Sorry I couldn't respond any sooner; quite a busy day I had today.
Thank you for your responses! Although we are supposedly still at rudimentary work, it's reassuring to know that the solution to a difficult problem is correct when I good critical reasoning of...
Homework Statement
A sphere of radius ## R ## carries charge density ## \rho = ar^5 ## where ## a ## is a constant. Find the flux ## \Phi ## of its electric field through a surface of a circle with radius ## R ## if the circle lies in a plane tangent to the sphere and its center touches the...
We are looking to illuminate a pre-mixed OH flame and collect the fluorescence emitted by the OH. This is a typical planar laser-induced fluorescence setup, which consists of an Nd:YAG, dye laser, and wavelength extender. The output is a ~300 nm laser beam with about. Take laser beam diameter is...
Ah, I have never heard of this set up, but it should be interesting enough to consider.
How wide is the image scanning range usually for an f{\theta} system? Does it only work for a specific wavelength as well as repetition rate of the laser, e.g. pulsed or CW?
Thank you so much for your input...
Hello,
I've been helping a graduate student out with his project which is planar laser-induced fluorescence. Part of the procedure involves generating a laser sheet with which to illuminate over the flame under study.
Now, typically a usual procedure involves using a beam expander to...
I wanted to point out, that nothing can go past the speed of light, in any inertial frame of reference. Either there must be something wrong in your calculations and the way you got to the correct answer or the correct answer you stated is wrong.
You don't have to. It only makes sense to do...
Indeed, I did not understand either when your professor said that there are contributions of acceleration from both rotation and gravity. Is rotation in this case not caused by the chimney's weight which is a gravitational force itself?
The lower segment of the chimney can be taken as a rod rotating about its fixed point on the ground. We have Newton's 2nd law for torques:
##Ʃτ=Iα##
##mgr=(\frac{1}{3}mL^2)α##
Assuming the center of gravity of the lower segment is taken to be at its geometrical center, ##r=\frac{L}{2}##...
Alright, so I set ##(Z-1)e=∫ρ_0e^{-\frac{r}{R}}(4πr^2dr)## from 0 → ∞ because I want to evaluate the total charge distribution.
I get ##ρ_0=\frac{(Z-1)e}{8πR^3}##.
Is this correct by far?
Yes, in part (b) when I apply Gauss' I will have to integrate the charge density with respect to ##dV## in order to obtain the total charge. And if I obtain the correct expression then my results should be consistent with part (c).
Nevertheless I still cannot be too sure for the expression...
Well, it's 4x smaller so what do you think?
I'm sure you can figure this out. Other than that there seems to be nothing else wrong with your calculations.