Recent content by FLms

Yes. The integral is actually:\int \Psi^{*} L_{z} \Psi \,d^{3}r = -i \hbar \int \Psi^{*} \frac{\partial \Psi}{\partial \phi} \,d^{3}r And, Y_{l,m}^{*} = (-1)^{m}Y_{l,-m} <L_{z}> = P_{m_{l} = 1} (m_{l} \hbar) + P_{m_{l} = -1} (m_{l} \hbar) <L_{z}> = \frac{2}{3}(1 \hbar) + \frac{1}{3} (0...

Homework Statement A particle is in a state described by the wave function: \Psi = \frac{1}{\sqrt{4}}(e^{i\phi} sin \theta + cos \theta) g(r), where \int\limits_0^\infty dr r^{2} |g(r)|^{2} = 1 and \phi and \theta are the azimuth and polar angle, respectively. OBS: The first...

Homework Statement What is the radius of convergence of the Taylor Series of the function f(z) = z cot(z), at the point z = 0? Homework Equations Taylor series is given by: \sum_{k=0}^{\infty} \frac{f^{(k)}(z_{0})}{k!} (z - z_{0}) And the radius R by: \lim_{n \to \infty}...

Yes, I found out that I was miss calculating the Divergence of the Electric Field. Too bad I was writing my other post while you answered it. Thanks anyway. By the way, about the Divergence of \frac{\hat{r}}{r^2}, it is equal to zero everywhere, except the origin. And in general form, it is...

How about this: \rho (\mathbf{r}) isn't exactly defined at the origin; it explodes at the point r = 0. So the integral would have to be from a small radius a to a general radius r \leq R Using this, I got: q_{enc} = 2 \pi A (r^{2} - a^{2}) The total charge, however, still is...

Anyone?

Taking r = a ; a < R, I get: q_{enc} = \iiint\limits_V \rho dV q_{enc} = 4 \pi A \int r dr q_{enc} = 2 \pi A a^{2} More generally: q_{enc} = 2 \pi A r^{2} And the total charge Q_{tot} is 2 \pi A R^{2} So: q_{enc} = \frac{Q_{tot} r^{2}}{R^{2}} Now: \oint \mathbf{E} \cdot...

Homework Statement Consider charge distribution \rho = \frac{A}{r} with spherical symmetry, for 0 \leq r \leq R, and \rho = 0 for r > R, and A is a constant. Find the Electric Field in all of space. Check your answer obtaining \rho from your answer. Homework Equations Gauss's law...

Ok, that was helpful. I guess I could use Newtonian Mechanics to find out that v^2 = 2 g R (1 - cos(\theta)), and then substituting in the equation, which gives me: \lambda = mg (3 cos(\theta) - 2) But, is there any other way to find v^2 or \ddot{\theta}^2? I.e., by using only the...

Homework Statement A heavy particle is placed at the top of a vertical hoop. Calculate the reaction of the hoop on the particle by means of the Lagrange's undetermined multipliers and Lagrange's Equations. Find the height at which the particle falls of. Homework Equations \frac{d}{dt}...

Homework Statement A particle of mass m1 and momentum p1 collides with a particle of mass m2 at rest. A reaction occurs, as a result giving two new particles, with masses m3 and m4, that are emitted at angles \theta_3 and \theta_4, in relation to the original direction of m1. Determine the...

I can't believe I missed a simple separated variables equation. Thanks for your help. The integral comes from the conservation of energy principle. T + V(x) = E \frac{1}{2}m v^2 + V(x) = E v = \frac{dx}{dt} = \sqrt{\frac{2}{m}} [E - V(x)]^\frac{1}{2} Then, you can find x(t) solving...

That integral came up wrong. x_0 and x should be the limits of integration.

Homework Statement What is the motion of a body thrown upwards from the Earth's surface, with escape velocity as it's initial velocity. Disregard the air resistance. Homework Equations v_e = \sqrt{\frac{2 G M}{x}} F_g = \frac{G M m}{x^2} The Attempt at a Solution I though this...

I used the trial x_{p2} = C e^{-a t} and got close to the correct answer. x_{p2} = C e^{-a t} \dot{x}_{p2} = -a C e^{-a t} \ddot{x}_{p2} = a^2 C e^{_a t} Substituting: a^2 C + 2 a^2 C - 2 a^2 C = \frac{-F_0}{m} C = \frac{F_0}{m a^2} So: x(t) = A e^{-at}cos(at + \theta) +...