Recent content by JHans

I'm sorry, but I'm having difficulty "seeing" this complete solution and how to work it. I understand what was said earlier, that: {\sqrt{\frac{\lambda}{\pi}} \int_{-\infty}^{+\infty} \! {x}{e^{-\lambda{(x-a)^2}}} \ dx} = {\sqrt{\frac{\lambda}{\pi}} \int_{-\infty}^{+\infty} \...

Is anyone able to help? This is seriously stumping me.

Now I'm a bit confused, though... I came across an integral on Wikipedia's list of integrals for exponential functions, and they have this: \int_{-\infty}^{\infty} \! {x}{e^{-a{(x-b)^2}}} \ dx={b\sqrt{\frac{\pi}{a}}} How does the integral we've broken into two parts become that?

I'm sorry! Here's the process I went through: {\sqrt{\frac{\lambda}{\pi}}\int_{-\infty}^{+\infty} \! {a}{e^{{-\lambda}{(x-a)^2}}} \, dx} Because a is a constant, it can be taken out of the integral: {{a}\sqrt{\frac{\lambda}{\pi}}\int_{-\infty}^{+\infty} \! {e^{{-\lambda}{(x-a)^2}}} \...

I hadn't thought of that! That's a clever trick. Am I correct in thinking the second integral simplifies to: {\sqrt{\frac{\lambda}{\pi}}\int_{-\infty}^{+\infty} \! {a}{e^{{-\lambda}{(x-a)^2}}} \, dx}={a} And so the integral: \sqrt{\frac{\lambda}{\pi}}\int_{-\infty}^{+\infty} \...

Hey, all. I need help computing a particular integral that's come up in my quantum mechanics homework (Griffith 1.3 for anyone who's interested). It involves integration of a gaussian function: \sqrt{\frac{\lambda}{\pi}}\int_{-\infty}^{+\infty} \! {x}{e^{{-\lambda}{(x-a)^2}}} \, dx Where...

Okay... I may have come to an equation for the acceleration of a particle in a cyclotron... I'm a little dubious of its validity, though. A charged particle in a cyclotron experiences a force (specifically, a centripetal one that does not alter speed, but does alter velocity) of magnitude...

The electromagnetic power radiated by a nonrelativistic particle with charge q moving with acceleration a is: P = \frac{q^2 a^2}{6 \pi \epsilon_0 c^3} If a proton is placed in a cyclotron with a radius of 0.500 m and a magnetic field of magnitude 0.350 T, what electromagnetic power...

Also, I apologize. The term that comes after e should be an exponent. That is, it should be e^((-Rt)/(2L)).

I'm sorry, can you explain a bit further? The equation that I have is: I = I_i e^(\frac{-R t}{2L}) (\omega_d sin(\omega_d t) + \frac{R}{2L} cos(\omega_d t)) How is it that I can convert the sine and cosine terms to a single cosine term? Is there a trigonometric identity that I'm...

Alright. I'm a bit confused, though. I've stated that the current is: I = I_i \frac{e^(\frac{-R t}{2L})}{2L} (2L \omega_d sin(\omega_d t) + cos(\omega_d t)) I could also have stated it as: I = I_i e^(\frac{-R t}{2L}) (\omega_d sin(\omega_d t) + \frac{R}{2L} cos(\omega_d t)) Which...

Electrical oscillations are initiated in a series circuit containing a capacitance C, inductance L, and resistance R. a) If R << \sqrt{\frac{4L}{C}} what time interval elapses before the amplitude of the current oscillation falls to 50.0% of its initial value? b)Over what time interval does...

Hmm... well, since the infinite sheet of current is along the yz-plane, with current moving in the positive z-direction, it would seem to make sense to construct a rectangular loop along the xz-plane. Then the vertical sides would have the magnetic field perpendicular to the direction of...

An infinite sheet of current lying in the yz plane carries a surface current of linear density Js. The current is in the positive z direction, and Js represents the current per unit length measured along the y axis. Prove that the magnetic field near the sheet is parallel to the sheet and...

Oh, I see now! Everything makes sense. I checked it, and my equation for modeling the velocity as a function of time works as well. It's a little bit messier, but it can be cleaned up at any point in time by substituting for B: B = \frac{2 \pi m}{T q} The end result is the same...