Recent content by ranger281

How can we determine the direction of dE (or E) in this example?

I have one more (related) question. In the attached exercise, to obtain ##E_{z}##, ##E## is multiplied by ##cos\psi##. Why is it so?

This I understand. But how can we obtain that by calculation? In this case, the total field (counting only the vertical components) is ##\vec{E} = \int_{S}^{}\frac{\lambda* cos\theta\hat{z}}{R^2+z^2}ds## (S being the ring), where we can simply multiply the integrated function by ##2\pi R##. Why...

Yes, the horizontal (radial) component of the field above the center.

The contribution coming from a little segment of the ring is ##d\vec{E}=\frac{dQ}{r^2}cos\theta \hat{z}##, assuming that the horizontal components cancel out. But how can we show that?

Where does ##do_{0}/4\pi## come from? Thank you for help.

I'm reading Mechanics by Landau and Lifshitz, chapter IV, and trying to understand how in a (closed) center of mass system, with randomly distributed and oriented particles that disintegrate, "the fraction of particles entering a solid angle element ##do_{0}## is proportional to ##do_{0}##, i.e...