Recent content by betel

No. Each field has a certain value at each point of space time. For simplicity take the value one to indicate that there is a particle and 0 no particle. Then the idea that at any point only one particle is possible does not exclude the existence of many different fields. And a field by...

I see. Well, a field usually will not "occupy" space. They only assign a physical quantity to point of spacetime. As a classical example at each point you could have a value for the temperature, humidity, wind, etc. each being a different field. Now a quantum field will describe a certain kind...

Why should it not be possible?

So can you express \sigma in terms of Q_{total}? For the \rho part you have to take Gauss surfaces inside the sphere as Matterwave said.

I think you missed to post the important point after "fact that". I don't get much sense out of your sentence this way.

\Phi=\frac{q}{\epsilon_0}=\frac{3.1\mu C}{8.85418782 \cdot 10^{-12} \frac{As}{Vm}}=0.35 \cdot 10^6 Vm=3.5\cdot 10^5 Vm = 3.5\cdot 10^2 kN m^2/C I had calculated with milli Coulomb before.

No it is 3.5\cdot 10^{11} Vm = 3.5\cdot 10^8 KN m^2/C. And from the data you gave (charges 8 mC and -4.9 mC) this is correct. I agree with sgd on that.

sgd is right. Seems i wasn't really awake when I tried to convert the units. But converting the units will only change factors of 10.

Does the numerical value of the answer fit? Look up SI units. There is no way to convert kiloNewton* meter^2 / Coloumb to Volt * meter

This is a hypothetical question, when you drill a hole through the Earth's core.

This data might help http://geophysics.ou.edu/solid_earth/prem.html" Using this I get 37.25 minutes for a trip.

Welcome to the forum. First you should sketch the situation. Also include all forces acting on the balls.

Sure. I always like to help.

You can never get the unit you gave. The Vm sgd gave is the correct unit.

x,y,z. You just have to find the right combination.