I figured it out!
I tried to do it without vectors and that just does not work ;)
If you do it with vectors and since it is symmetrical, you can calculate the forces acting on on point charge. The fifth point charge needs to be in the center.
Mh, my idea is that you can consider the four charges as one since the electric field lines between them equal out, so that only the lines on the outside of the square act on a point charge.
The force caused by the four charges that acts on the fifth charge has to be a great as the force that...
Hello,
I am doing electrostatics at the moment and have difficulties to solve the following problem. Any hint that helps me to find the answer will be appreciated!
"On every corner of a square are movable point charges with the charge of 'q'. Where does a fifth movable point charge have to...
Thank you very much for your help!
Because of you I could finally solve it! I was thinking too complicated I guess.
Here is how I did it:
\omega_p =220 s^{-1}
\omega_L =58.3 s^{-1}
The angular velocity of the loop is constant, the propeller's isn't. It depends on the plane's location within...
I have not been able to solve it. I just cannot see why the angular moment of the propeller should change since it has nothing to do with the circular motion (the loop). The only thing that changes, at least as far as I get it, is the plane's horizontal distance to the center of the loop. But I...
Hello!
This physics problem seems unsolvable to me. I have been trying to figure it out myself but without any success. Maybe I am missing something obvious or I misunderstand the problem. I would be very happy to have any help:
The motorshaft and the popeller of a single engine plane have...
Hello there!
This problem is giving me a hard time and I thought maybe one of you could give me a hint:
"A ship is going upstream with a constant power P. Its speed V is relative to the water and the water's speed U is relative to the shore. The ship needs tp overcome a frictional force...
I think I have got it!
Since the electric field is uniform, the equation to get the electric field is:
E= \frac{U}{r} where r is the distance to the particle.
r_{tube}=0,012m and r_{wire}=3 x 10^-5m
E_{tube}=\frac{1000V}{0,012m}= 83333.3 Vm and for the wire
E_{wire}=\frac{1000V}{3 x 10^-5m}=...
I am really sorry but I am not making any progress. If the field is uniform, it does not change its intensity in respect to distance, right ?
What am I missing?
mh, I guess that in the tube's center there is no field at all since all the field lines eliminate each other in the center. So only the wire's field is acting?
The wire can also being considert a point with the electric field of
E=\frac{q}{4\pi e} \frac{1}{r^2}
Am I getting closer?