Recent content by BSJ90

I just need to add 90 degrees to my theta so it restricts it to my hemisphere not a whole sphere. The ds stays the same except for that sin(θ) term which i added.

I get my |R- R'| = \sqrt{ a^2 +z^2 + 2azsin(θ)}. look right? Umm this isn't turning out to be a easy integral. I am thinking mabye go back to Cartesian.

ok so should my theta be from the z axis and i think that changes my whole set up oh wait i can use 90 + theta and it would work. Ok thanks time to go to work on it. Plus i only really care about the z direction.

oh yeah evenly distributed on surface. I made a paint object to help clarify. R' is to the ds and R is to the point on the z axis. I know R has to be z but I am having a problem figuring out R'. Thanks for this and the last problem you help me with.

Homework Statement Given that the hemisphere has a charge +Q distributed through its surface with radius a. Find the electric potential on any point on z axis (the plane of the hemisphere is oriented in positive z direction).Homework Equations \phi = \int\frac{kQ}{|R-R'|}*ds (surface integral)...

So how would i go about solving the problem from this point on. I mean i get a torque vector and obviously there is an answer (0.296 N * m) but I don't see how we can get from the vector field to this answer.

Now how am I supposed to work with that. It makes sense but now I'm integrating a vector field and the only thing i can think to do is take the magnitude of that. Shouldn't it be different because the of symmetry. Plus i tried taking the magnitude of that field and it was not fun by the time i...

Homework Statement A cylindrical coffee cup (8 cm in diameter and 10 cm tall) is filled to the brim with coffee. Neglecting the weight of the cup, determine the torque at the handle (2 cm from edge of cup 5 cm up from bottom of cup). The easy way would be to just use the center of mass of the...