Recent content by MissPenguins

Alright, thank you very much. My professor will probably explain it tomorrow. Thanks everyone.

Is it calculating the area of the cube? Can you please explain the concept? I am watching a youtube video on electric field now. Thanks.

So electric flux = E dot dA?? Can you please explain the concept? Thank you!

I figured the problem out. Thank you very much everyone!

You said E is different for part 2 since the charge density changed. So how do I find E? I attempted to do (0.25E-6)/(2*8.854E-12) = 14117.91281 for E, is this right for E? Then use (14117.91281)(2*8.854E-12) = 2.5E-7? It doesn't make sense. Can you give me more hints? Thanks.

The previous one wasn't a mistake. It's qE = mgtan(77) = 0.042448 I already did this \sigma = (E)(2Eo). So (0.042448)(2(8.85X10-12) = 7.5133X10-13. I converted back to uC = 7.51633X10-7. And it is wrong! :(

Homework Statement A closed surface with dimensions a = b = 0.294 m and c = 0.3528 m is located as in the figure. The electric field throughout the region is nonuniform and given by \vec{}E = (\alpha+\beta x2)ˆı where x is in meters, \alpha = 2 N/C, and \beta = 4 N/(Cm2). See figure...

Yea. tan(77) = (qE/mg) which is qE=mgtan(77) = 0.04244 I am not sure what other charge density do I have to use. E0is a constant, that's given. PLEASE HELP! Same question as this one https://www.physicsforums.com/showthread.php?t=109643 Except part 2 was not explained.

Homework Statement A charged mass on the end of a light string is attached to a point on a uniformly charged vertical sheet (with areal charge density 0.25 μC/m2) of infinite extent. Look at attachment for diagram Find the angle \theta the thread makes with the vertically charged...

I submitted the answer, and it's wrong! :(

So is my approach correct?

Homework Statement A wave pulse traveling along a string of linear mass density 0.0043 kg/m is described by the relationship y = A0 e −b x sin(k x − ω t) , where A0 = 0.0032 m, b = 0.68 m −1 , k = 0.57 m −1 and ω = 44 s −1 . What is the power carried by this wave at the point x = 2.2...

Can you please tell me where did you get 4.375 s from? thanks. Nvm, I got it w = sqrt of k/m. Hehe, I got the right answer too, thank you very much.

Alright, I figured it out and got the right answer. Thanks. ;)

This part is wrong. Remember 1 rev = 2pi. So you multiply by 2pi, and then 1 min = 60 seconds. So you divide by 60 seconds. I did not get 73 rev/s. I got another number.