Recent content by victorializ

i honestly have no clue what that is, i don't think we have gone over that yet. i was just assuming this was a gaussian surface of some kind maybe because it's enclosed?

i honestly have no clue what that is, i don't think we have gone over that yet. i was just assuming this was a gaussian surface of some kind maybe because it's enclosed?

Homework Statement A sealed metal can is placed in a uniform external electric field of 10 N/C . The field points along the +x direction. The can is 20cm in length and 10cm in diameter. What's the value of the electric field at the center of the can? Homework Equations The Attempt at...

Homework Statement A Gaussian Sphere with a radius of 1m surrounds an unknown charge at the center. At this surface a uniform outward directed electric field is 1 N/C. Use Gauss' Law to calculate the amount of charge enclosed by the sphere. Homework Equations E = q/4∏εor^2 The Attempt at a...

oh alright i thought that the permitivity of free space had to be used when a dielectric was present only but that works thanks!

so U =1/2 Q^2/C i'm still not sure how i would find capacitance without using the C=Q/V and not having a voltage? also does the distance between the plates changing not matter?

Homework Statement If you increase the charge on a parallel plate capacitor from 3μC to 9μC and increase the plate separation from 1mm to 3mm, the energy stored in the capacitor changes by a factor of... Homework Equations U = 1/2QV or 1/2CV^2 The Attempt at a Solution I'm not sure which...

oh my goodness so most of the information in this problem is irrelevant? all i need is to do a proportion between 1A and the 38 mA in the problem?

so i need to use the fact that 6.241 x 10^18 electrons/sec pass through a circuit which is 1 A  when i use that number though i still have it wrong so I'm not really sure where else my calculations are messed up.

Homework Statement The resistivity of gold is 2.44 x 10-8 Ω m at a temperature of 20°C. A gold wire, 0.5 mm in diameter and 44 cm long, carries a current of 380 ma. The number of electrons per second passing a given cross section of the wire, is closest to: A) 2.4 × 10^18 B)...

oh wow okay that makes so much more sense ! thank you so much for all your help!

yes it was sorry haha. that makes sense thank you! so why exactly don't you multiply it by the e constant?

so rather than multiplying it i would just leave my answer in joules but basically multiply by -1 because of -e which would leave me with about +920eV?

oh yeah that makes much more sense, thank you!

i just edited and added the rest of my calculations in the original thread post! i must be doing something wrong . the answer is supposed to be D) +920 eV