Recent content by Redfire66

Homework Statement A closed rectangular surface with dimensions a = b and c where the faces perpendicular to the field are a*b. The left edge of the closed surface is located at position x = a, for c > a.The electric field throughout the region is nonuniform and given by E = 3*x xhat N/C...

Yeah I sort of thought I didn't explain it properly. Anyway R would be the gas constant, nitrogen was used so using the table it was about .2968 I believe. I don't really care about the values, I know it's not calculations. I just want to know what this formula is PV = RT, since there's no...

Homework Statement I'm given a initial and final pressure and temperature of an ideal gas to solve for the work done after it expans in a polytropic process (n=1.2) Homework Equations W = integral of P*dV PV = nRT PV = RT* PV = mRT PV^n = Constant The Attempt at a Solution I get W = integral...

Oh okay, I understand. I misunderstood the whole thing. I assumed that there was a force with no magnetic field which seemed kind of weird to me. Thanks for clarifying.

I'm just really confused with this right hand rule even though it seems like it's supposed to be pretty simple and obvious. I know the rules, however it doesn't really work for me. Maybe I'm doing something wrong. I have three examples I'd like to cover which are related to each other, I hope...

Alright, that makes sense to what I also thought. Thanks for your help

Oh I see now, after thinking about it a bit, if the there is no charge inside then I can just take into account the Gaussian area and that would be the radius hence which you stated that r = R right? Then the equation would then cancel out and I would have KeQ/R^2 where R is the radius of the...

So let r be the radius of the Gaussian sphere And I have r < R. When R is the sphere of the ball. If I take the integral I have E∫dA = q/ε Then ∫dA = 4*pi*r^2 and p = qV1...... V1 in this case would be the volume of the sphere E (4pi*r^2) = pV/ε E = p(4/3)(pi*r^3)/[ε(4pi*r^2)] and we know that p...

Thanks I do understand now how the equation works out if r < R of the sphere.. however now it makes more sense to me than if the Gaussian radius is outside the sphere. Can you explain a bit more simply put why r = R if r > R? It doesn't seem right since I always thought the further away you are...

This link is where I saw it http://hyperphysics.phy-astr.gsu.edu/hbase/electric/elesph.html#c3 Also my textbook showed me how to derive it, so I do know how to derive the formula but I don't understand why it's right. "For a radius r < R, a Gaussian surface will enclose less than the total...

Thanks for the reply. This is starting to make sense to me. However when I'm reading through my textbook it states that if the electric field is outside the sphere (r > R) then the field is written as E = KeQ/r^2 However if r < R, meaning inside of a sphere the field would be E = KeQr/R^3 I...

Sorry I was thinking about something else; yes I meant outside the sphere where r > R

Homework Statement So I was given the equation that in a sphere, if at a distance r outisde a sphere of radius R (The sphere insulated, I guess it matters since it would prevent charge to leave?) The electric field E = kqR/r^3 I'm not sure how they got this Homework Equations Flux = Integral...

Okay thanks I'll look into it. Edit: yeah I actually meant 2pir/w instead. Mistyped

Homework Statement Two men with masses 70 kg and 120 kg rotate at 1 rpm on a frictionless surface and are attached by a 15 m rope. If they pull the rope so that only 10 m is between them when they rotate, how long does it take to make 1 revolution? Homework Equations Angular Momentum and...