Recent content by CARNiVORE

I'm interested in quantum physics (specifically quantum computing) lately

Oops, brain fart - Fm = iL x B, so Fm/L = i3*B(tot). B(tot) is equal to (μ0*√i1^2+i2^2) / (2PI*d), so Fm/L is equal to i3 times this. For c, if the angle is to be 45 degrees, then tan^-1(i2/i1) = 45 degrees. This would mean that the ratio of i2/i1 is one to one, so i1 would have to be the same...

Homework Statement Homework Equations B = μ0*i/(2PI*r) The Attempt at a Solution First of all, if you're looking at this, that means you intend to help me study for my upcoming physics exam on a Friday night. If you fit into this category of people, then god bless your soul. I should state...

Awesome! That was my last question, and now homework is finally over for the week... time to get started on next week's, lol. Thanks again, gneill!

I see what you mean - in this case, I get 3 independent equations: 0 = E1 + E2 + E3 2.8 * 10^5 = E1 - E2 - E3 8.4 * 10^5 = E1 + E2 - E3 Using these three equations, I found that E2 = 2.8 * 10^5 and E3 = -4.2 * 10^5. Is this correct?

All right, so I ended up with four equations. One for the space to the left of 1, one between 1 and 2, one between 2 and 3, and one for the space to the right of 3. Here they are: E1- = -E1 - E2 - E3 E12 = 1/3 * E23 = E1 - E2 - E3 E23 = E1 + E2 - E3 E3+ = E1 + E2 + E3 Combining equations 2 and...

Good idea - here's the sketch I drew: Now, I know the lengths of the arrows I drew don't necessarily reflect their magnitude, because of differences in charge density and E-field. They're just convenient to display the distance between them. It's useful to see them out like this, but I'm...

Not quite sure if I follow - I see that sheet 1 is pushed to the left by sheet 2 and slightly to the left by sheet 3, sheet 2 is pushed left by sheet 3 and right by sheet 1, and sheet 3 is pushed to the right by sheet 2 and slightly to the right by sheet 1. This, of course, is assuming that they...

Homework Statement Figure (a) shows three plastic sheets that are large, parallel, and uniformly charged. Figure (b) gives the component of the net electric field along an x axis through the sheets. The scale of the vertical axis is set by Es = 8.4 × 105 N/C. What is the ratio of the charge...

Understood - thanks so much, I got the question right and I really feel like I understand the subject more because of your help. It took me a long time to get this one, but thanks to your efforts, I finally got it. You're the man (or woman), gneill :)

Thanks so much for your help again! This was really difficult for me - I'll be sure to study up on questions similar to this in my textbook. I just have one question left, though - you say that, in some other situations, one of the fluxes could be entering while the other is leaving. If the...

So I had it backwards, then? The net charge should be positive 1.4768E-10 Coulombs rather than negative. Instead of negating the value at x=0, I should've negated the value at x=-1.8, which makes both values positive. That actually makes sense, since the flux comes from within the cube. I'm not...

I think I see what you mean - since the field leaves the cube from the yz square at x=0: Σq = ε0 * ((2.86x + 3.82)*(1.8)^2 - (3.82)*(1.8)^2) Σq = ε0 * (-4.30272 - 12.3768) Σq = ε0 * (-16.67952) Σq = -1.4768E-10 This makes more sense, actually - what do you think?

All right, I see what you mean, and I think I've finally got it: Σq = ε0 * E∫dA Σq = ε0 * ((2.86x + 3.82)*(1.8)^2 + (3.82)*(1.8)^2) Σq = ε0 * (-4.30272 + 12.3768) Σq = ε0 * (8.07408) Σq = 7.1488E-11 How's that look? I'm actually confident with this answer.

Will do: ε0 * ∫E*dA = Σq ε0 * (2.86x + 3.82) * ∫dA = Σq ε0 * (2.86x + 3.82) * x^2 = Σq At this point, I tried substituting my values in for x. I feel I'm missing something simple.