Recent content by mitleid

I am quite amazed that worked... and also that superposition predicted the same thing. wow. Thanks very much! I can see clearly now... :)

okay... i hope this is closer... I2 = I3 - I1 E2 - R(I2) - R(I3) - E3 - R(I3) - R(I2) = 0 Plug the first into the second and add it to the first equation... E2 - 2R(I3) + 2R(I1) - E3 = 0 + E1 - 2R(I1) - 2R(I3) - E3 = 0 so E1 + E2 - 2E3 - 4R(I3) = 0 and I3 = -E3/2R + E1/4R + E2/4R but...

So more like this... All right. So I will separate into initial stage, stage 1 (after traveling distance L) and stage 2 (after traveling distance 2L). I need to know\Deltay at stage 1, but to get this I also need E1. v1y = - (qE1/2m)(L/vi) E1 = - v1y (2m/q)(vi/L) v1x = vi...

okay, so... I simplified the dual equations to E1 - 2R(I3 - I2) - 2R*I3 - E3 = E2 - 2R(I3 - I1) - 2R*I3 - E3 2R*I3 and E3 cancel out. I expanded to get E1 - 2R*I3 - 2R*I2 = E2 - 2R*I3 - 2R*I1 Again, 2R*I3 cancels out to get E1 - 2R*I2 = E2 - 2R*I1 so I1 = I2 - (E1 +E2)/2R. If I sub...

There we go. Since it enters and exits at the same height, and the distance along each plate is the same (L), \Deltay MUST be the same for each side, but opposite polarity. It could go up and then down (and have a negative y component velocity), or down and then up, which applies here...

I'm not actually certain how to continue with the equations I was using (having corrected them). Even considering my mistake, I've ended up with the same thing which is E1 - I1R2 - I1R1 = E2 - I2R3 - I2R4 From there I'm not sure what the point of substitution is... since the R6 term...

Ah, damn it... you're right. I got jumbled up in my professors notes and my own. It's been a long day... went ahead and changed the drawing i made to have the right direction for the current.

Okay, tryin' to follow you... For E3, R + R refers to the resistors on either side of the battery which are in series, and can just be added together. (2R*2R/2R +2R) refers to the parallel equivalent resistors on the other two branches. That makes sense... I'm not so sure how you're getting...

So when the current is going 'forwards' across an element, delV = -IR. The opposite is true if its going backwards. Across the batteries (E1-E3), delV is positive when going from the negative to positive side of the battery, and vice-versa. Accordingly, I've set up a couple equations which...

All resistors, R1 through R6, have the same resistance R. What is the power delivered by the batteries to R6? I worked it like this first... I saw two junctions, the nodes at the left and right where all three connections meet. I figured I1, I2 and I3 would meet here, combine, then...

I did this next... delV = - (Qfree/(4*pi*Eps)) [ \int^{a}_{b} (dr/(k1*r^2)) + \int^{b}_{c} (dr/r^2) + \int^{c}_{d} (dr/(k2*r^2)) The general form of the solution for int(dr/r^2) is -(1/r). This let's the pre-existing negative cancel out, and I just have to substitute to get the expression...

A spherical capacitor is made of two insulating spherical shells with different dielectric strengths, k1 and k2, situated between two spherical metallic shells and separated by a vacuum gap. Calculate the capacitance of this system. Total C = Q(free) / \DeltaV where \LambdaV is the...

Two adjacent parallel plate capacitors are used to deflect charged particles as shown in the figure. The relevant dimensions l and h are shown. The lower capacitor plates are grounded while the upper plates are kept at some voltages V1 and V2, which are to be found in this problem given the...

Ahhh, I remember thinking almost that exact thing when I first started the problem, but I got a bit confused (obviously). Thanks for clearing that up.

I was thinking this was more complicated... C1/C2 is proportional to C4/C3, so I just plug in the numbers and cross-multiply to solve. I should probably get a firmer grasp on why that is before the exam, though...