Recent content by twisted079

Ok I figured it out using a cheap (but valid) math trick... in case anyone is wondering... xƩan(x+1)n = (x+1-1)Ʃan(x+1)n Now the x+1 can be distributed to give -Ʃan(x+1)n+1 ...anyone care to double check me on this?

So the differential equation I have to solve using power series is y''-xy=0 when x0 = -1 So i set it up Ʃ(n+2)(n+1)an+2(x+1)n - x Ʃ an(x+1)n I know how to generally solve equations like this, but I never solved one like this, where I have to distribute the x ... x(x+1)n ... I just...

Homework Statement An aluminum ring of radius r1 and resistance R is placed around the top of a long air-core solenoid with n turns per meter and smaller radius r2 as shown in Figure P31.7. Assume that the axial component of the field produced by the solenoid over the area of the end of the...

Right above where you enter these messages is a few options involving bold, italics, etc... if you look over more to the right you will see X2. Or you can type [.SUP.] "what you want supscripted" [./SUP] without the "." in the brackets.

Thank you so much! I knew I was using ln and e wrong (I was hoping it wouldn't come back to haunt me, but it found a way). Just so much to remember in so little time.

Here is what I did: 4 = 10(1-e(-3/RC)) 6/10 = e(-3/RC)) Here is where I think I lose it 6/10 = -ln(3) -ln(RC) My algebra is rusty :(

Homework Statement A 10.0 uF capacitor is charged by a 10-V battery through a resistance R. The capacitor reaches a potential difference of 4.00-V in a time interval of 3.00 s after charging begins. Find R. Homework Equations C = Q/ΔV q = εC(1-e(-t/RC) The Attempt at a Solution...

Thank you for taking the time out to reply, your answer helped clear my misconceptions.

I have a question about the ratio test. Suppose it proves inconclusive, we must than use another test to check for conditional convergence - 1) this test has to be associated with an alternating series, such as the Alternating Series Test, correct? (we wouldn't be able to use something like...

Homework Statement The electric potential immediately outside a charged conducting sphere is 250 V, and 10.0 cm farther from the center the magnitude of the electric field is 440 V/m. (c) Determine all possible values for the radius of the sphere. (Enter your answers from smallest to...

Ah okay, and the direction would be in the negative x-axis. Thank you so much for your help!

So I would have E = keλ(-1/x0 - 0) --> -(keλ / x0) ?

Here is what I have come up with so far: dq = λdx dE = keλdx / x2 E = keλ 0∫∞ dx/x2 E = keλ(-1/x) from 0 to ∞ Now, when I plus 0 in, I run into a problem

Ok, so this is my idea of what the solution would look like: (keQ) / L 0∫infinity dx/x2 ---> 1/x2 dx becomes -(1/x) which puzzles me because this leads to no solution except keQ/L (L or x) ... Please help!

But λ = Q/L, or in this case would it be Q/x? Edit: What about the constants that I could pull out of the integral?