Recent content by jumpman11372

bump please :]

I got an answer of 1 microFarad, or 1 x 10^-6 Farads. It seems that this is a quite simple answer, but WHY is this the answer? What leads the capacitance of this whole series to be 1 microF?

Ok, so we have one Capacitor with Ceq = C+D (since one capacitor and D are in parallel) Then, there are 3 capacitors which are in series with each other. Then here is where confusion steps in... (1/C) + (1/C) + (1/C+D) = 1/D or (1/C) + (1/C) + (1/C+D) = D I see that this will...

sorry for this... bump perhaps? :(

Thanks for the quick reply! Well, I've made this quick picture of what you've typed out: http://img158.imageshack.us/img158/3674/pic002311.jpg" So adding this link in this infinite chain will not affect the total capacitance D, where E=D. Correct me if I am wrong, but the capacitance...

Homework Statement The capacitance of each capacitor of the infinite series shown in the picture is C = 1\muF. Find the total capacitance between points a and b. IMAGE: http://img61.imageshack.us/img61/3674/pic002311.jpg" (continues to infinity) Homework Equations In series, (1/Ceq) =...

suggestions perhaps? :cry:

So what can be done about this problem then? I was so sure that was the right answer, it seemed right. I have no clue on how to come up with anything else.

According to my professor, that answer is wrong... He went on about how "the closest the particles will get is ALMOST d, and your answer is wrong" I had no clue what he was talking about, but he said by answer was wrong... He told me to try again.

v'(t) is acceleration, but I meant what is the formula for v(t)? According to the conservation of energy formula we had \frac{MVinf^2}{2} = \frac{KQ^2}{R} + \frac{MV(t)^2}{2}, v(t) = 0 when the potential energy of the two particles is equal to the kinetic energy of Particle1, which is moving...

I never thought this problem would be this difficult. I guess I'm just weak @ physics and calculus :/ \frac{d}{dt}\frac{2KQ^2}{M(Vinf^2 - V(t)^2)} = (I hope I did this right :X) \frac{-2KQ^2 *2V(t)*V'(t)}{M(Vinf^2 - V(t)^2)^2} What exactly is v(t)? How is this relating to finding the closest...

I feel so stupid right now...:/ I'm glad you're persistent and know your stuff :D r'(t) = 0 when the distance between the particles is a minimum. I found r(t)= \frac{2KQ^2}{m(Vinf^2 - V(t)^2)} How exactly would I go about to find the derivative of r(t), which is r'(t)? What exactly am I...

Yeah man, <-- only in calc2 here. No idea on ODE and all this higher order stuff :( The total energy of the particle @ t=0 is U+K = 0 + (1/2)M(Vinf^2) At sometime later the energy is: U'+K' = \frac{kQ^2}{sqrt(x^2 + y^2)} + (1/2)M(v^2) So, to find when the distance is the smallest between...

[SIZE="6"]V\frac{dV}{dx} = \frac{kQ^2}{(X^2 + Y^2){3/2}(*M)} I got that to yield V = \sqrt{-2kQ^2 / [m SQRT(X^2 + Y^2)] +2C} Where could I go from there? (This is x component for now, but I can just change the variables at the end to change into Y)

So I would have to go about integrating the functions...twice...correct? That would then yield the particle's trajectory, x(t) and y(t), only one of which is relevant since we are trying to find the particle's displacement above the origin...right? :( integrate the x component in terms of dx...