Recent content by misho

Thanks a lot! Answers my question perfectly.

Let's say I have a field (electric or magnetic or something) and it's time-varying so I choose to represent its components as phasors. Say the field is: \vec{F} = X\hat{x} + Y\hat{y} + Z\hat{z} where, X, Y and Z are complex numbers. Now, I want to find the amplitude of the field. If...

ok, "floating" and "ungrounded" mean the same thing, in case that wasn't clear (sorry). My point was that if you have an electric field, and the source of the field is WITHIN a floating sphere, the sphere will not shield its surroundings (although, it may alter the field if the situation isn't...

It doesn't. The easiest way to think of it is by Gauss' Law (for electrostatics, obviously). Take the Gaussian surface to be a sphere outside the conducting sphere. If the charge in the middle is Q, and the charge on the conducting sphere is 0, then the flux (\Psi) through the Gaussian sphere...

This can generally be shown with Gauss' Law (or Divergence Theorem), or in this 2-D case, with Green's Theorem, which implies that what happens along a closed contour is dependent on what happens in the area within the contour. If you are comfortable with vector calculus, take a look at the...

Yeah, it's zero assuming everything's ideal. Mathematically even, you have a wire (R=0) in parallel with a resistor (R=R_0). Then the parallel combination is: R_{eq} = \frac{1}{ \frac{1}{0} + \frac{1}{R_0} } = \frac{1}{\infty + \frac{1}{R_0}} = 0 I should probably have put some limits in there...

For this, I'm going to take up the ramp to be the positive direction and down the ramp to be the negative direction. Let F be the net force (this should end up being positive, since the initial velocity is negative, based on the definition above, and the car is slowing down). Let f be the...

Yes.

2h+( \frac{16}{h-8} + 2 ) ( \frac{-8}{(h-8)^2} ) let x = \frac{-8}{(h-8)^2} , then: 2h+( \frac{16}{h-8} + 2 ) x 2h+ \frac{16}{h-8} \times x + 2 \times x sub x = \frac{-8}{(h-8)^2} back in: 2h+ \frac{16}{h-8} \times \frac{-8}{(h-8)^2} + 2 \times \frac{-8}{(h-8)^2} 2h+ \frac{16 \times...

I'm going to have \overline{XY} denote the length of some line segment XY. let: a = \overline{KM} = \overline{ML}. By similar triangles (KEM and JEU), \overline{JU} = 2a. By similar triangles (MUL and EUN), \overline{EN} = 2a. It follows that \overline{JU} = \overline{EN}...

That's really neat. I didn't know about that theorem. Weird that in 6 or more years of geometry, it's never come up. It's a fairly easy one to prove, too. Then again, here in Ontario, they didn't even teach us that medians cut each other in a 2 to 1 ratio, and that little fact has come in useful...

1. use KCL to get initial current through L and voltage across C (maybe find an equivalent cct for the left part, although, I doubt it would be any easier to do that) 2. make supernode around inductor 3. use V=L dI/dt for inductor 4. solve the differential equations 5. the voltage across R4...

V = L di/dt <- the V here is the voltage across the inductor, which could be zero. Having said that, dI/dt is not zero, I was wrong about that. I decided to look at this in terms of conservation of power: Source power: P_s (t) = VI(t) Inductor power: P_L (t) = V_L I_L = L I(t)...

I think you're confused just because the diagram isn't very good. I think that there is a wire that goes to the middle of the spinner and then the current goes down to the brush (please tell me if I'm wrong about that).

Well, the current shouldn't change with time. I'm pretty sure about that.