Recent content by VU2

So +3q charge doesn't play any roll at all right?

Yes, its E=q/[ε*area], where q is the charge enclosed.

An uncharged spherical conducting shell surrounds charge -q at the center of the shell. Then a charge+ +3q is placed on the outside of the shell. When static equilibrium is reached, the charges on the inner and outer surfaces of the she are respecteively... +q,-q is the answer. Does the +3q...

Yeah, its much different. Thanks!

Im sorry, I meant 9 times more.

So at the magnitude of the electric field is actually 4 times more for radius R than 2R?

E=kq/x^2

Thanks Tanya for replying. But is the magnitude of E the same for both radius's?

Two concentric imaginary spherical surface of radius R and 3R, respectively, surrounds a point charge -Q, located at the center of the surface. When compared to the electric flux I1 through the surface of radius R, the electric flux I2 through the surface 3R is. I know the answer is that...

Yeah, me too. But the problem on the paper clearly states that. I'll bring it up to my professor. Thanks!

let f(xy)=e^(2y)sin(pix), is f a solution to (fxy)^2 -fxx(fyy)=4pi^2 e^(4x) So I found fxx and fyy and fxy, which are -pi(e^(2y))sin(pix), 4sin(pix)e^(2y), 2pi(e^(2y))cos(pix) respectively, When i reduced everything i got e^4y(sin^2(pix))+pie^4ycos^2(pix)=pie^(4x). I am assuming it is...

I think I understand what you meant. In general, because the the source is moving with the medium, the air, it will have a higher frequency than when the source is static and receiver moving at the same velocity? Btw, thanks for being patient with me.

It is more than one. But I' still don't understand why they produce different numbers, unfortunately.

So fs/fd=va^2/(va^2-v^2)

Okay, so fs=fo(va)/(va-v) and fd=fo(va+v)/va and if fo and v are constants, let's say f=1 hz and v=2. I see now, so fs is larger. How can fs be larger though? Don't fs and fd have the same fo and v in the first place? This is a conceptual problem by the way.