Recent content by VU2

1. Uncharged spherical conducting shell!

So +3q charge doesn't play any roll at all right?
2. Uncharged spherical conducting shell!

Yes, its E=q/[ε*area], where q is the charge enclosed.
3. Uncharged spherical conducting shell!

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...
4. Gauss Theory Question !

Yeah, its much different. Thanks!
5. Gauss Theory Question !

Im sorry, I meant 9 times more.
6. Gauss Theory Question !

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

E=kq/x^2
8. Gauss Theory Question !

Thanks Tanya for replying. But is the magnitude of E the same for both radius's?
9. Gauss Theory Question !

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...
10. Calculus 3 hw help

Yeah, me too. But the problem on the paper clearly states that. I'll bring it up to my professor. Thanks!
11. Calculus 3 hw help

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...
12. Frequency Problem

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.
13. Frequency Problem

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

So fs/fd=va^2/(va^2-v^2)
15. Frequency Problem

Okay, so fs=fo(va)/(va-v) and fd=fo(va+v)/va and if fo and v are constants, lets 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.