# 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.