{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

20092ee1_1_HW2_Solutions

20092ee1_1_HW2_Solutions - EE 1 Homework#2 Solutions Spring...

Info iconThis preview shows pages 1–4. Sign up to view the full content.

View Full Document Right Arrow Icon
EE 1 - Homework #2 Solutions Spring 2009 (1) (1pt each) 1
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
(2) (10 points) The gradient of a scalar function is given by: V = ˆ ze - 2 z If V = 20 V at z = 0, find V ( z )? The gradient of V is essentially the derivative of V . Integrating the gradient will result in the general form of V ( z ): V 2 - V 1 = Z P 2 P 1 V · dl = Z P 2 P 1 ˆ ze - 2 z · xdx + ˆ ydy + ˆ zdz ) Integrating, and choosing P 1 = 0 and P 2 = z the general result is: V ( z ) - V (0) = Z z 0 e - 2 z dz = - 1 2 e - 2 z | z 0 V ( z ) = 1 - e - 2 z 2 + V (0) = 41 - e - 2 z 2 ( V ) 2
Background image of page 2
(3) (15 Points) For the vector field: ~ E = ˆ r 5 e - r - ˆ z 6 z verify the divergence theorem for the cylindrical region enclosed by r = 2, z = 0, and z = 4. To verify the divergence theorem, we merely have to show that the surface integral of the vector field is equal to the volume integral of the divergence of the same vector field: Z S ~ E · d ~ S = Z V ∇ · ~ EdV First, we set-up the surface integral. For a cylinder, there are three surfaces to worry about, the top and bottom ( S 1 and S 2, respectively) of the cylinder and the side surface of the cylinder ( S 3).
Background image of page 3

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Image of page 4
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}