{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

prelim1sol08

# prelim1sol08 - ECE 3030 Prelim 1 Solutions Problem 1 Image...

This preview shows pages 1–2. Sign up to view the full content.

ECE 3030 Prelim 1 Solutions 2008/10/30 Problem 1: Image problem with a 1 m thick perfect conductor. a. Refer to your workshop notes where I sketched the problem. b. vector E ( vector r = (0 , 8 , 0)) = Q 4 πε 0 bracketleftbig 1 4 2 1 12 2 bracketrightbig ˆ r = 1 . 499 × 10 9 ˆx V/m c. vector E ( vector r = (8 , 90 , 90 )) = 1 . 499 × 10 9 vector r V/m d. vector E ( vector r = (0 , 8 , 0)) = 0ˆx V/m e. vector E ( vector r = (8 , 90 , 270 )) = 0 vector r V/m Problem 2: Battery with concentric spheres. a. (1) Φ 1 ( r ) = C 0 , Φ 2 ( r ) = C 1 + C 2 r , Φ 3 ( r ) = C 3 , Φ 4 ( r ) = C 4 + C 5 r , Φ 5 ( r ) = C 6 , Φ 6 ( r ) = C 7 + C 8 r . b. The magnitude of the total charge on the inside and outside surfaces of the middle shell is the same, but have the opposite sign. So 2 πc 2 σ c = 2 πb 2 σ b and σ c = b 2 c 2 σ b . c. Boundary conditions: Φ 2 ( a ) = Φ 1 ( a ), Φ 3 ( b ) = Φ 4 ( b ), Φ 4 ( c ) = Φ 5 ( c ), Φ 5 ( d ) = Φ 6 ( d ), Φ 6 ( e ) = Φ 5 ( e ), Φ 6 ( ) = 0. If we surround the system with a Gaussian surface, the total charge inclosed is zero. So the electric field and potential outside the outer shell is therefore zero. The potentials in-

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

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### Page1 / 2

prelim1sol08 - ECE 3030 Prelim 1 Solutions Problem 1 Image...

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online