# 329sp09hw5sol - ECE-329 Spring 2009 Homework 5 — Solution...

This preview shows pages 1–3. Sign up to view the full content.

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.

Unformatted text preview: ECE-329 Spring 2009 Homework 5 — Solution February 17, 2009 1. Verifying vector calculus identities, ∇ × ( ∇ Φ) = and ∇ · ( ∇ × A ) = 0 . a) The gradient of a scalar field Φ is defined as ∇ Φ = ∂ Φ ∂x ˆ x + ∂ Φ ∂y ˆ y + ∂ Φ ∂z ˆ z. Taking the curl, we obtain ∇ × ( ∇ Φ) = ˆ x ˆ y ˆ z ∂ ∂x ∂ ∂y ∂ ∂z ∂ Φ ∂x ∂ Φ ∂y ∂ Φ ∂z = ∂ ∂y ∂ Φ ∂z- ∂ ∂z ∂ Φ ∂y ˆ x + ∂ ∂z ∂ Φ ∂x- ∂ ∂x ∂ Φ ∂z ˆ y + ∂ ∂x ∂ Φ ∂y- ∂ ∂y ∂ Φ ∂x ˆ z = . b) The curl of a vector field A = A x ˆ x + A y ˆ y + A z ˆ z is defined as ∇ × A = ˆ x ˆ y ˆ z ∂ ∂x ∂ ∂y ∂ ∂z A x A y A z = ∂A z ∂y- ∂A y ∂z ˆ x + ∂A x ∂z- ∂A z ∂x ˆ y + ∂A y ∂x- ∂A x ∂y ˆ z. Taking the divergence, we obtain ∇ · ( ∇ × A ) = ∂ ∂x ∂A z ∂y- ∂A y ∂z + ∂ ∂y ∂A x ∂z- ∂A z ∂x + ∂ ∂z ∂A y ∂x- ∂A x ∂y = ∂ 2 A z ∂x∂y- ∂ 2 A y ∂x∂z + ∂ 2 A x ∂y∂z- ∂ 2 A z ∂y∂x + ∂ 2 A y ∂z∂x- ∂ 2 A x ∂z∂y = 0 . 2. Let us verify if A = ( x- y )ˆ x + ( x + y )ˆ y satisfies the following vector identity ∇ × ( ∇ × A ) = ∇ ( ∇ · A )- ∇ 2 A . The left-hand side gives ∇ × ( ∇ × A ) = ∇ × ˆ x ˆ y ˆ z ∂ ∂x ∂ ∂y ∂ ∂z x- y x + y = ∇ × (2ˆ z ) = , 1 ECE-329 Spring 2009 and the right hand side gives ∇ ( ∇ · A )- ∇ 2 A = ∇ ( ∇ · (( x- y )ˆ x + ( x + y )ˆ y ))- ∇ 2 (( x- y )ˆ x + ( x + y )ˆ y ) = ∇ (2)- = , thus, the identity is verified. 3. The charge distribution associated with a point charge Q = 4 π o C can be expressed as ρ ( x,y,z ) = Qδ ( x ) δ ( y ) δ ( z ) where δ ( · ) is the delta function. a) Calculating the closed volume integral ¸ V ρdV. i. If V contains the origin, we have ˛ V ρdV = ˛ V Qδ ( x ) δ ( y ) δ ( z ) dxdydz = Q. ii. If V excludes the origin, we simply have ˛ V ρdV = 0 . b) Above, we found that no matter what the shape or size of the volume V are, as long as V contains the charge distribution ρ, the charge inside the volume is Q. Notice, that we would have found the same, if we were considered instead a point charge Q placed at the origin. Thus, we can argue that the charge density function ρ ( x,y,z ) = Qδ ( x ) δ ( y ) δ ( z ) is a valid description of a point charge.point charge....
View Full Document

## This note was uploaded on 11/16/2009 for the course ECE 329 taught by Professor Franke during the Spring '08 term at University of Illinois at Urbana–Champaign.

### Page1 / 6

329sp09hw5sol - ECE-329 Spring 2009 Homework 5 — Solution...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online