{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

# 329lect03 - 3 Gausss law and static charge densities We...

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

3 Gauss’s law and static charge densities We continue with examples illustrating the use of Gauss’s law in macroscopic field calculations: ρ S E x ( x ) = ρ s 2 o A z S x y E x ( x ) ρ s 2 o sgn( x ) x Example 1: Point charges Q are distributed over x = 0 plane with an average surface charge density of ρ s C/m 2 . Determine the macroscopic electric field E of this charge distribution using Gauss’s law. Solution: First, invoking Coulomb’s law, we convince ourselves that the field produced by surface charge density ρ s C/m 2 on x = 0 plane will be of the form E = ˆ xE x ( x ) where E x ( x ) is an odd function of x because y - and z -components of the field will cancel out due to the symmetry of the charge distribution. In that case we can apply Gauss’s law over a cylindrical integration surface S having circular caps of area A parallel to x = 0 , and obtain S D · d S = Q V o E x ( x ) A - o E x ( - x ) A = A ρ s , which leads, with E x ( - x ) = - E x ( x ) , to E x ( x ) = ρ s 2 o for x > 0 . Hence, in vector form E = ˆ x ρ s 2 o sgn ( x ) , where sgn ( x ) is the signum function, equal to ± 1 for x 0 . Note that the macroscopic field calculated above is discontinuous at x = 0 plane containing the surface charge ρ s , and points away from the same surface on both sides. 1

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

View Full Document
ρ E x ( x ) = ρ x o A z x W 2 - W 2 E x ( x ) ρ W 2 o - W 2 W 2 x y Example 2: Point charges Q are distributed throughout an infinite slab of width W located over - W 2 < x < W 2 with an average charge density of ρ C/m 3 . Determine the macroscopic electric field E of the charged slab inside and outside. Solution: Symmetry arguments based on Coulomb’s law once again indicates that we expect a solution of the form E = ˆ xE x ( x ) where E x ( x ) is an odd function of x . In that case, applying Gauss’s law with a cylindrical surface S having circular caps of area A parallel to x = 0 extending between - x and x < W 2 , we obtain S D · d S = Q V o E x ( x ) A - o E x ( - x ) A = ρ 2 xA, which leads, with E x ( - x ) = - E x ( x ) , to E x ( x ) = ρ x o for 0 < x < W 2 .
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}