FWLec6 - A. F. Peterson: Notes on Electromagnetic Fields &...

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A. F. Peterson: Notes on Electromagnetic Fields & Waves 10/04 Fields & Waves Note #6 Gauss’ Law in Integral Form Objectives: Introduce the concept of electric flux to complement the electric flux density and electric field. Present the integral form of Gauss’ Law and use it to solve for fields in highly symmetrical situations. Electric flux One of the uses of vector surface integrals is the definition of the total electric flux Y e passing through some surface S : Y e S Dd S = ÚÚ (6.1) where DE = e is the electric flux density introduced in Note #5, and where eee = 0 r is the effective permittivity of the medium. For the moment, assume that ee = 0 . The differential surface vector dS points normal to the surface S , as discussed in Note #3. (For an open surface, the direction of dS determines the direction in which the flux is measured.) From the units of D (C/m 2 ), we observe that the electric flux has units of Coulombs (C). This suggests that the flux has a direct connection with the charge that produces the electric field under consideration. The link between these two is the focus of Gauss’ Law. Gauss’ Law In 1813, Gauss formulated the following law: S Q S = ÚÚ enclosed (6.2) Equation (6.2) states that the total electric flux out of a closed surface S is equal to the total amount of charge contained within S . If this charge is in the form of a volume charge density, Qd v v V enclosed = ÚÚÚ r (6.3) V denotes the volume contained within S . However, this charge may also be in the form of discrete point sources, surface charge densities, etc., that can be summed or integrated to obtain Q enclosed . To illustrate the computation of the total charge, we consider the following examples.
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A. F. Peterson: Notes on Electromagnetic Fields & Waves 10/04 Example: A uniform surface charge density r s is distributed along the x - y plane. Determine the total charge contained within the volume 0 < x < 1, 0 < y < 2, and –1 < z < 3. Solution: The portion of the surface charge density that is contained within the specified volume is that residing within the range 0 < x < 1 and 0 < < 2. Therefore, Q dxdy s y x s enclosed == = = Ú Ú rr 0 2 0 1 2 (6.4) Example: A volume charge density v is given by the function v xyz xy z (,,) ( ) =- 0 2 1 (6.5) Determine the total charge contained within the volume 0 < x < 1, and 0 < z < 1. Solution: The integral to be evaluated is Q dxdydz z dxdydz xdx ydy z dz v z y x z y x xz y enclosed = = Ê Ë Á ˆ ¯ ˜ Ê Ë Á ˆ ¯ ˜ Ê Ë Á ˆ ¯ ˜ = = = = = = = = Ú Ú Ú Ú Ú Ú ÚÚ Ú 0 1 0 1 0 1 0 2 0 1 0 1 0 1 0 2 0 1 0 1 0 1 0 0 1 1 1 3 1 2 1 2 12 () (6.6) Field calculation using Gauss’ Law Gauss’ Law provides an alternative way to calculate the electric field (or force, or flux density) produced by a distribution of charge. This procedure only works for situations where there is considerable symmetry in the charge distribution. To illustrate the process, we initially consider the three fundamental charge distributions considered in Note #4: the point charge, the infinite line charge density, and the infinite sheet of charge.
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A. F. Peterson: Notes on Electromagnetic Fields & Waves 10/04 Example:
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This note was uploaded on 01/27/2011 for the course ECE 3025 taught by Professor Citrin during the Spring '08 term at Georgia Institute of Technology.

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FWLec6 - A. F. Peterson: Notes on Electromagnetic Fields &...

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