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lecture3

# lecture3 - Lecture 3 Electrostatics In this lecture you...

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1 ECE 303 – Fall 2007 – Farhan Rana – Cornell University Lecture 3 Electrostatics In this lecture you will learn: • Three ways to solve problems in electrostatics: a) Application of the Superposition Principle (SP) b) Application of Gauss’ Law in Integral Form (GLIF) c) Application of Gauss’ Law in Differential Form (GLDF) ECE 303 – Fall 2007 – Farhan Rana – Cornell University Field of a Point Charge (GLIF) Consider a point charge of q Coulombs sitting at 0 = r r q By symmetry , electric field can only point in the radial direction Surround the charge by a Gaussian surface in the form of a spherical shell of radius r r q By symmetry , the E-field magnitude on the surface must be uniform and pointing in the radial direction Using Gauss’ Law: ( ) q r E r o = 2 4 π ε r r q E r q E o o r ˆ 4 or 4 2 2 πε πε = = r dV a d E o ∫∫∫ = ∫∫ ρ ε r r . E

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2 ECE 303 – Fall 2007 – Farhan Rana – Cornell University The Principle of Superposition in Electromagnetics Maxwell’s equations are LINEAR and allow for the superposition principle to hold • Suppose for some charge and current densities, , one has found the E and H fields, 1 1 and H E r r • Suppose for some other charge and current densities, , one has also found the E and H fields, 2 2 and H E r r The superposition principle says that the sums, , are the solution for the charge and current densities, ( ) ( ) 2 1 2 1 and H H E E r r s r + + ( ) ( ) 2 1 2 1 and J J r r + + ρ ρ 1 1 and J r ρ A Simple Proof t E J H t H E H E o o o o + = × = × = = 1 1 1 1 1 1 1 1 0 . . r r r r r r r ε µ µ ρ ε + t E J H t H E H E o o o o + = × = × = = 2 2 2 2 2 2 2 2 0 . . r r r r r r r ε µ µ ρ ε = ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) t E E J J H H t H H E E H H E E o o o o + + + = + × + = + × = + + = + 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 0 . . r r r r r r r r r r r r r r ε µ µ ρ ρ ε 2 2 and J r ρ ECE 303 – Fall 2007 – Farhan Rana – Cornell University Two Point Charges (SP) We know that a single charge produces a radial field given by: r r q E o ˆ 4 2 πε = r Consider Two Charges 1 q 2 q 2 2 2 2 2 1 2 1 1 1 ˆ 4 ˆ 4 r r q E r r q E o o πε πε = = r r 1 r 2 r 1 E r 2 E r
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