Chapter2.revised

# Chapter2.revised - Chapter 2 Introduction to electrostatics...

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Chapter 2 Introduction to electrostatics 2.1 Coulomb and Gauss’ Laws We will restrict our discussion to the case of static electric and magnetic f elds in a homogeneous, isotropic medium. In this case the electric f eld satis f es the two equations, Eq. 1.59a with a time independent charge density and Eq. 1.77 with a time independent magnetic F ux density, · D ( r )= ρ 0 ( r ) , (1.59a) × E ( r 0 . (1.77) Because we are working with static f elds in a homogeneous, isotropic medium the constituent equation is D ( r ε E ( r ) . (1.78) Note : D is sometimes written : (1.78b) D = ² o E + P .... SI units D = E +4 π P in Gaussian units in these cases ε = [ 1+4 π P/E ] Gaussian The solution of Eq. 1.59 is D ( r 1 4 π ZZZ ρ 0 ( r 0 )( r r 0 ) | r r 0 | 3 d 3 r 0 + D 0 ( r ) , SI units (1.79) with · D 0 ( r )=0 If we are seeking the contribution of the charge density, ρ 0 ( r ) , to the electric displacement vector then D 0 ( r 0 . The given charge density generates the electric f eld E ( r 1 4 πε ρ 0 ( r 0 r r 0 ) | r r 0 | 3 d 3 r 0 SI units (1.80) 18

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Section 2.2 The electric or scalar potential 2.2 Theelectricorscalarpotential Faraday’s law with static f elds, Eq. 1.77, is automatically satis f ed by any electric f eld E ( r ) which is given by E ( r )= φ ( r ) (1.81) The function φ ( r ) is the scalar potential for the electric f eld. It is also possible to obtain the difference in the values of the scalar potential at two points by integrating the tangent component of the electric f eld along any path connecting the two points Z path r a r b E ( r ) · d` = Z path r a r b φ ( r ) · d` (1.82) = Z path r a r b · dx ∂φ ( r ) x + dy ( r ) y + dz ( r ) z ¸ = Z path r a r b d φ ( r φ ( r b ) φ ( r a ) The result obtained in Eq. 1.82 is independent of the path taken between the points r a and r b . It follows that the integral of the tangential component along a closed path is zero, I E ( r ) · d` = I d φ ( r )=0 . (1.83) This last result actually follows from the requirement that × E ( r 0 and the application of Stoke’s theorem. To obtain the scalar potential due to the charge density ρ 0 ( r ) we note that 1 q ( x x 0 ) 2 +( y y 0 ) 2 z z 0 ) 2 (1.84) = ( x x 0 ) i y y 0 ) j z z 0 ) k h ( x x 0 ) 2 y y 0 ) 2 z z 0 ) 2 i 3 / 2 = r r 0 | r r 0 | 3 . Comparing the expression on the right hand side of Eq. 1.84 to the integrand in Eq. 1.80 we f nd that can write that the scalar potential due to the charge density ρ o ( r ) is φ ( r 1 4 πε ZZZ ρ o ( r 0 ) | r r 0 | d 3 r 0 + φ 0 , (1.85) where φ 0 is a constant which f xes the (arbitrary) location of the zero for the scalar potential. Since the observed quantity is the electric force and, therefore, the electric f eld, only the difference in the values of the scalar potential at any two different points is signi f cant. (See Eq. 1.82) 19
Section 2.3 Surface charges and charge dipoles 2.3 Surface charges and charge dipoles A surface charge is a charge density which is ‘restricted to lie on a surface’.

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## This note was uploaded on 12/19/2010 for the course PHYS 411 taught by Professor G during the Spring '10 term at Missouri S&T.

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Chapter2.revised - Chapter 2 Introduction to electrostatics...

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