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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 fi elds in a homogeneous, isotropic medium. In this case the electric fi eld satis fi es the two equations, Eq. 1.59a with a time independent charge density and Eq. 1.77 with a time independent magnetic fl ux density, · D ( r ) = ρ 0 ( r ) , (1.59a) × E ( r ) = 0 . (1.77) Because we are working with static fi 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 fi eld E ( r ) = 1 4 πε ZZZ 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 The electric or scalar potential Faraday’s law with static fi elds, Eq. 1.77, is automatically satis fi ed by any electric fi eld E ( r ) which is given by E ( r ) = φ ( r ) (1.81) The function φ ( r ) is the scalar potential for the electric fi 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 fi 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 ∂φ ( r ) y + dz ∂φ ( 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 fi 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 fi xes the (arbitrary) location of the zero for the scalar potential. Since the observed quantity is the electric force and, therefore, the electric fi eld, only the difference in the values of the scalar potential at any two different points is signi fi cant. (See Eq. 1.82) 19
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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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