three surfaces are \u03c6 \u03c1 \u03c1 d d \u02c6 d 1 z a r \u03c6 \u03c1 d d \u02c6 d 2 z \u03c1 a r and \u03c6 \u03c1 \u03c1 d d \u02c6

Three surfaces are φ ρ ρ d d ˆ d 1 z a r φ ρ d

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three surfaces are ( )( ) φ ρ ρ d d ˆ d 1 z = a r , ( ) φ ρ d d ˆ d 2 z ρ = a r and ( )( ) φ ρ ρ d d ˆ d 3 z - = a r . Then Figure 2.10 : A Gaussian disc straddling a uniform planar distribution of electric scalar charges E E 2 0 0 2 0 0 3 0 2 0 1 0 e 2 d d 2 d d d 3 2 1 ρ πε φ ρ ρ ε ε ε ε ρ ρ π φ = = + + = Φ ∫ ∫ = = a a a a a a r r r r r r E E E The portion of the charged plane enclosed by the Gaussian cylinder is a circular surface of area a . Hence the net charge within this cylinder is 2 0 2 0 face circular enc d d d πσρ φ ρ ρ σ σ ρ ρ π φ = = = ∫ ∫ = = a q (2.41) Substituting these results into Gauss’ law gives . . . . . . . . + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + 1 d a r 2 d a r 3 d a r E r E r E r σ l 2 d a r E r
17 < - = - > + = + = 0 z , 4 ˆ 2 ˆ 0 z , 4 ˆ 2 ˆ 2 0 cps 0 2 0 cps 0 z q z q πε ε σ πε ε σ z z z z E r where, for any field point at a distance z from the charged plane, a σ = cps q arises from a circular area 2 2 z π = a that extends to a variable radius of z 2 1 2 within the plane. Exercise 5 : A thin spherical shell of radius a r has a uniform surface charge density of σ . Find the electric field E r and potential V due to this charge distribution. Exercise 6 : Two infinitely wide planar distributions of electric scalar charges cut the z –axis at positions a z 2 1 ± = and have uniform surface charges densities of σ ± . Find their combined electric field E r everywhere. Sketch the field E r against position z . 2.6.4. Differential form of Gauss’ law and Poisson’s equation In equations (2.39) and (2.40) the integrals are over the source scalar charges to remind ourselves that our interest is to determine the E r due to these sources only. Using the divergence theorem to the LHS of (2.32), and expressing the RHS as a volume integral over the source distribution, Gauss’ law becomes ( ) = = = Φ v v v v v v d d d enc 0 e ρ ρ ε q E r (2.42) The equivalence is valid for v v if there are no charges in the rest of the Gaussian volume v . Then the integrands over the same volume on the two sides of the equation must be equal, that is: ρ ε = E r 0 (2.43) This is the differential form of Gauss’ law. Exercise 7 : An electric scalar charge q is deposited on a conducting sphere of radius 0 r . The potential at a radial distance s r from the centre of this conducting sphere is s V . Express the charge density everywhere in terms of s V , s r and 0 r . Find the electric field E r everywhere. 2.7 Work and energy in electrostatics If we have a collection of electric scalar point charges n i q i , . . . , 3 , 2 , 1 , = , the minimum work done to assemble them is ( ) = = n i i i V q W 1 2 1 r r where ( ) i V r r is the potential at the position of the charge i q due to all the other 1 - n charges in the system. For a continuous electric scalar charge distribution the work done is = q V W d 2 1 [ J ] (2.44)
18 where V is the potential at the position of the elemental charge q d . The work represents the energy stored in the assembly of charges. Thus the electric potential energy is = q V U d 2 1 [ J ] (2.45) We can also rewrite this potential energy in terms of the electric field at the position of the charge distribution as ( ) = = space All 2 0 2 1 space All 0 2 0 d d 2 v v E ε ε ε E U (2.46) where the integration is over the entire space. It follows that the electric potential energy density

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