P25_011 - 11. (a) For r > r2 the field is like that...

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Unformatted text preview: 11. (a) For r > r2 the field is like that of a point charge and V= 1Q , 4πε0 r where the zero of potential was taken to be at infinity. (b) To find the potential in the region r1 < r < r2 , first use Gauss’s law to find an expression for the electric field, then integrate along a radial path from r2 to r. The Gaussian surface is a sphere of radius r, concentric with the shell. The field is radial and therefore normal to the surface. Its magnitude is uniform over the surface, so the flux through the surface is Φ = 4πr2 E . The volume 3 3 of the shell is (4π/3)(r2 − r1 ), so the charge density is ρ= 3Q 3 3, 4π (r 2 − r 1 ) and the charge enclosed by the Gaussian surface is 4π 3 q= 3 (r 3 − r 1 )ρ = Q 3 r 3 − r1 3 − r3 r2 1 . Gauss’ law yields 3 r 3 − r1 3 − r3 r2 1 4πε0 r2 E = Q =⇒ E = 3 r 3 − r1 Q . 2 (r 3 − r 3 ) 4πε0 r 2 1 If Vs is the electric potential at the outer surface of the shell (r = r2 ) then the potential a distance r from the center is given by r V = Vs − E dr = Vs − r2 = Vs − r 1 Q 3 − r3 4πε0 r2 1 r− r2 r2 r2 r3 r3 − 2+ 1− 1 2 2 r r2 Q 1 3 3 4πε0 r2 − r1 3 r1 r2 dr . The potential at the outer surface is found by placing r = r2 in the expression found in part (a). It is Vs = Q/4πε0 r2 . We make this substitution and collect terms to find V= Q 1 3 3 4πε0 r2 − r1 2 3r 2 r2 r3 − −1 2 2 r . 3 3 Since ρ = 3Q/4π (r2 − r1 ) this can also be written V= 2 r2 r3 3r 2 − −1 2 2 r ρ 3 ε0 . (c) The electric field vanishes in the cavity, so the potential is everywhere the same inside and has the same value as at a point on the inside surface of the shell. We put r = r1 in the result of part (b). After collecting terms the result is V= 2 2 Q 3(r2 − r1 ) 3 3, 4πε0 2(r2 − r1 ) or in terms of the charge density V= (d) The solutions agree at r = r1 and at r = r2 . ρ 2 (r 2 − r 1 ) . 2 ε0 2 ...
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This note was uploaded on 11/12/2011 for the course PHYS 2001 taught by Professor Sprunger during the Fall '08 term at LSU.

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