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Unformatted text preview: A. If it were otherwise, it would imply an electric field within the solid sphere and that is not possible for a conductor in static equilibrium. 0!
ˆ
We can find the electric potential at the origin by calculating: !V = " % E # r dr . We $ can break this out into the region r > rC, rA < r < rB. %1 1(
rC kQ
rA kQ
kQ
2
1
!V = V (0 ) = " $
dr " $
dr = 2 + kQ1 ' " * #
rB
r2
r2
rC
& rA rB )
We can think of this situation in terms of superposition of electric potentials, but we must be carful! There are three charged spherical surfaces with respective charges, Q1,
Q1, and Q2. The potential on any of the 3 surfaces is that of a point charge at the origin. Once inside a given surface, the potential due to that surface charge does not change. Therefore, we can simply write down the solution: Potential at the origin due to charge Q2 is VQ2 (0 ) =
so V (0 ) = VQ1 (0 ) + V!Q1 (0 ) + VQ2 (0 ) = "1 1%
kQ2
+ kQ1 $ ! ' rC
# rA rB & kQ2
kQ
kQ
, V!Q1 (0 ) = ! 1 , VQ1 (0 ) = 1 , rC
rB
rA...
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 Spring '12
 DavidBradfordBlasing
 Charge

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