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Unformatted text preview: R V ‡ ~ ∇ · ~ A · d 3 r = H S ~ A · d ~ S , substitute ~ A = φ ~ ∇ φ . This then leads to Z V ‡ ~ ∇ · ‡ φ ~ ∇ φ ·· d 3 r = I S ‡ φ ~ ∇ φ · · d ~ S Z V ( φ ∇ 2 φ + k∇ φ k 2 ) d 3 r = I S ± φ ∂φ ∂n ¶ dS But, since ∇ 2 φ = 0, we get Z V k∇ φ k 2 d 3 r = 0 . However, k∇ φ k 2 is semipositive deﬁnite quantity and integration over the region can be zero only if ∇ φ = 0. Thus φ is a constant function and then from continuity it must be zero at all points inside....
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This note was uploaded on 01/28/2010 for the course PHYS 351 taught by Professor Gangopashyaya during the Spring '09 term at Loyola Chicago.
 Spring '09
 Gangopashyaya
 Charge, Magnetism

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