**Unformatted text preview: **24.68: a)â€‚ r < R : u = 1 Îµ0 E 2 = 0. 2
2 1 2 2 1 Q Q2 = . b)â€‚ r > R : u = Îµ0 E = Îµ0 2 4Ï€Îµ0 r 2 32Ï€ 2 Îµ0 r 4 Q 2 dr Q2 = c)â€‚ U = âˆ« udV = 4Ï€ âˆ« r udr = 8Ï€Îµ0 âˆ« r 2 8Ï€Îµ0 R â‹… R R
2 âˆž âˆž d)â€‚This energy is equal to 1 Q 2 4 Ï€Îµ 0 R 2 which is just the energy required to assemble all the charge into a spherical distribution. (Note, being aware of double counting gives the factor of 1 2 in front of the familiar potential energy formula for a charge Q a distance R from another charge Q.) Q2 Q2 e)â€‚From Equation (24.9): U = 2C = 8Ï€Îµ 0 R from part (c) â‡’ C = 4Ï€Îµ0 R, as in Problem (24.67). ...

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