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**Unformatted text preview: **24.68: a) r < R : u = 1 ε0 E 2 = 0. 2
2 1 2 2 1 Q Q2 = . b) r > R : u = ε0 E = ε0 2 4πε0 r 2 32π 2 ε0 r 4 Q 2 dr Q2 = c) U = ∫ udV = 4π ∫ r udr = 8πε0 ∫ r 2 8πε0 R ⋅ R R
2 ∞ ∞ d) This energy is equal to 1 Q 2 4 πε 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) From Equation (24.9): U = 2C = 8πε 0 R from part (c) ⇒ C = 4πε0 R, as in Problem (24.67). ...

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