ch23-p055

# ch23-p055 - 55. (a) We integrate the volume charge density...

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(d) For r = R , the electric field is 2 92 2 1 2 3 00 2 1 (8.99 10 N m C ) (14.1 10 C/m )(0.0560 m) 44 2.23 10 N/C. ss RR E R πρ π πε == = × × 55. (a) We integrate the volume charge density over the volume and require the result be equal to the total charge: 2 0 4 . R dx dy dz dr r Q ρρ = π= ∫∫∫ Substituting the expression ρ = s r / R , with s = 14.1 pC/m 3 , and performing the integration leads to 4 4 4 s R Q R ⎛⎞ π = ⎜⎟ ⎝⎠ or 31 2 3 3 1 5 (14.1 10 C/m )(0.0560 m) 7.78 10 C. s QR −− ==× = × (b) At r = 0, the electric field is zero ( E = 0) since the enclosed charge is zero. At a certain point within the sphere, at some distance r from the center, the field (see Eq. 23-8 through Eq. 23-10) is given by Gauss’ law:

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## ch23-p055 - 55. (a) We integrate the volume charge density...

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