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Unformatted text preview: SOLUTIONS, ASSIGNMENT 5 21.55. I DENTIFY : For a ring of charge, the electric field is given by Eq. (21.8). q F = E r r . In part (b) use Newton's third law to relate the force on the ring to the force exerted by the ring. S ET U P : 9 0.125 10 C, Q- = × 0.025 m a = and 0.400 m x = . E XECUTE : (a) 2 2 3/ 2 1 ˆ ˆ (7.0 N/C) 4 ( ) Qx x a π + E = i = i r P . (b) 6 5 on ring on q ˆ ˆ ( 2.50 10 C) (7.0 N/C) (1.75 10 N) q--- = - = - - × = × F = F E i i r r r E VALUATE : Charges q and Q have opposite sign, so the force that q exerts on the ring is attractive. 21.57. I DENTIFY : By superposition we can add the electric fields from two parallel sheets of charge. S ET U P : The field due to each sheet of charge has magnitude /2 σ P and is directed toward a sheet of negative charge and away from a sheet of positive charge. (a) The two fields are in opposite directions and 0. E = (b) The two fields are in opposite directions and 0. E = (c) The fields of both sheets are downward and 2 2 E σ σ = = P P , directed downward....
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