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HOMEWORK-CH 25

# HOMEWORK-CH 25 - 5 d 2 from P V P = k 5.0 q 5 d 2-2.0 q...

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HOMEWORK ASSIGNMENT 3 4. The electrostatic potential difference is given by Eq. 25-8 as V = - W elec q , where W elec is the work done by the field as the charge q moves in the electric field. (a) V B - V A =- W elec - e ( 29 =- 3.94 × 10 - 19 J (29 - 1.60 × 10 -19 C (29 = 2.46 V. (b) V B = V C since points B and C are on the same equipotential line. Therefore, V C - V A = V B - V A = 2.46 V. (c) V B = V C since points B and C are on the same equipotential line. Therefore, V C - V B = 0 V. 20. The net electric potential at point P is the sum of those due to the six charges: V P ( 29 = V i P ( 29 i = 1 6 = k q i r i i = 1 6 , where the r i are the distances from each charge to the point P . The charges above and below the point P are a distance d 2 from P , while those at the corners are a distance
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Unformatted text preview: 5 d 2 from P . V ( P ) = k 5.0 q 5 d 2 +-2.0 q d /2 +-3.0 q 5 d 2 + 3.0 q 5 d 2 +-2.0 q d /2 + 5.0 q 5 d 2 = kq d 20.0 5-8.0 = 0.94 kq d . 24. The electric potential energy of this configuration is ( 29 [ 2 2 2 3 1 4 1 2 1 3 2 4 3 4 all pairs N m 9 C 6 2 2 8.99 10 (12nC)( 24nC) (12nC)(31nC) ( 24nC)(17nC) 1.3m (12nC)(17nC) ( 24nC)(31nC) (31nC)(17nC) 2 2 1.2 10 J. i j i j q q q q q q k U k q q q q q q q q r d → ⋅- = = + + + + + × =-+ + --+ + + = -× ∑...
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