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HW02_Solutions

# HW02_Solutions - HOMEWORK 2 ELECTRIC CHARGE AND ELECTRIC...

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21-1 H OMEWORK 2: E LECTRIC C HARGE AND E LECTRIC F IELD Q21.14. ANSWER: The initial force on each charge is the same and depends only on the product of charges and initial distance. Therefore, the acceleration will still be a 0 . 21.26. IDENTIFY: For a point charge, 2 q Ek r . SET UP: E is toward a negative charge and away from a positive charge. EXECUTE: (a) The field is toward the negative charge so is downward. 9 9 2 2 2 3.00 10 C (8.99 10 N m /C ) 432 N/C (0.250 m) E . (b) 9 2 2 9 (8.99 10 N m /C )(3.00 10 C) 1.50 m 12.0 N/C kq r E EVALUATE: At different points the electric field has different directions, but it is always directed toward the negative point charge. 21.31. IDENTIFY: For a point charge, 2 . q r The net field is the vector sum of the fields produced by each charge. A charge q in an electric field E experiences a force . q FE SET UP: The electric field of a negative charge is directed toward the charge. Point A is 0.100 m from q 2 and 0.150 m from q 1 . Point B is 0.100 m from q 1 and 0.350 m from q 2 . EXECUTE: (a) The electric fields due to the charges at point A are shown in Figure 21.31a. 9 1 9 2 2 3 1 22 1 6.25 10 C (8.99 10 N m /C ) 2.50 10 N/C (0.150 m) A q r 9 2 9 2 2 4 2 2 12.5 10 C (8.99 10 N m /C ) 1.124 10 N/C (0.100 m) A q r Since the two fields are in opposite directions, we subtract their magnitudes to find the net field. 3 21 8.74 10 N/C, E E E to the right. (b) The electric fields at points B are shown in Figure 21.31b. 9 1 9 2 2 3 1 1 (8.99 10 N m /C ) 5.619 10 N/C (0.100 m) B q r 9 2 9 2 2 2 2 2 (8.99 10 N m /C ) 9.17 10 N/C (0.350 m) B q r Since the fields are in the same direction, we add their magnitudes to find the net field. 3 12 6.54 10 N/C, E E E to the right. (c) At A , 3 8.74 10 N/C E  , to the right. The force on a proton placed at this point would be 19 3 15 (1.60 10 C)(8.74 10 N/C) 1.40 10 N, F qE  to the right. EVALUATE: A proton has positive charge so the force that an electric field exerts on it is in the same direction as the field.

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21-2 Chapter 21 Figure 21.31 21.33. IDENTIFY: Eq. (21.3) gives the force on the particle in terms of its charge and the electric field between the plates. The force is constant and produces a constant acceleration. The motion is similar to projectile motion; use constant acceleration equations for the horizontal and vertical components of the motion.
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HW02_Solutions - HOMEWORK 2 ELECTRIC CHARGE AND ELECTRIC...

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