PHYS-102hw1soln

# PHYS-102hw1soln - × 10 3 N C ˆ j(b r F = q r E = 5.00 ×...

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PHYS-102 Homework 1 Solutions Spring 06/07 ____________________________________________________________ Chapter 19 Electric Forces and Electric Fields P19.7 (a) The force is one of attraction . The distance r in Coulomb’s law is the distance between centers. The magnitude of the force is F = k e q 1 q 2 r 2 = 8.99 × 10 9 N m 2 C 2 ( ) 12.0 × 10 9 C ( ) 18.0 × 10 9 C ( ) 0.300 m ( ) 2 = 2.16 × 10 5 N . (b) The net charge of 6.00 × 10 9 C will be equally split between the two spheres, or 3.00 × 10 9 C on each. The force is one of repulsion , and its magnitude is F = k e q 1 q 2 r 2 = 8.99 × 10 9 N m 2 C 2 ( ) 3.00 × 10 9 C ( ) 3.00 × 10 9 C ( ) 0.300 m ( ) 2 = 8.99 × 10 7 N . P19.13 (a) r E 1 = k e q 1 r 1 2 ˆ j ( ) = 8.99 × 10 9 ( ) 3.00 × 10 9 ( ) 0.100 ( ) 2 ˆ j ( ) = − 2.70 × 10 3 ( r E 2 = k e q 2 r 2 2 ˆ i ( ) = 8.99 × 10 9 ( ) 6.00 × 10 9 ( ) 0.300 ( ) 2 ˆ i ( ) = − 5.99 × 10 ( r E = r E 2 + r E 1 = − 5.99 × 10 2 N C ( ) ˆ i 2.70

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Unformatted text preview: × 10 3 N C ( ) ˆ j (b) r F = q r E = 5.00 × 10 − 9 C ( ) − 599 ˆ i − 2 700 ˆ j ( ) N C r F = − 3.00 × 10 − 6 ˆ i − 13.5 × 10 − 6 ˆ j ( ) N = − 3.00 ˆ i − 13.5 ˆ j ( ) μ N (b) r F = q r E = 5.00 × 10 − 9 C ( ) − 599 ˆ i − 2 700 ˆ j ( ) N C r F = − 3.00 × 10 − 6 ˆ i − 13.5 × 10 − 6 ˆ j ( ) N = − 3.00 ˆ i − 13.5 ˆ j ( ) N FI P19.16 The electric field at any point x is E = k e q x − a ( ) 2 − k e q x − − a ( ) ( ) 2 = k e q 4 ax ( ) x 2 − a 2 ( ) 2 . When x is much, much greater than a , we find E ≈ 4 a k e q ( ) x 3 ....
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## This note was uploaded on 06/02/2008 for the course PHYS 102 taught by Professor N/a during the Spring '08 term at Drexel.

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PHYS-102hw1soln - × 10 3 N C ˆ j(b r F = q r E = 5.00 ×...

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