P22_042.PDF

42. (a) Charge Q 1 =+80

10 − 9 Cisonthe y axis at y =0 . 003 m, and charge Q 2 =+80

10 − 9 Cison

10 − 9 C) is

due to the vector sum of the repulsive forces from Q 1 and Q 2 . In symbols, F 31 + F 32 = F 3net ,

where

−

0 . 003 m. The force on particle 3 (which has a charge of q =+18

= k q 3 |

q 1 |

r 31

= k q 3 q 2

r 32

F 31 |

and

F 32 |

Using the Pythagorean theorem, we have r 31 = r 32 =0 . 005 m. In magnitude-angle notation

(particularly convenient if one uses a vector capable calculator in polar mode), the indicated vector

addition becomes

(0 . 518

−

37 ◦ )+(0 . 518

37 ◦ )=(0 . 829

0 ◦ ) .

Therefore, the net force is 0 . 829 N in the + x direction.

(b) Switching the sign of Q 2 amounts to reversing the direction of its force on q .Consequently,wehave

(0 . 518

−

37 ◦ )+(0 . 518

−

143 ◦ )=(0 . 621

−

90 ◦ ) .

Therefore, the net force is 0 . 621 N in the

−

y direction.

the y axis at y =