12_Chapter 15 HomeworkCH15 Electric Forces and Electric Fields

12_Chapter 15 HomeworkCH15 Electric Forces and Electric Fields

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
12 CHAPTER 15 Thus, Σ= and The resultant force on the 6.00 nC charge is then () 7 23 cos45.0 3.81 10 N x FFF + ° = × 8 sin45.0 7.63 10 N y Σ= − ° = × ()() 2 2 7 3.89 10 N Rx y =Σ + Σ = × at 1 tan 11.3 y x F F θ Σ  = =− °  Σ  or 7 3.89 10 N at 11.3 below + ax R x ° F G is 15.13 The forces on the 7.00 µ C charge are shown at the right. ( )( ) ( 66 2 9 1 2 2 2 9 2 2 2 7.00 10 C 2.00 10 C Nm 8.99 10 C 0.500 m 0.503 N 7.00 10 C 4.00 10 C 8.99 C 0.500 m 1.01 N F F −− ×× = = ( Thus, ) 12 cos60.0 0.755 N x Σ= + ° = sin60.0 0.436 N y ° = and The resultant force on the 7.00
Background image of page 1
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

Ask a homework question - tutors are online