3 charges are lined up in a horizontal cm to the east of Q Qt has a charge of +1
ID: 776973 • Letter: 3
Question
3 charges are lined up in a horizontal cm to the east of Q Qt has a charge of +10, has a charge of 2 uc and is located at a distance of 5 line : O has a charge of +30pC and is located at a distance of 6 cm to the east of Q2. a. How many excess electrons are on Q2? b. Find the value of each of the forces acting on Q2. (-720 N, 1500 N) c. Find the net force (magnitude and direction) on Q2.(780 N) d. d is now moved to a location of 6 cm to the North of Q2. Find the new net Force (magnitude and direction) on Q (1663N@ 64 N of E)Explanation / Answer
a)
N=Q2/e =(-20*10-6)/(-1.6*10-19)=1.25*1014 electrons
b)
Force due to Q2 on Q1 is
F1=-K|Q1||Q2|/r12 =-(9*109)(10*10-6)(20*10-6)/0.052=-720 N
Force due to Q2 on Q3 is
F3=K|Q3||Q2|/r22 =(9*109)(20*10-6)(30*10-6)/0.062=1500 N
c)
Net force on Q2 is
F=1500-720=780 N/C
Direction is towards east or positive x-axis
d)
Now Here
F2x=-720 N
F2y=1500 N
Magnitude
|F|=sqrt[7202+15002]
|F|=1663 N
Direction
o=tan-1(1500/-720) =-64o =64o N of E
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.