Homework Answers
answer) a) from the diagram we have
the sum of forces in x direction
Fx=qEx-Tsin37=0............1)
the sum of forces in y direction
Fy=qEy-Tcos37-mg=0.........2)
equating both the equations we have
q=mg/(Ey+Ex/tan37o)=0.9*10-3kg*9.8/(4.40*105+3.80*105/tan37o)=0.934*10-8C
so the answer is 0.93*10-8C or 0.934*10-8C or 9.34*10-9C
b) from eqn 1 we have
qEx-Tsin37=0
0.934*10-8*3.8*105-Tsin37o=0
T=5.897*10-3N
so the answer is 5.897*10-3N or 5.9*10-3N or 6*10-3N
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