Chapter 16: We are given 3 charges Q1 Q2-+(1/9)x10°C and Q3-4 С Name (Print) Q3 2...

Question

Question

Answer 1

Answer #1

1)Force on Q3 due to Q1 =F1= 9*10⁹*(1/9)*10^-9*4/2²= 1 N (directed from Q1 to Q3)

Force on Q3 due to Q2=F2=9*10⁹*(1/9)*10^-9*4/1²= 4 N (directed from Q1 to Q3)

2)horizontal component of F1=1*cos20=0.94 N

vertical component of F1=1*sin20=0.34 N

horizontal component of F2=4*cos50=2.57 N

vertical component of F2=4*sin50 = 3.06 N

total horizontal force = 0.94+2.57=3.51 N

total vertical force = 0.34+3.06=3.4 N

Resultant force = sqrt(3.51²+3.4²)=4.89 N

Answer 2

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