Let, 
k=4πϵo1=9×109 We know that 
F12=kd122q1q2 
(a). 
b).
Q1=250μC,Q2=−150μC,d12=5m⟹F12=k25250×150×10−12=15×10−10k=13.5N c).
Q1=250μC,Q3=−200μC,d13=8m⟹F13=k64250×200×10−12=7.81×10−10k=7.03N d).
Q2=−150μC,Q3=−200μC,d23=m⟹F23=k9150×200×10−12=30.0N e).
From the free body diagram,
FnetQ1=F12+F13=20.53N acting along the direction away from Q1 to Q2 
f).
 
FnetQ2=F12+F23=43.5N along the direction left of Q2 
g).
FnetQ3=F23−F13=22.97N acting along the direction right of Q3 
                             
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