Two charged particles+3C and -9C are placed at two points A and B, then magnitude of electric force at third charge particle +2C at point C is Distance between A and B is 20 cm and B and C is 10 cm
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Fac= k×qa×qc\rac2 = [9×(10)9 × 3×2]÷0.1=54×(10)10
Fbc= [9×(10)9 × 9×2]÷0.1= 162×(10)10
net force= Fbc- Fac= 108 × (10)10 newton
Fbc= [9×(10)9 × 9×2]÷0.1= 162×(10)10
net force= Fbc- Fac= 108 × (10)10 newton
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