(a) Potential energy (U) is given by
U=k(x12q1q2+x13q1q3+x23q2q3)k=8.9×109Nm2/C2q1=4nC=4×10−9Cq2=−3nC=−3×10−9Cq3=2nC=2×10−9Cx12=20cm=0.2mx13=10cm=0.1mx23=10cm=0.1mU=8.9×109(0.24×10−9×−3×10−9+0.14×10−9×2×10−9+0.1−3×10−9×2×10−9)U=8.9×109(−6×10−17+8×10−17−6×10−17)U=8.9×109×10−17(−4)U=−35.6×10−8U=−356nJ
(b)
U=k(x12q1q2+x13q1q3+x23q2q3)0=10−18(0.24×−3+x4×2+0.2−x−3×2)0=−60+x8−0.2−x660=x8−0.2−x660=x(0.2−x)8(0.2−x)−6(x)60=0.2x−x21.6−8x−6x60=0.2x−x21.6−14x60(0.2x−x2)=1.6−14x12x−60x2=1.6−14x60x2−30x+1.6=0
Solving the quadratic equation
x=0.06, 0.44
The distance is 0.06m(6cm) since it is between 0 and 20cm
System
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