Question #162856

Three charges lie in the plane in a vacuum as shown. The distance between q1 and q2 is 5.0 cm while the distance between q2 and q3 is 7.5cm. If q1= +3.0 nC, q2= -1.5nC and q3= +2.0 nC, find the magnitude and direction of the net electrostatic force on q2.


1
Expert's answer
2021-02-11T14:07:17-0500

Assume that the charges are placed on one line. So, we have


F1=kq1q2r12=910931091.51090.052=0.0000162(N)F_1=k\frac{q_1q_2}{r_1^2}=9\cdot10^9\cdot \frac{3\cdot10^{-9}\cdot1.5\cdot10^{-9}}{0.05^2}=0.0000162(N)


F1=kq3q2r22=910921091.51090.0752=0.00000048(N)F_1=k\frac{q_3q_2}{r_2^2}=9\cdot10^9\cdot \frac{2\cdot10^{-9}\cdot1.5\cdot10^{-9}}{0.075^2}=0.00000048(N)


F=0.000000480.0000162=0.00001572(N)F=0.00000048-0.0000162=-0.00001572(N) to the charge q1q_1 . Answer

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