Question #317352

Two point charges are placed as follows: charge q_{1} = - 1.50nC is at y = 6.00 m and charge q_{2} = 3.20nC is at the origin. What is the total force (magnitude and direction) exerted by these two charges on a negative point charge q_{3} = - 5.00nC located at (2.00 m, -4.00 m)?

1
Expert's answer
2022-03-25T15:02:12-0400

Force

F=F1+F2F=F_1+F_2

F=kq31[q1r2+q2r2]F=\frac{kq_3}{1}[\frac{q_1}{r^2}+\frac{q_2}{r^2}]

F=9×109×5×1091[1.50×109(104)2+3.20×109(20)2]F=-\frac{9\times10^9 \times 5\times10^{-9}}{1}[\frac{1.50\times10^{-9}}{(\sqrt{104})^2}+\frac{3.20\times10^{-9}}{(\sqrt20)^2}]

F=45×(0.01442+0.16)×109=7.8489×109NF=-45\times(0.01442+0.16)\times10^{-9}=-7.8489\times10^{-9}N


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