Chapter 15 × \\ e Chegg Study l Guided S 0 A molecule of DNA (dec Three point ch
ID: 1652584 • Letter: C
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Chapter 15 × e Chegg Study l Guided S 0 A molecule of DNA (dec Three point charges are × × .> clwww.webassign.net/web/Student/Assignment-Responses/submit?dep:16668630 Need Help?Read It 4. 15/30 points | Previous Answers SerCP11 15.P.010 My Notes Ask Your Teac Calculate the magnitude and direction of the Coulomb force on each of the three charges shown in the figure belovw 6.00 uC charge magnitude 6.00 pC ch arge direction toward the left 1.50 HC charge magnitude 1.50 C charge direction IN toward the right magnitude -2.00 uC charge -2.00 uC charge direction toward the left 6.00 C 1.50 C -2.00 C 3.00 cm 2.00 cm Need Help?Read It Submit Answer Save Progress 5. +-120 points SerCP11 15.P.011 MI My Notes Ask Your T Three charges are arranged as shown in the figure below. Find the magnitude and direction of the electrostatic force on the charge q = 5.04 nC at the origin. (Let r12 = 0.305 m.) magnitudeExplanation / Answer
Force on charge 6 uC due to 1.5 uC = kq1q2 / r^2 = 9*10^9*6*10^-6*1.5*10^-6 / 0.03^2
= 90 N ( to the left)
Force on charge 6 uC due to -2uC uC = kq1q2 / r^2 = 9*10^9*6*10^-6*2*10^-6 / 0.05^2
= 43.2 N ( to the right)
so net force on 6uC = (90 - 43.2) = 46.8 N (to the left)
Force on charge 1.5 uC due to 6 uC = kq1q2 / r^2 = 9*10^9*6*10^-6*1.5*10^-6 / 0.03^2
= 90 N ( to the right)
Force on charge 1.5 uC due to -2 uC = kq1q2 / r^2 = 9*10^9*1.5*10^-6*3*10^-6 / 0.02^2
= 67.5 N ( to the right)
net force on 1.5 uC = (90+67.5) = 157.5 N ( to the right)
Force on charge -2 uC due to 1.5 uC = kq1q2 / r^2 = 9*10^9*1.5*10^-6*3*10^-6 / 0.02^2
= 67.5 N ( to the left)
Force on charge -2 uC due to 6uC uC = kq1q2 / r^2 = 9*10^9*6*10^-6*2*10^-6 / 0.05^2
= 43.2 N ( to the left)
net force on -2 uC = (67.5 + 43.2) = 110.7 N ( to the left)
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