Find the net electric force that the two charges would exert on an electron plac

Question

Find the net electric force that the two charges would exert on an electron placed at point on the z-axis at 0.200 m. Express your answer with the appropriate units. Enter positive value if the force is in the positive r-direction and negative value if the force is in the negative z-direction. Constants A-4.00 nC point charge is at the origin, and a second 6.50 nC point charge is on the z-axis at z 0.800 m. B+I Value Units Submit Request Answer Part E Find the net electric force that the two charges would exert on an electron placed at point on the x-axis at ¢ 1.20 m. Express your answer with the appropriate units. Enter positive value if the force is in the positive -direction and negative value if the force is in the negative r-direction. F.-Value Units Submit Request Answer Part F Find the net electric force that the two charges would exert on an electron placed at point on the z-axis at x =-0.200 m.

Explanation / Answer

Given

two charges are say q1 = -4.0 nc is at origin and  

q2 = -6.50 nC at x = 0.80 m

Part D

the net electric force on the electron q3, placed at x = 0.200m by the charges q1,q2 is

from coulomb's law the electric force between two charges q1,q2 separated by a distance r is F = k*q1*q2/r^2

F13 = kq1*q3/r13^2 , F23 = kq2*q3/r23^2

here r13 = 0.2 m ,r23 = 0.6 m

the directions of the forces on the electron by q1 is to right and by q2 is to the left  

F13 +ve , F23 -ve

F13 = (9*10^9*4*10^-9*1.6*10^-19)/(0.2)^2 N = + 1.44*10^-16 N

F23 = (9*10^9*6.5*10^-9*1.6*10^-19)/(0.6)^2 N = - 2.6*10^-17 N

the net force acting on the electron is F = F13+F23 = 1.44*10^-16 -2.6*10^-17 N = 1.18*10^-16 N

Part E

the electron at x = 1.2 m

F13 = kq1*q3/r13^2 , F23 = kq2*q3/r23^2

here r13 = 1.2 m ,r23 = 1.2-0.8 = 0.4 m

the directions of the forces on the electron by q1 is to right and by q2 is also to the right

F13 +ve , F23 +ve

F13 = (9*10^9*4*10^-9*1.6*10^-19)/(1.2)^2 N = 4*10^-18 N

F23 = (9*10^9*6.5*10^-9*1.6*10^-19)/(0.4)^2 N = 5.85*10^-17 N

the net force acting on the electron is F = F13+F23 = 4*10^-18 +5.85*10^-17 N = 6.25*10^-17 N

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