Two point charges are placed on the x-axis as follows: one positive charge, q_1

Question

Two point charges are placed on the x-axis as follows: one positive charge, q_1 = 3.99 nC, is located to the right of the origin at x = 0.200 m, and a second positive charge, q_2 = 4.97 nC, is located to the left of the origin at x = - 0 303 m. What is the total force (magnitude and direction) exerted by these two charges on a negative point charge, q_3 = - 5.96 nC, that is placed at the origin? Use 8.85 times 10^-12 C^2/(N middot m^2) for the permittivity of free space. Take positive forces to be along the positive x-axis. Do not use unit vectors.

Explanation / Answer

Electrostatic force is given by:

F = kq1q2/r^2

k = 9*10^9

Now total force on q3 will be

Fnet = F23 + F13

force between q1 and q3

F13 = 9*10^9*3.99*5.96*10^-18/0.200^2 = 5.351*10^-6 in +x direction

force between q2 and q3

F23 = 9*10^9*4.97*5.96*10^-18/0.303^2 = 2.903*10^-6 in -x direction

Since both forces are in opposite direction So

Fnet = 5.351*10^-6 - 2.903*10^-6

Fnet = 2.448*10^-6 N

Direction = towards +x axis

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