4. -12 points SerPSE10 22.3.0P.003. My NoteAsk Your Teacher Two charged particle

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4. -12 points SerPSE10 22.3.0P.003. My NoteAsk Your Teacher Two charged particles, q1 and 92 are located on the x axis, with q1 at the origin and 92 initially at x1 = 14.5 mm. In this configuration, 91 exerts a repulsive force of 2.62 ?? on 92 Particle q2 is then moved to x2-17.4 mm. what is the force (magnitude and direction) that q2 exerts on q1 at this new location? (Give the magnitude in ? magnitude Select Need Help?Read It Show My Work Required) What steps or reasoning did you use? Your work counts towards your score. You can submit show my work an unlimited number of times. Uploaded File (10 file maximum) No Files to Display Upload File

Explanation / Answer

Electrostatic force is given by
F = k*q1*q2/r^2
r = distance between charges
q1 and q2 = charge
k = 9*10^9

In first configuration,

F1 = 2.62*10^-6 N = K*q1*q2/(r1)^2

here, r1 = 14.5 mm

So, k*q1*q2 = 2.62*10^-6*(14.5*10^-3)^2 = 5.51*10^-10

Now, r2 = 17.4 mm

therefore, F2 = k*q1*q2/(r2)^2 = 5.51*10^-10/(17.4*10^-3)^2 = 1.82*10^*-6 N

F2 = 1.82 mu*N

So, magnitude = 1.82

force is replusive. So,direction of force on q1 in negative x-axis..

please upvote.

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