Three charged particles form a triangle: particle 1 with charge Q1 = 88.0 nC is

Question

Three charged particles form a triangle: particle 1 with charge Q1 = 88.0 nC is at xy coordinates (0, 6.00 mm), particle 2 with charge Q2 is at (0, -6.00 mm), and particle 3 with charge q = 24.0 nC is at (4.10 mm, 0). What are (a) the x component and (b) the y component of electrostatic force on particle 3 due to the other two particles if Q2 is equal to 88.0 nC? What are (c) the x component and (d) the y component of electrostatic force on particle 3 due to the other two particles if Q2 is equal to -88.0 nC, ?

Explanation / Answer

Q1 = 88.0 nC at (0, 6 mm)

Q2 = 88.0 nC at (0, -6 mm)

Q3 = 24.0 nC at (4.10 mm, 0)

F3 = F31 + F32

electrostatic force is given by

F = k*q1*q2/r^2

Now,

r13 = r23 = sqrt (6^2 + 4.1^2) mm

r13 = r23 = 7.27 mm

F13 = k*Q1*Q3/r13^2

F13 = 9*10^9*88*10^-9*24*10^-9/(7.27*10^-3)^2

F13 = 0.3596 N

force will be directed in the

theta = arctan (6/4.1) = 55.65 deg

F23 will be same as F13

F23 = 0.3596 N in negative 55.65 deg

Now in vector mode since this triangles's base is on y-axis, y-component of force will cancel each other and total force will be only in x-direction

F3 = F13 + F23

F3 = 0.3596*cos 55.65 deg + 0.3596*cos 55.65 deg

F3 = 0.406 N i

A.

F3x = 0.406 N in x-direction

B.

F3y = 0 N in y-direction

C.

WQ2 = -88.0 nC

in this part magnitude of F13 and F23 will be same but force between Q2 and Q3 will become attractive force so in this part total force will be in negative y-direction

F3 = F13 + F23

F13 = 0.3596 in negative 55.65 deg

F23 = 0.3596 in negative 124.35 deg

in the x-direction force will cancel each other so

F3x = 0 N

D.

F3y = 0.3596*sin (-55.65 deg) + 0.3596*sin (-124.35 deg)

F3y = - 0.594 N j

Let me know if you doubt in any step.

Please Recheck now. Initially my mistake was that I calculated the r12 = r13 = 7.27 mm, but using to calculate the force, In hurry I used 7.67 mm.

Let me know if it's still wrong.

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