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Particles q1, 92, and q3 are in a straight line.
Particles q1 = -5.00 x 10-6 C,q2 = -5.00 x 10-6 C,
and q3 = -5.00 x 10-6 C. Particles q1 and q2 are
separated by 0.500 m. Particles q2 and q3 are
separated by 0.250 m. What is the net force on 92?
Remember: Negative forces (-F) will point Left
Positive forces (+F) will point Right
-5.00 x 10-6 C
-5.00 x 10-6
-5.00 x 10-6 C
91
92
93
0.500 m
0.250 m

Particles q1 92 and q3 are in a straight line Particles q1 500 x 106 Cq2 500 x 106 C and q3 500 x 106 C Particles q1 and q2 are separated by 0500 m Particles q2 class=

Respuesta :

ryuttek

q1 = -5.00 x 10-6 C

q2 = -5.00 x 10-6 C

q3 = -5.00 x 10-6 C

E1 = kq/r^2 = ( 9 x 10^9)( 5 x 10^-6)/(0.5^2) = 180000 N/C to the left

E2 = kq/r^2 = ( 9 x 10^9)( 5 x 10^-6)/(0.25^2) = 720000 N/C to the right

E net = 720000 - 180000 = 540000 N/C to the right

F = qE

F = (-5 x 10^6 C)(540000 N/C) = - 2.7 N

The force on q2 is 2.7 N to the left.

The net electrostatic force on the q2 is 2.7N owards left

The equation for electrostatic force is

        [tex]F= k\frac{q_{1}q_{2} }{r^{2} }[/tex]

where k = [tex]9*10^{9} Nm^{2}/C^{2}[/tex] and r is the distance separating charges q1 and q2.

the force has to be calculated on a charge q2 = -5.0 ×[tex]10^{-6}[/tex] C by the charges q1=  -5.0 ×[tex]10^{-6}[/tex] C and q3=  -5.0 ×[tex]10^{-6}[/tex] C

distance between q1 and q2 is 0.5 m = 5×[tex]10^{-1}[/tex]m

distance between q2 and q3 is 0.25 m = 25×[tex]10^{-2}[/tex]m

force due to charge q1

           [tex]F_{1}[/tex] = 9×[tex]10^{9}[/tex]×(-5)×(-5)×[tex]10^{-12}[/tex]/25×[tex]10^{-2}[/tex] N = +0.9N =  0.9N towards right

           [tex]F_{2}[/tex] = 9×[tex]10^{9}[/tex]×(-50)×(-4)×[tex]10^{-12}[/tex]/625×[tex]10^{-4}[/tex] N = -3.6N = 3.6N towards left

hence net force F = [tex]F_{1}+F_{2}[/tex]

                              = 0.9N - 3.6N = -2.7N

                           F = 2.7 N towards left

Learn more about electrostatic force:

https://brainly.com/question/11375177

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