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Particles q1, 92, and q3 are in a straight line.
Particles q1 = -28.1 μC, q2 = +25.5 µC, and
93-47.9 μC. Particles q₁ and q2 are separated
by 0.300 m. Particles q2 and q3 are separated by
0.300 m. What is the net force on 93?
Remember: Negative forces (-F) will point Left
Positive forces (+F) will point Right
+25,5 μC
-28.1 μC
-47.9 μC

Particles Q1 92 And Q3 Are In A Straight Line Particles Q1 281 ΜC Q2 255 ΜC And 93479 ΜC Particles Q And Q2 Are Separated By 0300 M Particles Q2 And Q3 Are Sepa class=

Sagot :

The net force on q₃ will be 17.51 N. The net force is the algebraic sum of the two forces on the pleading q₃

What is Columb's law?

The force of attraction between two charges, according to Coulomb's law, is directly proportional to the product of the charges and inversely proportional to the square of the distance between them.

The force,by the charge q₁ on the q₃;

[tex]\rm F_{31}} = \frac{Kq_1q_3}{r^2} \\\\ \rm F_{31}} = \frac{9 \times 10^9 \times -28.1 \times 10^{-6}\times \times -47.9 \times 10^{-6}}{(0.600)^2} \\\\ F_{31}} =33.64 \ N[/tex]

The force,by the charge q₂ on the q₃;

[tex]\rm F_{32}} = \frac{Kq_2q_3}{r^2} \\\\ \rm F_{32}} = \frac{9 \times 10^9 \times 25.5 \times 10^{-6}\times \times -74.9\times 10^{-6}}{(0.300)^2} \\\\ F_{32}} =-19.09 \ N[/tex]

The net force is the sum of the two forces;

[tex]\rm F_{net}=F_{32}+F_{31}\\\\\ \rm F_{net}=36.6-19.9 \\\\ \rm F_{net}=17.51 \ N[/tex]

Hence, the net force on q₃ will be 17.51 N.

To learn more about Columb's law, refer to the link;

brainly.com/question/1616890

#SPJ1

alliebitz

I think its -88 on acellus

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