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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 0.500 m. What is net force on 93?
Remember: Negative forces (-F) will point Left
Positive forces (+F) will point Right
-5.00 x 10-6 C
91
0.500 m-
-5.00 x 10-6 C
92
-5.00 x 10-6 C
93
0.500 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 Q And Q2 Are Separated By 0500 M Particles Q2 class=

Sagot :

The answer to the question is 1.125  Newton

Formula for electrostatic force is F = ( K q1 q2 )/ r²

where q1 and q2 represent the charges and r represents the distance between them and the value of K is 9 × 10⁹.

Total net force on q3 will be the summation of electrostatic force between q1 and q3 and electrostatic force between q2 and q3, as all the three charges are of same sign and lie in the same line.

Electrostatic force between q1 and q3

r will be 0.500 + 0.500 = 1 m

F = ( K q1 q3 )/ r²

F = ( (9 × 10⁹ ) x (5.00 x 10⁻⁶) x (5.00 x 10⁻⁶) )/ 1²

F₁₃ = 2.25  × 10⁻¹ N

Electrostatic force between q2 and q3

r will be 0.500 m

F = ( K q1 q3 )/ r²

F = ( (9 × 10⁹ ) x (5.00 x 10⁻⁶) x (5.00 x 10⁻⁶) )/ 0.5²

F = (225 × 10⁻³) / (25 × 10⁻²)

F₂₃ = 9 × 10⁻¹ N

Total force on q3 will be F₁₃ + F₂₃

Total force on q3 =  ( 2.25  × 10⁻¹ ) + (9 × 10⁻¹ ) N

Total force on q3 =  ( 2.25  × 10⁻¹ ) + (9 × 10⁻¹ ) N

Total force on q3 =  ( 11.25  × 10⁻¹ ) N

Total force on q3 =  1.125  N

Thus after solving we got the net force on q3 as 1.125  Newton

Learn more about Electrostatic Force here:

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