Question

Forces in a Three-Charge System 5 of 14 ConstantsI Periodic Table Part A Coulomb's law for the magnitude of the force F between two particles with charges Q and Q separated by a distance d is What is the net force exerted by these two charges on a third charge 52.0 nC placed between q and g2 at Zg=-1.160 m? Your answer may be positive or negative, depending on the direction of the force Express your answer numerically in newtons to three significant figures. QQ'1 where K =-, and 4neo View Available Hint(s) 8.854 x 1012 C2 /(N m2) is the permittivity of free space Consider two point charges located on the x axis one charge, q1 17.5 nC, is located at 11.650 m the second charge, 92-31.0 nC is at the origin (z = 0.0000) Force on 3.11.10 Submit Incorrect: Try Again: 2 attempts remaining








Hello, I am not sure what part I miss.

Do you what to fix to get the correct answer?

Since there is only two chances remained, it would be very helpful if you provide me with the answer.

Thank you.

***Answer is neither -3.11*10^-5 nor 3.11*10^-5.

Answer 1

1.650m q1 -17.5nC q3 52.0nC q2-31.0nC 0.49m F1 F2

Like charges repel and unlike charges attracts

so q₃ will be moving towards q₁ and away from q₂

$F=KQQ'/d^{2}$

$F_{1}=KQ_{1}Q_{3}/d^{2}$

$F_{1}=8.99*10^{9}*17.5*10^{-9}C*{\color{Red} 52}*10^{-9}C/(0.49m)^{2}$

(In written answer, mistake is made on values of F₃₁. Red is the correction )

$F_{1}=3.41*10^{-5}N$(towards negative side)

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$F_{2}=KQ_{2}Q_{3}/d^{2}$

$F_{2}=8.99*10^{9}*31*10^{-9}C*52*10^{-9}C/(1.160m)^{2}$

$F_{2}=1.077*10^{-5}$(towards negative side)

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$F_{3}=F_{1}+F_{2}$

$F_{3}=3.41*10^{-5}+1.077*10^{-5}N$

$F_{3}=4.487*10^{-5}N$(towards negative side)

ANSWER: $F_{3}=-4.487*10^{-5}N$

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