Three charges (q1 = 7.9 ?C, q2 = -5.7 ?C, and q3 = 1.6?C) are located at the ver

ID: 1433198 • Letter: T

Question

Three charges (q1 = 7.9 ?C, q2 = -5.7 ?C, and q3 = 1.6?C) are located at the vertices of an equilateral triangle with side d = 6.9 cm as shown.

1) What is F3,x, the value of the x-component of the net force on q3?

2) What is F3,y, the value of the y-component of the net force on q3?

A charge q4 = 1.6 ?C is now added as shown.

3) What is F2,x, the x-component of the new net force on q2?

4) What is F2,y, the y-component of the new net force on q2?

5) What is F1,x, the x-component of the new net force on q1?

6) How would you change q1 (keeping q2,q3 and q4 fixed) in order to make the net force on q2 equal to zero?

- Increase its magnitude and change its sign

- Decrease its magnitude and change its sign

- Increase its magnitude and keep its sign the same

- Decrease its magnitude and keep its sign the same

- There is no change you can make to q1 that will result in the fet force on q2 being equal to zero.

There are lot of parts into this equation, I will not be able tto show all of the calculations, Will provide fromulas and answers after calculatuons.

F3 ( x component) = k ( q1q3)/ d^2 cos 60 - k( q2q3)/d^2 cos60 = 20.338 N apprx

f3 ( Y componnet) = k ( q1q3)/ d^2 sin 60 + k( q2q3)/d^2 sin 60 = 5.698 N apprx

f2 ( x componnet) = K { ( q1q2)/d^2 + q2q3/d^2 cos60 + q2q4/d^2 cos60)= -101.225 N apprx

f2 ( y componet) = 0 newtons ( as the forces will vancel each other to maintain symmetry)

F1 ( x component) = K { [ q1q2/d^2 + q11q3/ d^2 cos 60 + q1q4 / d^2 cos 60) = 60.548 N apprx

6) Increase its magnitude and keep its sign the same

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