Three charges (q1 = 5.4 ?C, q2 = -4.9 ?C, and q3 = 2.7?C) are located at the ver

ID: 1430415 • Letter: T

Question

Three charges (q1 = 5.4 ?C, q2 = -4.9 ?C, and q3 = 2.7?C) are located at the vertices of an equilateral triangle with side d = 8.3 cm

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

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

3)A charge q4 = 2.7 ?C is now added as shown.What is F2,x, the x-component of the new net force on q2?(N)

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

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

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

a-Increase its magnitude and change its sign

b-Decrease its magnitude and change its sign

c-Increase its magnitude and keep its sign the same

d-Decrease its magnitude and keep its sign the same

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

1) force on 3 due 1

F31 = (kq1q3/d^2) ( cos60i + sin60 j)

k = 1/4pie0 = 9 x 10^9

F31 = 9.52i + 16.50 j N

force on 3 due 2

F32 = (kq2q3/d^2) ( cos60i - sin60 j)

F32 = 8.64i - 14.97j N

F3 = F32 + F31

F3x = 9.52 + 8.64 = 18.16 N

2) F3y = 16.50 - 14.97 = 1.8 N

3) F21 = (kq1q2 / d^2) (-i) = - 34.57 i N

F23 = - F32 = - 8.64i + 14.97j N

F24 = (kq2q4 / d^2) ( -cos60 i - sin60)

F24 = - 8.64i - 14.97j N

F2 = F21 + F23 + F24

F2x = -34.57 -8.64 - 8.64 = - 51.85 N

4) F2y = 14.97 - 14.97 = 0

5) f1 = 34.57 i - 9.52i - 16.50 j - 9.52i + 16.50 j

F1 = 15.53 N

6) first we have to change the signe

so force due to it will be + 34.57.

but force due to other charge is -(8.64+8.64) =- 17.28

so we have to reduce the magnitude also.

Decrease its magnitude and change its sign

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