Three charges (q 1 = 6 ?C, q 2 = -5.6 ?C, and q 3 = 3.8?C) are located at the ve
ID: 250679 • Letter: T
Question
Three charges (q1 = 6 ?C, q2 = -5.6 ?C, and q3 = 3.8?C) are located at the vertices of an equilateral triangle with side d = 9.8 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?
3)
A charge q4 = 3.8 ?C is now added as shown.
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?
Three charges (q1 = 6 pC, q2 =-5.6 pC and q3-3.8pC) are located at the vertices of an equilateral triangle with side d = 9.8 cm as shown 93 1 1) What is Fs,x, the value of the x-component of the net force on q3? N Submit You currently have 3 submissions for this question. Only 5 submission are allowed. You can make 2 more submissions for this question. 2) What is Fs,y, the value of the y-component of the net force on qs? N Submit You currently have 2 submissions for this question. Only 5 submission are allowed You can make 3 more submissions for this question. 3) 93 A charge q4 3.8 HC is now added as shown What is F2,x, the x-component of the new net force on q2? N Submit You currently have 2 submissions for this question. Only 5 submission are allowed You can make 3 more submissions for this question. 4) What is F2.y, the y-compone nt of the new net force on q2? N Submit You currently have 1 submissions for this question. Only 5 submission are allowed. You can make 4 more submissions for this question. 5) What is F1,x, the x-component of the new net force on q1? N Submit You currently have 0 submissions for this question. Only 5 submission are allowed You can make 5 more submissions for this question. 6) How would you chae q keeping q2,q3 and q, 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 O Increase its magnitude and keep its sign the same O Decrease its magnitude and keep its sign the same O There is no change you can make to q, that will result in the fet force on q2 being equal to zero. Submit You currently have 3 submissions for this question. Only 5 submission are allowed You can make 2 more submissions for this question.Explanation / Answer
1) "F3x" = [(kq1q3/d^2)+(kq2q3/d^2)]cos60 =kq3/d^2*cos60 [q1+q2]
=[(9*10^9*3.8*10^-6)/0.098^2]*0.5*(6-5.6)*10^-6
= 0.712 N
2) "F3y" = [(kq1q3/d^2)-(kq2q3/d^2)]sin60 =kq3/d^2*sin60 [q1-q2]
=[(9*10^9*3.8*10^-6)/0.098^2]*0.866*(6+5.6)*10^-6
= 35.77 N
3) "F2x" = -(9*10^9)*[{(q1*q2)/d^2} + {((q3 + q4)*q2)/d^2}*cos 60]
= -(9*10^9)*{q2/d^2}*[{(q1+ {(q3 + q4)/2}]
= -(9*10^3)*{(5.6)/(0.098)^2}*[6 + 0.5*(3.8+3.8)]
= -(9*10^3)*{(5.6)/(0.098)^2}*9.8 = -5.143*10^7 N
4) "F2y" = 0 as q3 = q4
5) "F1x" = (9*10^9)*[{(q1*q2)/d^2} - {((q3 + q4)*q1)/d^2}*cos 60]
= (9*10^9)*{q1/d^2}*[{(q2-{(q3 + q4)/2}]
= (9*10^3)*{(6)/(0.098)^2}*[5.6 - 0.5*(3.8+3.8)]
= (9*10^3)*{(6)/(0.098)^2}*1.8 = 1.012*10^7 N
6) Decrease the magnitude and change the sign.
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