Three charges (q 1 = 5.1 C, q 2 = -3.8 C, and q 3 = 2.2C) are located at the ver

ID: 1579356 • Letter: T

Question

Three charges (q1 = 5.1 C, q2 = -3.8 C, and q3 = 2.2C) are located at the vertices of an equilateral triangle with side d = 7.6 cm as shown.

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

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

A charge q4 = 2.2 C is now added as shown.

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

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

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

(1)

the x-component of the net force on q3

F3x = (kq1*q3*cos60 / d2) + ( kq2*q3*cos60/d2)

F3x = kq3*(q1+q2)*cos60 / d2

F3x = 9*109 * 2.2*10-6 *(5.1*10-6 + 3.8*10-6)*cos60 / (7.6*10-2)2

F3x = 15.25 N

(2)

y-component of the new net force on q3

F3x = (kq1*q3*sin60 / d2) - ( kq2*q3*sin60/d2)

F3x = kq3*(q1-q2)*sin60 / d2

F3x = 9*109 * 2.2*10-6 *(5.1*10-6 - 3.8*10-6)*sin60 / (7.6*10-2)2

F3x = 3.85 N

(3)

F2x = (kq1q2/d2 ) + ( kq2q3/d2 ) + (kq2q4/d2 )

F2x = kq2*(q1 + q3 + q4 ) / d2

F2x = 9*109 *3.8*10-6 *( 5.1*10-6 + 2.2*10-6 +2.2*10-6) / (7.6*10-2)2

F2x = 56.25 N (in -x direction)

(4)

F2y = ( kq2q3/d2 ) - (kq2q4/d2)

F2y = k*q2*(q3 -q4)/d2

F2y = 9*109 * 3.8*10-6 *(2.2*10-6 - 2.2*10-6) /(7.6*10-2)2

F2y = 0

(5)

x component of net force on q1

F1x = - kq1q4*cos60/d2 - kq1q3*cos60/d2

F1x = -kq1*cos60*(q3 + q4) /d2

F1x = - 9*109*5.1*10-6 *cos60*(2.2*10-6 + 2.2*10-6) / (7.6*10-2)2

F1x = -17.48 N

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