, q1= -8.0uc,q2=3.0uc and q3=-4.0uc. the distance between q1 and q2 is. 30m while the q1q3 distance is 0.50m whats the net force on q3..
serious answers please I've done this like 10 times
serious answers please I've done this like 10 times
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Line up the charges so q1 is on the left and q3 is on the right. The 13 distance is .5m and the 23 distance is .2m.
The 13 force is repulsive (like charges), so the force on q3 , from q1, is to the "Right". The magnitude of that force is;
F13 = kq1q3/r13^2 = (9x10^9)(8x10^-6)(4x10^-6)/.25 = 1.15N Right.
The 23 force is attractive (unlike charges). So the force on q3 ,from q2, is to the Left. The magnitude of that force is;
F23 = kq2q3/r23^2 = (9x10^9)(3x10^-6)(4x10^-6)/.04 = 2.70N Left
Sincethese two forces are in opposite directions their "Net" will be the difference between them and in the direction of the larger force;
Fnet = F23 - F13 = 1.55 N Left
The 13 force is repulsive (like charges), so the force on q3 , from q1, is to the "Right". The magnitude of that force is;
F13 = kq1q3/r13^2 = (9x10^9)(8x10^-6)(4x10^-6)/.25 = 1.15N Right.
The 23 force is attractive (unlike charges). So the force on q3 ,from q2, is to the Left. The magnitude of that force is;
F23 = kq2q3/r23^2 = (9x10^9)(3x10^-6)(4x10^-6)/.04 = 2.70N Left
Sincethese two forces are in opposite directions their "Net" will be the difference between them and in the direction of the larger force;
Fnet = F23 - F13 = 1.55 N Left