A charge q!--1.2 C is placed at the origin. A second charge q,-1.5 pCis placed a

Question

A charge q!--1.2 C is placed at the origin. A second charge q,-1.5 pCis placed at x 2.5 m. A third charge q3 2.5 uC is brought to x 5.2 m to the same system. Please answer the following. a) b) c) Identify the correct equations to solve part b Draw a correct diagram for the problem. Find the magnitude and direction of the force F acting on the charge q2 A system has the following constraints; q,-4 C at (0, 0) and q,--1 C at (5,0). A third charge q35 C is brought at a distance 3 m from qi at an angle 620. Find the following: a) The magnitude of the net force F exerted on charge qi b) The direction of the net force on charge qi

Explanation / Answer

net force of q2 = Force on it due to q1 + Force on it due to q3

Fnet on q2 = F12 + F23

Fnet on q2 = Kq1q2/2.5^2 + K q2q3/(2.5+ 5.2)^2

Fnet on q2 = 9*10^9*(-1.2*1.5/2.5^2 + 2.5 *1.5/7.7^2) *10^-6

Fnet on q2 = -2022.764 N

force on q2 duw to q1 acts towards right side

force on q2 ue to 2.5 uC acts in left side

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componets of q3 are q3x = q3cos theta

q3xx = 4uc * cos 62

q3x = 1.877 uC

q3x = 4uC * sin 62 = 3.51 uC

FOrce of q1 = Force on it due to q2 and q3 along x axis

Force on q1 = kq1q2/5^2 + Kq1q3x/3^2

Force on q1 = 9*10^9 * (4*10^-6 * -1*10^-6)/5^2 + (9*10^9 * 4*10^-6*1.877*10^-6)/3^2

Force on q1 along x axis = 0.006068 N

Fon q1 along y axis is = 9*10^9 * 4*10^-6 * 3.51*10^-6 /3^2

Fon q1 along y axis = 0.01404 N

FNet on q1 ^2   = Fx^2 + Fy^2

Fnet on q1^2 = 0.006068^2 + 0.01404^2

Fnet of q1 = 1.53*10^-2 N

direction tan theta = 0.01404/0.006068

theta = 66.62 deg

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