a) P3 initially accelerating in the + x direction means the net force on P3, in the y direction, is 0:
[(kq1q3)/(r13^2)]sin45 + [(kq2q3)/(r23^2)]sin53.36 = 0
Divide out common factors k and q3, and sub in known quantities to solve for q2:
[(70.5 * 10^-6)/(28.5*^2 *2]sin45 + [q2/(21.2^2 + 28.5^2)]sin 53.36 = 0
q2= -4.83 x 10^-5 C
(Draw a FBD of P3 to see why q2 must be negative for this case).
b) net force in the x direction is 0, so sum the forces on P3 in this direction to find q2. (which will be positive this time; again a FBD is useful for understanding this)
Two particles are fixed on an x axis. Particle 1 of charge 70.5 μC is located at x = -28.5 cm; particle 2 of charge Q is located at x = 21.2 cm. Particle 3 of charge magnitude 37.4 μC is released from rest on the y axis at y = 28.5 cm. What is the value of Q if the initial acceleration of particle 3 is in the positive direction of (a) the x axis and (b) the y axis?
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