Two identical point charges, each of charge +2.00 times 10^-6 C and mass 7.00 ti

Question

Two identical point charges, each of charge +2.00 times 10^-6 C and mass 7.00 times 10^-6 kg, arc fixed on the y-axis at the points y = +3.00 m and y = - 3.00 m. Suppose a negative charged particle of mass 8.00 times l0^-9 kg and of charge - 6.00 C is released from rest on the x-axis at the point x = - 4.00 m. What will be the speed of the negative charge at the instant it passes though the origin of the coordinate system? {Note that k_e = 9.00 times 10^9 N m^2 C^-2, and epsilon_0 = 8.85 times l0^-12 C^2 NT^-1 m^-2}.

Explanation / Answer

WE compute the total energy of the system of 3 charges when the third charge(-6.0C0 is at x=-4.0m and at the origin. The difference in the energy is equal to the KE of the particle

initial energy(x=-4, y=0)

Ui = 9.00e+9(2.0*(-6.0)/5 + 2.0.*(-6.0)/5 +2.0*2.0/6)e-12) = -37.2e-3

Uf (x=0,y=0)

    =9.00e+9(2.0*(-6.0)/3 + 2.0.*(-6.0)/3 +2.0*2.0/6)e-12) = -65.97e-3

Change in energy of the system = (65.97-37.2)e-3 = 28.77e-3

= (1/2)Mv2

speed of the particle v = sqrt(2*28.77e-3/8.0e-9) = 2682m/s

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