As illustrated in the figure, a negatively charged particle is released from res

Question

As illustrated in the figure, a negatively charged particle is released from rest at point B and accelerates until it reaches point A. The mass and charge of the particle are 1.28 x 10-4 kg and -3.48 x 10-5 C, respectively. Only the gravitational force and the electrostatic force act on the particle, which moves on a horizontal straight line without rotating. The electric potential at A is 36.0 V greater than that at B; in other words, VA - VB = 36.0 V. What is the translational speed of the particle at point A?

Explanation / Answer

apply change of KE = change of PE

0.5 m(Vf^2) - 0 = -Vq = -36* *-3.48*10^-5

Vf^2 = 2*125.28 *10^-5/(1.28*10^-4)

Vf^2 = 19.75

Vf = 4.424 m/s

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