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 6.44 x 10-4 kg and -1.07 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

The kinetic energy of the particle at point A is 36 V times the charge.

Solve for velocity in the equation E = ½mv²

E = -2qV

here charge of the particle is -1.07 x 10-5 C

V = 36.0V

the translational speed of the particle at point A is v = (2E / m)1/2

v = (-2qV / m)1/2

v =  ((-2 * (-1.07 x 10-5 C) * 36.0 V) / (6.44 x 10-4 kg))1/2

= (1.19)1/2 m/s

v = 1.09 m/s

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