Question
?? = m/s
As illustrated in the figure below, a negatively charged particle is released from rest at point Band accelerates until it reaches point A. The mass and charge of the particle are 5.00 10-6 kg and -8.00 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 31 V greater than that at B; in other words, VA ? VB = 31 V. What is the translational speed of the particle at point A?
Explanation / Answer
potential enenrgy U = transational KE
i.e
q(V2-V1)= 0.5 mV^2
-8*10^-5 * (31) = 0.5 * 5*10^-6*V^2
V^2 = 992
V = 31.49 m/s