?? = m/s As illustrated in the figure below, a negatively charged particle is re

Q: = m/s As illustrated in the figure below, a negatively charged particle is re

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

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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

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?? = m/s As illustrated in the figure below, a negatively charged particle is re

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